Abdulatipov J.
Materiallar · Amaliyot · ML / AI

MAE va MSE oddiy misollarda

Batafsil yechilgan 20 ta masala — doskadagi misoldan chetga chiquvchi qiymatlar va model tanlashgacha. Har bir qadam darsdagidek yozilgan: $y_i$ va $\hat y_i$ juftlari, xatolar, ularning modullari va kvadratlari, yig‘indi, $n$ ga bo‘lish. Sonlar oddiy: hammasi og‘zaki yoki qog‘ozda hisoblanadi. Xatoni teskari tarqatish uchun esa — ikkala metrikaning har bir bashorat bo‘yicha hosilalari, gradiyent tushish qadami va neyron vaznlarigacha zanjir qoidasi.

Javharbek Abdulatipov · 50 daq o‘qish
20 ta misol8 ★☆☆9 ★★☆3 ★★★
$$\begin{gathered}\mathrm{MAE}=\frac{1}{n}\sum_{i=1}^{n}\bigl|y_i-\hat{y}_i\bigr|\\[6pt]\mathrm{MSE}=\frac{1}{n}\sum_{i=1}^{n}\bigl(y_i-\hat{y}_i\bigr)^2\end{gathered}$$

Formulalarni qanday o‘qish kerak

Ikkala metrika ham «model qanchalik adashadi?» degan savolga javob beradi. $n$ ta obyekt bor; har biri uchun to‘g‘ri javob $y_i$ va model bashorati $\hat y_i$ («igrek qalpoqcha» deb o‘qiladi) ma’lum. $e_i = y_i - \hat y_i$ ayirma — $i$-obyektdagi xato.

$y_i$
$i$-obyekt uchun to‘g‘ri javob (haqiqat)
$\hat y_i$
$i$-obyekt uchun model bashorati
$e_i = y_i - \hat y_i$
xato: plyus — model kam ko‘rsatdi, minus — oshirib yubordi
$n$
obyektlar soni
$\sum_{i=1}^{n}$
$1$ dan $n$ gacha barcha obyektlar bo‘yicha yig‘indi

To‘rt qadamli retsept

  1. $(y_i,\ \hat y_i)$ juftlarini yozing va $e_i = y_i - \hat y_i$ xatolarni hisoblang.
  2. MAE uchun har bir xatoning modulini $|e_i|$, MSE uchun kvadratini $e_i^2$ oling.
  3. Barcha modullarni (yoki barcha kvadratlarni) qo‘shing.
  4. Yig‘indini $n$ ga bo‘ling — bu MAE (yoki MSE).
Javobni qanday o‘qish kerak

MAE — $y$ bilan bir xil birlikdagi o‘rtacha adashish kattaligi: «o‘rtacha 0,83 ga adashamiz». MSE — adashishning o‘rtacha kvadrati, «kvadrat» birliklarda; u katta xatolarga kuchliroq ta’sir qiladi. Ikkala metrika uchun ham: qancha kichik bo‘lsa, shuncha yaxshi, nol — ideal.

Teskari tarqatish uchun hosilalar

Neyron tarmoq shunday o‘rganadi: yo‘qotishni (masalan, MSE ni) hisoblaydi, keyin orqaga — yo‘qotishdan vaznlarga qarab — yuradi va har bir qatlamda hosilalarni zanjir qoidasi bo‘yicha ko‘paytiradi. Bu o‘tish xatoni teskari tarqatish (backpropagation) deb ataladi. Zanjirning eng birinchi bo‘g‘ini — yo‘qotishning $\hat y_i$ bashorat bo‘yicha hosilasi: u har bir bashoratni qaysi tomonga va qanchalik kuchli siljitishni aytadi. Agar MAE va MSE bilan endigina tanishayotgan bo‘lsangiz, bu bo‘limni o‘tkazib yuborib, keyinroq qaytishingiz mumkin.

$g_i$
yo‘qotishning $\hat y_i$ bashorat bo‘yicha hosilasi, $\nabla_{\hat y} L = (g_1, \dots, g_n)$ gradiyentning komponenti

MSE ning hosilasi

$$\begin{aligned}\frac{\partial\,\mathrm{MSE}}{\partial \hat y_i} &= \frac{1}{n}\cdot 2\,(y_i-\hat y_i)\cdot(-1)\\ &= \frac{2}{n}\,(\hat y_i - y_i)\end{aligned}$$

Butun yig‘indidan $\hat y_i$ ga faqat bitta qo‘shiluvchi bog‘liq: $(y_i - \hat y_i)^2$. Kvadratning hosilasi — «ikki karra ifoda», $y_i - \hat y_i$ ifodaning $\hat y_i$ bo‘yicha hosilasi esa $-1$ (zanjir qoidasi). Shuning uchun ishora: $(\hat y_i - y_i)$, $(y_i - \hat y_i)$ emas.

MAE ning hosilasi

$$\frac{\partial\,\mathrm{MAE}}{\partial \hat y_i} = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$$
$$\operatorname{sign}(x)=\begin{cases}\phantom{-}1, & x>0\\\phantom{-}0, & x=0\\-1, & x<0\end{cases}$$

$|x|$ modulning hosilasi noldan o‘ngda $1$ ga, chapda $-1$ ga teng. $x = 0$ nuqtaning o‘zida modul sinadi va hosila yo‘q; amalda $0$ olinadi — bu joiz qiymatlardan biri (subgradiyent), PyTorch va NumPy shunday qiladi.

Ishorani qanday o‘qish kerak

Plyus — bashorat oshirilgan, tushish uni kamaytiradi; minus — kam ko‘rsatilgan, tushish uni oshiradi; nol — obyekt joyida qoladi. Gradiyent tushish qadami: $\hat y_i \leftarrow \hat y_i - \eta\, g_i$, bu yerda $\eta$ — o‘rganish tezligi (learning rate). Metrikalarning asosiy farqi: MSE da signal kuchi xatoga mutanosib, MAE da esa har doim $\tfrac{1}{n}$.

Doskadagi misol: gradiyent va bitta tushish qadami

$y = (3;\ 5;\ 8)$ va $\hat y = (2{,}5;\ 5;\ 10)$ uchun har bir bashorat bo‘yicha hosilalar:

$i$$y_i$$\hat y_i$MAE uchun $g_i$MSE uchun $g_i$
132,5≈ −0,33≈ −0,33
25500
3810≈ 0,33≈ 1,33

To‘g‘ridan-to‘g‘ri bashoratlar bo‘yicha $\eta = 0{,}3$ bilan bitta tushish qadamini qo‘yamiz — avval MSE gradiyenti, keyin MAE gradiyenti bo‘yicha:

$$\text{MSE:}\quad \begin{aligned}\hat y_{1} &\leftarrow 2{,}5 - 0{,}3\cdot (-\tfrac{1}{3}) = 2{,}6\\\hat y_{2} &\leftarrow 5 - 0{,}3\cdot 0 = 5\\\hat y_{3} &\leftarrow 10 - 0{,}3\cdot (\tfrac{4}{3}) = 9{,}6\end{aligned}$$
$$\text{MAE:}\quad \begin{aligned}\hat y_{1} &\leftarrow 2{,}5 - 0{,}3\cdot (-\tfrac{1}{3}) = 2{,}6\\\hat y_{2} &\leftarrow 5 - 0{,}3\cdot 0 = 5\\\hat y_{3} &\leftarrow 10 - 0{,}3\cdot (\tfrac{1}{3}) = 9{,}9\end{aligned}$$

Ikkala metrika kamaydi: MSE ≈ 1,42 → ≈ 0,91, MAE ≈ 0,83 → ≈ 0,77. MSE uchinchi bashoratni eng ko‘p siljitdi — xato eng katta bo‘lgan joyni.

Haqiqiy tarmoqda bashoratlar to‘g‘ridan-to‘g‘ri siljitilmaydi — vaznlar siljitiladi. Bashoratlar bo‘yicha gradiyent — vaznlarga qaytib boradigan signal; qanday qilib — «Bashoratdan vaznlarga» bo‘limida ko‘rsatilgan. Quyidagi har bir misolda hosilalar «Teskari tarqatish uchun hosilalar» blokida yozilgan.

Doskadagi misol

Darsda yechilgan misoldan boshlaymiz: uchta obyekt va uchta bashorat. Quyidagi har bir qadam — doskada yozilganining aynan o‘zi, faqat batafsilroq.

Misol 1★☆☆

Doskadagi misol

$y = (3;\ 5;\ 8)$,   $\hat y = (2{,}5;\ 5;\ 10)$. MAE va MSE ni toping.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 3-2{,}5\rvert = \lvert 0{,}5\rvert = 0{,}5\\\lvert e_{2}\rvert &= \lvert 5-5\rvert = \lvert 0\rvert = 0\\\lvert e_{3}\rvert &= \lvert 8-10\rvert = \lvert -2\rvert = 2\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 0{,}5 + 0 + 2 = 2{,}5$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 2{,}5 = \frac{2{,}5}{3} \approx 0{,}83$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (3-2{,}5)^2 = 0{,}5^2 = 0{,}25\\e_{2}^2 &= (5-5)^2 = 0^2 = 0\\e_{3}^2 &= (8-10)^2 = (-2)^2 = 4\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 0{,}25 + 0 + 4 = 4{,}25$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 4{,}25 = \frac{4{,}25}{3} \approx 1{,}42$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
132,50,50,50,25
255000
3810−224
Yig‘indi2,54,25
O‘rtacha (÷ 3)≈ 0,83≈ 1,42
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(2{,}5-3)}{3} \approx -0{,}33\\g_{2} &= \frac{\operatorname{sign}(5-5)}{3} = 0\\g_{3} &= \frac{\operatorname{sign}(10-8)}{3} \approx 0{,}33\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(2{,}5-3)}{3} \approx -0{,}33\\g_{2} &= \frac{2\cdot(5-5)}{3} = 0\\g_{3} &= \frac{2\cdot(10-8)}{3} \approx 1{,}33\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (-0{,}33;\ 0;\ 0{,}33)\\[4pt]\nabla\mathrm{MSE} \approx (-0{,}33;\ 0;\ 1{,}33)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE 3-obyektni eng kuchli tortadi: uning signali modul bo‘yicha 1,33 — unda eng katta xato (2).

Hosilalar nimani ko‘rsatadi

3-obyekt (xato 2) MSE dan 1,33 signal oladi — 0,5 xatoli 1-obyektdan (0,33) to‘rt baravar kuchli. MAE da ikkala signal modul bo‘yicha bir xil — 0,33: modul «ozgina» va «ko‘p»ni farqlamaydi.

Javob: $\mathrm{MAE} \approx 0{,}83$,   $\mathrm{MSE} \approx 1{,}42$

Misol nimani ko‘rsatadi

Uchta xatodan ikkitasi kichik, bittasi 2 ga teng. MAE ≈ 0,83 — «o‘rtacha birdan kamroqqa adashamiz». MSE ≈ 1,42 MAE dan katta: 2 xatoning kvadrati 4 ga teng va o‘rtachani o‘ziga tortadi.

Qizish

Retsept odatga aylanishi uchun oddiy sonlar to‘plamlari. Birinchi masala ochiq, qolganlarini o‘zingiz yechib ko‘ring.

Misol 2★☆☆

Mukammal bashorat

$y = (2;\ 4;\ 6)$,   $\hat y = (2;\ 4;\ 6)$. MAE va MSE ni toping.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 2-2\rvert = \lvert 0\rvert = 0\\\lvert e_{2}\rvert &= \lvert 4-4\rvert = \lvert 0\rvert = 0\\\lvert e_{3}\rvert &= \lvert 6-6\rvert = \lvert 0\rvert = 0\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 0 + 0 + 0 = 0$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 0 = \frac{0}{3} = 0$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (2-2)^2 = 0^2 = 0\\e_{2}^2 &= (4-4)^2 = 0^2 = 0\\e_{3}^2 &= (6-6)^2 = 0^2 = 0\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 0 + 0 + 0 = 0$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 0 = \frac{0}{3} = 0$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
122000
244000
366000
Yig‘indi00
O‘rtacha (÷ 3)00
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(2-2)}{3} = 0\\g_{2} &= \frac{\operatorname{sign}(4-4)}{3} = 0\\g_{3} &= \frac{\operatorname{sign}(6-6)}{3} = 0\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(2-2)}{3} = 0\\g_{2} &= \frac{2\cdot(4-4)}{3} = 0\\g_{3} &= \frac{2\cdot(6-6)}{3} = 0\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} = (0;\ 0;\ 0)\\[4pt]\nabla\mathrm{MSE} = (0;\ 0;\ 0)\end{gathered}$$

Barcha hosilalar nolga teng: bashorat allaqachon mukammal, teskari tarqatishda tuzatadigan narsa yo‘q.

Javob: $\mathrm{MAE} = 0$,   $\mathrm{MSE} = 0$

Misol nimani ko‘rsatadi

Barcha xatolar nolga teng, demak MAE = 0 va MSE = 0. Ikkala metrika ham noldan kichik bo‘lmaydi: modul va kvadrat manfiy emas.

Misol 3★☆☆

Faqat bitta nuqtadagi xato

$y = (1;\ 2;\ 3;\ 4)$,   $\hat y = (1;\ 2;\ 3;\ 6)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 1-1\rvert = \lvert 0\rvert = 0\\\lvert e_{2}\rvert &= \lvert 2-2\rvert = \lvert 0\rvert = 0\\\lvert e_{3}\rvert &= \lvert 3-3\rvert = \lvert 0\rvert = 0\\\lvert e_{4}\rvert &= \lvert 4-6\rvert = \lvert -2\rvert = 2\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{4}\lvert e_i\rvert = 0 + 0 + 0 + 2 = 2$$

3. MAE — yig‘indini $n = 4$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{4}\cdot 2 = \frac{2}{4} = 0{,}5$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (1-1)^2 = 0^2 = 0\\e_{2}^2 &= (2-2)^2 = 0^2 = 0\\e_{3}^2 &= (3-3)^2 = 0^2 = 0\\e_{4}^2 &= (4-6)^2 = (-2)^2 = 4\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{4}e_i^2 = 0 + 0 + 0 + 4 = 4$$

6. MSE — yig‘indini $n = 4$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{4}\cdot 4 = \frac{4}{4} = 1$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
111000
222000
333000
446−224
Yig‘indi24
O‘rtacha (÷ 4)0,51
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(1-1)}{4} = 0\\g_{2} &= \frac{\operatorname{sign}(2-2)}{4} = 0\\g_{3} &= \frac{\operatorname{sign}(3-3)}{4} = 0\\g_{4} &= \frac{\operatorname{sign}(6-4)}{4} = 0{,}25\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(1-1)}{4} = 0\\g_{2} &= \frac{2\cdot(2-2)}{4} = 0\\g_{3} &= \frac{2\cdot(3-3)}{4} = 0\\g_{4} &= \frac{2\cdot(6-4)}{4} = 1\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} = \begin{pmatrix}0\\0\\0\\0{,}25\end{pmatrix}\\[8pt]\nabla\mathrm{MSE} = \begin{pmatrix}0\\0\\0\\1\end{pmatrix}\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,25 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Javob: $\mathrm{MAE} = 0{,}5$,   $\mathrm{MSE} = 1$

Misol nimani ko‘rsatadi

To‘rtta obyektda bitta 2 xato: MAE = 2 / 4 = 0,5. MSE ga esa u 2² = 4 bo‘lib kiradi, shuning uchun MSE = 1 — MAE dan ikki baravar katta.

Misol 4★☆☆

Hamma joyda bir xil xato

$y = (5;\ 7;\ 9)$,   $\hat y = (6;\ 8;\ 10)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 5-6\rvert = \lvert -1\rvert = 1\\\lvert e_{2}\rvert &= \lvert 7-8\rvert = \lvert -1\rvert = 1\\\lvert e_{3}\rvert &= \lvert 9-10\rvert = \lvert -1\rvert = 1\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 1 + 1 + 1 = 3$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 3 = \frac{3}{3} = 1$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (5-6)^2 = (-1)^2 = 1\\e_{2}^2 &= (7-8)^2 = (-1)^2 = 1\\e_{3}^2 &= (9-10)^2 = (-1)^2 = 1\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 1 + 1 + 1 = 3$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 3 = \frac{3}{3} = 1$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
156−111
278−111
3910−111
Yig‘indi33
O‘rtacha (÷ 3)11
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(6-5)}{3} \approx 0{,}33\\g_{2} &= \frac{\operatorname{sign}(8-7)}{3} \approx 0{,}33\\g_{3} &= \frac{\operatorname{sign}(10-9)}{3} \approx 0{,}33\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(6-5)}{3} \approx 0{,}67\\g_{2} &= \frac{2\cdot(8-7)}{3} \approx 0{,}67\\g_{3} &= \frac{2\cdot(10-9)}{3} \approx 0{,}67\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (0{,}33;\ 0{,}33;\ 0{,}33)\\[4pt]\nabla\mathrm{MSE} \approx (0{,}67;\ 0{,}67;\ 0{,}67)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Javob: $\mathrm{MAE} = 1$,   $\mathrm{MSE} = 1$

Misol nimani ko‘rsatadi

Model hamma joyda 1 ga oshirib yuboradi. Agar barcha |eᵢ| bir xil va a ga teng bo‘lsa, MAE = a, MSE = a². Bu yerda a = 1, shuning uchun ikkala metrika ham 1 ga teng.

Misol 5★☆☆

Xatoning ishorasi muhim emas

$y = (10;\ 20)$,   $\hat y = (12;\ 18)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 10-12\rvert = \lvert -2\rvert = 2\\\lvert e_{2}\rvert &= \lvert 20-18\rvert = \lvert 2\rvert = 2\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{2}\lvert e_i\rvert = 2 + 2 = 4$$

3. MAE — yig‘indini $n = 2$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{2}\cdot 4 = \frac{4}{2} = 2$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (10-12)^2 = (-2)^2 = 4\\e_{2}^2 &= (20-18)^2 = 2^2 = 4\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{2}e_i^2 = 4 + 4 = 8$$

6. MSE — yig‘indini $n = 2$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{2}\cdot 8 = \frac{8}{2} = 4$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
11012−224
22018224
Yig‘indi48
O‘rtacha (÷ 2)24
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(12-10)}{2} = 0{,}5\\g_{2} &= \frac{\operatorname{sign}(18-20)}{2} = -0{,}5\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(12-10)}{2} = 2\\g_{2} &= \frac{2\cdot(18-20)}{2} = -2\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} = (0{,}5;\ -0{,}5)\\[4pt]\nabla\mathrm{MSE} = (2;\ -2)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,5 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Javob: $\mathrm{MAE} = 2$,   $\mathrm{MSE} = 4$

Misol nimani ko‘rsatadi

Birinchi xato −2 (model oshirib yubordi), ikkinchisi +2 (kam ko‘rsatdi). Modul va kvadrat ishorani olib tashlaydi: ikkala xato ham MAE da 2, MSE da 4 beradi. Yuqoriga va pastga adashishlar bir-birini yo‘qqa chiqarmaydi.

Misol 6★☆☆

Birdan kichik xatolar

$y = (1;\ 1)$,   $\hat y = (1{,}5;\ 0{,}5)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 1-1{,}5\rvert = \lvert -0{,}5\rvert = 0{,}5\\\lvert e_{2}\rvert &= \lvert 1-0{,}5\rvert = \lvert 0{,}5\rvert = 0{,}5\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{2}\lvert e_i\rvert = 0{,}5 + 0{,}5 = 1$$

3. MAE — yig‘indini $n = 2$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{2}\cdot 1 = \frac{1}{2} = 0{,}5$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (1-1{,}5)^2 = (-0{,}5)^2 = 0{,}25\\e_{2}^2 &= (1-0{,}5)^2 = 0{,}5^2 = 0{,}25\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{2}e_i^2 = 0{,}25 + 0{,}25 = 0{,}5$$

6. MSE — yig‘indini $n = 2$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{2}\cdot 0{,}5 = \frac{0{,}5}{2} = 0{,}25$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
111,5−0,50,50,25
210,50,50,50,25
Yig‘indi10,5
O‘rtacha (÷ 2)0,50,25
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(1{,}5-1)}{2} = 0{,}5\\g_{2} &= \frac{\operatorname{sign}(0{,}5-1)}{2} = -0{,}5\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(1{,}5-1)}{2} = 0{,}5\\g_{2} &= \frac{2\cdot(0{,}5-1)}{2} = -0{,}5\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} = (0{,}5;\ -0{,}5)\\[4pt]\nabla\mathrm{MSE} = (0{,}5;\ -0{,}5)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,5 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Javob: $\mathrm{MAE} = 0{,}5$,   $\mathrm{MSE} = 0{,}25$

Misol nimani ko‘rsatadi

Bu yerda MSE = 0,25 MAE = 0,5 dan kichik: 1 dan kichik sonning kvadrati yanada kichrayadi (0,5² = 0,25). Shuning uchun MAE va MSE ni kattaligi bo‘yicha solishtirish ma’nosiz — ularning birliklari va tabiati har xil.

Misol 7★☆☆

Beshta obyekt — 5 ga bo‘lamiz

$y = (2;\ 4;\ 6;\ 8;\ 10)$,   $\hat y = (3;\ 3;\ 7;\ 7;\ 10)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 2-3\rvert = \lvert -1\rvert = 1\\\lvert e_{2}\rvert &= \lvert 4-3\rvert = \lvert 1\rvert = 1\\\lvert e_{3}\rvert &= \lvert 6-7\rvert = \lvert -1\rvert = 1\\\lvert e_{4}\rvert &= \lvert 8-7\rvert = \lvert 1\rvert = 1\\\lvert e_{5}\rvert &= \lvert 10-10\rvert = \lvert 0\rvert = 0\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{5}\lvert e_i\rvert = 1 + 1 + 1 + 1 + 0 = 4$$

3. MAE — yig‘indini $n = 5$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{5}\cdot 4 = \frac{4}{5} = 0{,}8$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (2-3)^2 = (-1)^2 = 1\\e_{2}^2 &= (4-3)^2 = 1^2 = 1\\e_{3}^2 &= (6-7)^2 = (-1)^2 = 1\\e_{4}^2 &= (8-7)^2 = 1^2 = 1\\e_{5}^2 &= (10-10)^2 = 0^2 = 0\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{5}e_i^2 = 1 + 1 + 1 + 1 + 0 = 4$$

6. MSE — yig‘indini $n = 5$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{5}\cdot 4 = \frac{4}{5} = 0{,}8$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
123−111
243111
367−111
487111
51010000
Yig‘indi44
O‘rtacha (÷ 5)0,80,8
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(3-2)}{5} = 0{,}2\\g_{2} &= \frac{\operatorname{sign}(3-4)}{5} = -0{,}2\\g_{3} &= \frac{\operatorname{sign}(7-6)}{5} = 0{,}2\\g_{4} &= \frac{\operatorname{sign}(7-8)}{5} = -0{,}2\\g_{5} &= \frac{\operatorname{sign}(10-10)}{5} = 0\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(3-2)}{5} = 0{,}4\\g_{2} &= \frac{2\cdot(3-4)}{5} = -0{,}4\\g_{3} &= \frac{2\cdot(7-6)}{5} = 0{,}4\\g_{4} &= \frac{2\cdot(7-8)}{5} = -0{,}4\\g_{5} &= \frac{2\cdot(10-10)}{5} = 0\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} = \begin{pmatrix}0{,}2\\-0{,}2\\0{,}2\\-0{,}2\\0\end{pmatrix}\\[8pt]\nabla\mathrm{MSE} = \begin{pmatrix}0{,}4\\-0{,}4\\0{,}4\\-0{,}4\\0\end{pmatrix}\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,2 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Javob: $\mathrm{MAE} = 0{,}8$,   $\mathrm{MSE} = 0{,}8$

Misol nimani ko‘rsatadi

Har doim obyektlar soni n ga bo‘lamiz, noldan farqli xatolar soniga emas: xato to‘rtta, obyekt esa beshta, shuning uchun MAE = 4 / 5 = 0,8.

Manfiy sonlar, kasrlar va bitta obyekt

Xuddi shu retsept, lekin sonlar murakkabroq: minuslar, kasrlar va obyekt bittagina bo‘lgan hol.

Misol 8★★☆

Manfiy sonlar: harorat

$y = (-2;\ 0;\ 3)$,   $\hat y = (1;\ -1;\ 3)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert -2-1\rvert = \lvert -3\rvert = 3\\\lvert e_{2}\rvert &= \lvert 0-(-1)\rvert = \lvert 1\rvert = 1\\\lvert e_{3}\rvert &= \lvert 3-3\rvert = \lvert 0\rvert = 0\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 3 + 1 + 0 = 4$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 4 = \frac{4}{3} \approx 1{,}33$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (-2-1)^2 = (-3)^2 = 9\\e_{2}^2 &= (0-(-1))^2 = 1^2 = 1\\e_{3}^2 &= (3-3)^2 = 0^2 = 0\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 9 + 1 + 0 = 10$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 10 = \frac{10}{3} \approx 3{,}33$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
1−21−339
20−1111
333000
Yig‘indi410
O‘rtacha (÷ 3)≈ 1,33≈ 3,33
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(1-(-2))}{3} \approx 0{,}33\\g_{2} &= \frac{\operatorname{sign}(-1-0)}{3} \approx -0{,}33\\g_{3} &= \frac{\operatorname{sign}(3-3)}{3} = 0\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(1-(-2))}{3} = 2\\g_{2} &= \frac{2\cdot(-1-0)}{3} \approx -0{,}67\\g_{3} &= \frac{2\cdot(3-3)}{3} = 0\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (0{,}33;\ -0{,}33;\ 0)\\[4pt]\nabla\mathrm{MSE} \approx (2;\ -0{,}67;\ 0)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE 1-obyektni eng kuchli tortadi: uning signali modul bo‘yicha 2 — unda eng katta xato (3).

Javob: $\mathrm{MAE} \approx 1{,}33$,   $\mathrm{MSE} \approx 3{,}33$

Misol nimani ko‘rsatadi

Asosiysi — ehtiyotkorlik bilan ayirish: 0 − (−1) = 0 + 1 = 1, (−3)² esa 9 ga teng, −9 ga emas. Modul va kvadrat hech qachon manfiy bo‘lmaydi.

Misol 9★★☆

O‘nli kasrlar

$y = (1{,}2;\ 2{,}5;\ 3{,}1)$,   $\hat y = (1;\ 2{,}9;\ 3{,}1)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 1{,}2-1\rvert = \lvert 0{,}2\rvert = 0{,}2\\\lvert e_{2}\rvert &= \lvert 2{,}5-2{,}9\rvert = \lvert -0{,}4\rvert = 0{,}4\\\lvert e_{3}\rvert &= \lvert 3{,}1-3{,}1\rvert = \lvert 0\rvert = 0\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 0{,}2 + 0{,}4 + 0 = 0{,}6$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 0{,}6 = \frac{0{,}6}{3} = 0{,}2$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (1{,}2-1)^2 = 0{,}2^2 = 0{,}04\\e_{2}^2 &= (2{,}5-2{,}9)^2 = (-0{,}4)^2 = 0{,}16\\e_{3}^2 &= (3{,}1-3{,}1)^2 = 0^2 = 0\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 0{,}04 + 0{,}16 + 0 = 0{,}2$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 0{,}2 = \frac{0{,}2}{3} \approx 0{,}0667$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
11,210,20,20,04
22,52,9−0,40,40,16
33,13,1000
Yig‘indi0,60,2
O‘rtacha (÷ 3)0,2≈ 0,0667
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(1-1{,}2)}{3} \approx -0{,}33\\g_{2} &= \frac{\operatorname{sign}(2{,}9-2{,}5)}{3} \approx 0{,}33\\g_{3} &= \frac{\operatorname{sign}(3{,}1-3{,}1)}{3} = 0\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(1-1{,}2)}{3} \approx -0{,}13\\g_{2} &= \frac{2\cdot(2{,}9-2{,}5)}{3} \approx 0{,}27\\g_{3} &= \frac{2\cdot(3{,}1-3{,}1)}{3} = 0\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (-0{,}33;\ 0{,}33;\ 0)\\[4pt]\nabla\mathrm{MSE} \approx (-0{,}13;\ 0{,}27;\ 0)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE 2-obyektni eng kuchli tortadi: uning signali modul bo‘yicha 0,27 — unda eng katta xato (0,4).

Javob: $\mathrm{MAE} = 0{,}2$,   $\mathrm{MSE} \approx 0{,}0667$

Misol nimani ko‘rsatadi

Kasrlar bilan ham xuddi shunday, faqat kvadratlarda ehtiyot bo‘ling: 0,2² = 0,04 va 0,4² = 0,16. MSE ≈ 0,0667 MAE = 0,2 dan ancha kichik — barcha xatolar 1 dan kichik.

Misol 10★☆☆

Bitta obyekt, n = 1

$y = (7)$,   $\hat y = (4)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 7-4\rvert = \lvert 3\rvert = 3\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{1}\lvert e_i\rvert = 3$$

3. MAE — yig‘indini $n = 1$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{1}\cdot 3 = 3$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (7-4)^2 = 3^2 = 9\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{1}e_i^2 = 9$$

6. MSE — yig‘indini $n = 1$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{1}\cdot 9 = 9$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
174339
Yig‘indi39
O‘rtacha (÷ 1)39
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(4-7)}{1} = -1\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(4-7)}{1} = -6\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} = (-1)\\[4pt]\nabla\mathrm{MSE} = (-6)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±1 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Javob: $\mathrm{MAE} = 3$,   $\mathrm{MSE} = 9$

Misol nimani ko‘rsatadi

n = 1 da o‘rtacha olishga hojat yo‘q: MAE — xatoning moduli (3), MSE — uning kvadrati (9).

MAE MSE dan nimasi bilan farq qiladi

Bu yerda misollar metrikalar «bahslashadigan» qilib tanlangan: bir xil MAE va har xil MSE, chetga chiquvchi qiymat, ikki baravar xatolar, o‘rtacha va mediana.

Misol 11★★☆

Bir xil MAE — har xil MSE

$y = (3;\ 5;\ 7;\ 9)$. Model A: $\hat y = (4;\ 6;\ 6;\ 8)$; Model B: $\hat y = (3;\ 5;\ 7;\ 5)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

Model A

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 3-4\rvert = \lvert -1\rvert = 1\\\lvert e_{2}\rvert &= \lvert 5-6\rvert = \lvert -1\rvert = 1\\\lvert e_{3}\rvert &= \lvert 7-6\rvert = \lvert 1\rvert = 1\\\lvert e_{4}\rvert &= \lvert 9-8\rvert = \lvert 1\rvert = 1\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{4}\lvert e_i\rvert = 1 + 1 + 1 + 1 = 4$$

3. MAE — yig‘indini $n = 4$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{4}\cdot 4 = \frac{4}{4} = 1$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (3-4)^2 = (-1)^2 = 1\\e_{2}^2 &= (5-6)^2 = (-1)^2 = 1\\e_{3}^2 &= (7-6)^2 = 1^2 = 1\\e_{4}^2 &= (9-8)^2 = 1^2 = 1\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{4}e_i^2 = 1 + 1 + 1 + 1 = 4$$

6. MSE — yig‘indini $n = 4$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{4}\cdot 4 = \frac{4}{4} = 1$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
134−111
256−111
376111
498111
Yig‘indi44
O‘rtacha (÷ 4)11
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(4-3)}{4} = 0{,}25\\g_{2} &= \frac{\operatorname{sign}(6-5)}{4} = 0{,}25\\g_{3} &= \frac{\operatorname{sign}(6-7)}{4} = -0{,}25\\g_{4} &= \frac{\operatorname{sign}(8-9)}{4} = -0{,}25\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(4-3)}{4} = 0{,}5\\g_{2} &= \frac{2\cdot(6-5)}{4} = 0{,}5\\g_{3} &= \frac{2\cdot(6-7)}{4} = -0{,}5\\g_{4} &= \frac{2\cdot(8-9)}{4} = -0{,}5\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} = \begin{pmatrix}0{,}25\\0{,}25\\-0{,}25\\-0{,}25\end{pmatrix}\\[8pt]\nabla\mathrm{MSE} = \begin{pmatrix}0{,}5\\0{,}5\\-0{,}5\\-0{,}5\end{pmatrix}\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,25 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Model B

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 3-3\rvert = \lvert 0\rvert = 0\\\lvert e_{2}\rvert &= \lvert 5-5\rvert = \lvert 0\rvert = 0\\\lvert e_{3}\rvert &= \lvert 7-7\rvert = \lvert 0\rvert = 0\\\lvert e_{4}\rvert &= \lvert 9-5\rvert = \lvert 4\rvert = 4\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{4}\lvert e_i\rvert = 0 + 0 + 0 + 4 = 4$$

3. MAE — yig‘indini $n = 4$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{4}\cdot 4 = \frac{4}{4} = 1$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (3-3)^2 = 0^2 = 0\\e_{2}^2 &= (5-5)^2 = 0^2 = 0\\e_{3}^2 &= (7-7)^2 = 0^2 = 0\\e_{4}^2 &= (9-5)^2 = 4^2 = 16\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{4}e_i^2 = 0 + 0 + 0 + 16 = 16$$

6. MSE — yig‘indini $n = 4$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{4}\cdot 16 = \frac{16}{4} = 4$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
133000
255000
377000
4954416
Yig‘indi416
O‘rtacha (÷ 4)14
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(3-3)}{4} = 0\\g_{2} &= \frac{\operatorname{sign}(5-5)}{4} = 0\\g_{3} &= \frac{\operatorname{sign}(7-7)}{4} = 0\\g_{4} &= \frac{\operatorname{sign}(5-9)}{4} = -0{,}25\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(3-3)}{4} = 0\\g_{2} &= \frac{2\cdot(5-5)}{4} = 0\\g_{3} &= \frac{2\cdot(7-7)}{4} = 0\\g_{4} &= \frac{2\cdot(5-9)}{4} = -2\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} = \begin{pmatrix}0\\0\\0\\-0{,}25\end{pmatrix}\\[8pt]\nabla\mathrm{MSE} = \begin{pmatrix}0\\0\\0\\-2\end{pmatrix}\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,25 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Hosilalar nimani ko‘rsatadi

MSE gradiyentlari xuddi shu farqni ko‘rsatadi: A modelda to‘rtala signal 0,5 dan, B modelda butun −2 signal bitta nuqtaga to‘g‘ri keladi. MAE da bu nuqtaning signali bor-yo‘g‘i −0,25 — boshqa har qanday xato kabi.

Solishtirish

MAEMSE
Model A11
Model B14

Javob: Model A: $\mathrm{MAE} = 1$,   $\mathrm{MSE} = 1$; Model B: $\mathrm{MAE} = 1$,   $\mathrm{MSE} = 4$

Misol nimani ko‘rsatadi

Ikkala modelda ham MAE = 1. Lekin A hamma joyda ozgina adashadi, B esa uchta nuqtada aniq va bittasida qattiq adashadi. MSE buni farqlaydi: A da 1, B da 4. MSE katta adashishlarni «jazolaydi».

Misol 12★★☆

Chetga chiquvchi qiymat

$y = (2;\ 3;\ 4;\ 20)$,   $\hat y = (3;\ 3;\ 3;\ 4)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 2-3\rvert = \lvert -1\rvert = 1\\\lvert e_{2}\rvert &= \lvert 3-3\rvert = \lvert 0\rvert = 0\\\lvert e_{3}\rvert &= \lvert 4-3\rvert = \lvert 1\rvert = 1\\\lvert e_{4}\rvert &= \lvert 20-4\rvert = \lvert 16\rvert = 16\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{4}\lvert e_i\rvert = 1 + 0 + 1 + 16 = 18$$

3. MAE — yig‘indini $n = 4$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{4}\cdot 18 = \frac{18}{4} = 4{,}5$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (2-3)^2 = (-1)^2 = 1\\e_{2}^2 &= (3-3)^2 = 0^2 = 0\\e_{3}^2 &= (4-3)^2 = 1^2 = 1\\e_{4}^2 &= (20-4)^2 = 16^2 = 256\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{4}e_i^2 = 1 + 0 + 1 + 256 = 258$$

6. MSE — yig‘indini $n = 4$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{4}\cdot 258 = \frac{258}{4} = 64{,}5$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
123−111
233000
343111
42041616256
Yig‘indi18258
O‘rtacha (÷ 4)4,564,5
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(3-2)}{4} = 0{,}25\\g_{2} &= \frac{\operatorname{sign}(3-3)}{4} = 0\\g_{3} &= \frac{\operatorname{sign}(3-4)}{4} = -0{,}25\\g_{4} &= \frac{\operatorname{sign}(4-20)}{4} = -0{,}25\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(3-2)}{4} = 0{,}5\\g_{2} &= \frac{2\cdot(3-3)}{4} = 0\\g_{3} &= \frac{2\cdot(3-4)}{4} = -0{,}5\\g_{4} &= \frac{2\cdot(4-20)}{4} = -8\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} = \begin{pmatrix}0{,}25\\0\\-0{,}25\\-0{,}25\end{pmatrix}\\[8pt]\nabla\mathrm{MSE} = \begin{pmatrix}0{,}5\\0\\-0{,}5\\-8\end{pmatrix}\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,25 signal beradi: faqat xatoning yo‘nalishi muhim. MSE 4-obyektni eng kuchli tortadi: uning signali modul bo‘yicha 8 — unda eng katta xato (16).

Hosilalar nimani ko‘rsatadi

MSE bo‘yicha o‘qitishda chetga chiquvchi qiymat −8 signal oladi — boshqa nuqtalardan (±0,5) 16 baravar kuchli — va modelni o‘ziga tortadi. MAE bo‘yicha uning signali −0,25 — oddiy xatoniki bilan bir xil. Shuning uchun ma’lumotlarda chetga chiquvchilar bo‘lsa, MAE yoki uning silliq qarindoshi — Huber bo‘yicha o‘qitiladi.

Javob: $\mathrm{MAE} = 4{,}5$,   $\mathrm{MSE} = 64{,}5$

Misol nimani ko‘rsatadi

Bitta nuqta (y = 20) uzoqqa ketgan. MAE da uning 16 xatosi 18 lik yig‘indining 16 tasini beradi, MSE da esa 258 ning 256 tasini — deyarli butun metrika bitta nuqtadan. MSE chetga chiquvchi qiymatlarga juda sezgir, MAE ancha xotirjam.

Misol 13★★☆

Ikki baravar katta xatolar

$y = (1;\ 2;\ 3)$. Model A: $\hat y = (2;\ 3;\ 4)$; Model B: $\hat y = (3;\ 4;\ 5)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

Model A

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 1-2\rvert = \lvert -1\rvert = 1\\\lvert e_{2}\rvert &= \lvert 2-3\rvert = \lvert -1\rvert = 1\\\lvert e_{3}\rvert &= \lvert 3-4\rvert = \lvert -1\rvert = 1\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 1 + 1 + 1 = 3$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 3 = \frac{3}{3} = 1$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (1-2)^2 = (-1)^2 = 1\\e_{2}^2 &= (2-3)^2 = (-1)^2 = 1\\e_{3}^2 &= (3-4)^2 = (-1)^2 = 1\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 1 + 1 + 1 = 3$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 3 = \frac{3}{3} = 1$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
112−111
223−111
334−111
Yig‘indi33
O‘rtacha (÷ 3)11
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(2-1)}{3} \approx 0{,}33\\g_{2} &= \frac{\operatorname{sign}(3-2)}{3} \approx 0{,}33\\g_{3} &= \frac{\operatorname{sign}(4-3)}{3} \approx 0{,}33\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(2-1)}{3} \approx 0{,}67\\g_{2} &= \frac{2\cdot(3-2)}{3} \approx 0{,}67\\g_{3} &= \frac{2\cdot(4-3)}{3} \approx 0{,}67\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (0{,}33;\ 0{,}33;\ 0{,}33)\\[4pt]\nabla\mathrm{MSE} \approx (0{,}67;\ 0{,}67;\ 0{,}67)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Model B

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 1-3\rvert = \lvert -2\rvert = 2\\\lvert e_{2}\rvert &= \lvert 2-4\rvert = \lvert -2\rvert = 2\\\lvert e_{3}\rvert &= \lvert 3-5\rvert = \lvert -2\rvert = 2\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 2 + 2 + 2 = 6$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 6 = \frac{6}{3} = 2$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (1-3)^2 = (-2)^2 = 4\\e_{2}^2 &= (2-4)^2 = (-2)^2 = 4\\e_{3}^2 &= (3-5)^2 = (-2)^2 = 4\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 4 + 4 + 4 = 12$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 12 = \frac{12}{3} = 4$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
113−224
224−224
335−224
Yig‘indi612
O‘rtacha (÷ 3)24
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(3-1)}{3} \approx 0{,}33\\g_{2} &= \frac{\operatorname{sign}(4-2)}{3} \approx 0{,}33\\g_{3} &= \frac{\operatorname{sign}(5-3)}{3} \approx 0{,}33\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(3-1)}{3} \approx 1{,}33\\g_{2} &= \frac{2\cdot(4-2)}{3} \approx 1{,}33\\g_{3} &= \frac{2\cdot(5-3)}{3} \approx 1{,}33\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (0{,}33;\ 0{,}33;\ 0{,}33)\\[4pt]\nabla\mathrm{MSE} \approx (1{,}33;\ 1{,}33;\ 1{,}33)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Hosilalar nimani ko‘rsatadi

Xatolarni ikki baravar oshirish MSE gradiyentini ham ikki baravar oshiradi (0,67 → 1,33), MAE gradiyenti esa o‘zgarmaydi (0,33): u faqat xatoning ishorasiga bog‘liq.

Solishtirish

MAEMSE
Model A11
Model B24

Javob: Model A: $\mathrm{MAE} = 1$,   $\mathrm{MSE} = 1$; Model B: $\mathrm{MAE} = 2$,   $\mathrm{MSE} = 4$

Misol nimani ko‘rsatadi

B modeli A dan ikki baravar ko‘p adashadi (1 o‘rniga 2 ga). MAE ham ikki baravar oshadi (1 → 2), MSE esa to‘rt baravar (1 → 4): ikki baravar oshgan sonning kvadrati 4 baravar katta.

Misol 14★★★

Hammaga bitta son: o‘rtacha yoki mediana?

$y = (2;\ 4;\ 9)$. Model A: $\hat y = (5;\ 5;\ 5)$; Model B: $\hat y = (4;\ 4;\ 4)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

Model A

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 2-5\rvert = \lvert -3\rvert = 3\\\lvert e_{2}\rvert &= \lvert 4-5\rvert = \lvert -1\rvert = 1\\\lvert e_{3}\rvert &= \lvert 9-5\rvert = \lvert 4\rvert = 4\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 3 + 1 + 4 = 8$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 8 = \frac{8}{3} \approx 2{,}67$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (2-5)^2 = (-3)^2 = 9\\e_{2}^2 &= (4-5)^2 = (-1)^2 = 1\\e_{3}^2 &= (9-5)^2 = 4^2 = 16\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 9 + 1 + 16 = 26$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 26 = \frac{26}{3} \approx 8{,}67$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
125−339
245−111
3954416
Yig‘indi826
O‘rtacha (÷ 3)≈ 2,67≈ 8,67
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(5-2)}{3} \approx 0{,}33\\g_{2} &= \frac{\operatorname{sign}(5-4)}{3} \approx 0{,}33\\g_{3} &= \frac{\operatorname{sign}(5-9)}{3} \approx -0{,}33\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(5-2)}{3} = 2\\g_{2} &= \frac{2\cdot(5-4)}{3} \approx 0{,}67\\g_{3} &= \frac{2\cdot(5-9)}{3} \approx -2{,}67\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (0{,}33;\ 0{,}33;\ -0{,}33)\\[4pt]\nabla\mathrm{MSE} \approx (2;\ 0{,}67;\ -2{,}67)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE 3-obyektni eng kuchli tortadi: uning signali modul bo‘yicha 2,67 — unda eng katta xato (4).

Model B

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 2-4\rvert = \lvert -2\rvert = 2\\\lvert e_{2}\rvert &= \lvert 4-4\rvert = \lvert 0\rvert = 0\\\lvert e_{3}\rvert &= \lvert 9-4\rvert = \lvert 5\rvert = 5\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 2 + 0 + 5 = 7$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 7 = \frac{7}{3} \approx 2{,}33$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (2-4)^2 = (-2)^2 = 4\\e_{2}^2 &= (4-4)^2 = 0^2 = 0\\e_{3}^2 &= (9-4)^2 = 5^2 = 25\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 4 + 0 + 25 = 29$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 29 = \frac{29}{3} \approx 9{,}67$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
124−224
244000
3945525
Yig‘indi729
O‘rtacha (÷ 3)≈ 2,33≈ 9,67
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(4-2)}{3} \approx 0{,}33\\g_{2} &= \frac{\operatorname{sign}(4-4)}{3} = 0\\g_{3} &= \frac{\operatorname{sign}(4-9)}{3} \approx -0{,}33\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(4-2)}{3} \approx 1{,}33\\g_{2} &= \frac{2\cdot(4-4)}{3} = 0\\g_{3} &= \frac{2\cdot(4-9)}{3} \approx -3{,}33\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (0{,}33;\ 0;\ -0{,}33)\\[4pt]\nabla\mathrm{MSE} \approx (1{,}33;\ 0;\ -3{,}33)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE 3-obyektni eng kuchli tortadi: uning signali modul bo‘yicha 3,33 — unda eng katta xato (5).

Hosilalar nimani ko‘rsatadi

Agar model bitta $c$ son bo‘lsa, u bo‘yicha hosila gradiyent komponentlari yig‘indisiga teng. MSE da u $c = 5$ da (o‘rtacha) nolga teng: $2 + 0{,}67 - 2{,}67 = 0$; MAE da — $c = 4$ da (mediana): $0{,}33 + 0 - 0{,}33 = 0$. Hosilaning nolga tengligi — minimum: shuning uchun eng yaxshi o‘zgarmaslar — o‘rtacha va mediana.

Solishtirish

MAEMSE
Model A≈ 2,67≈ 8,67
Model B≈ 2,33≈ 9,67

Javob: Model A: $\mathrm{MAE} \approx 2{,}67$,   $\mathrm{MSE} \approx 8{,}67$; Model B: $\mathrm{MAE} \approx 2{,}33$,   $\mathrm{MSE} \approx 9{,}67$

Misol nimani ko‘rsatadi

A modeli har doim o‘rtacha (2 + 4 + 9) / 3 = 5 bilan, B modeli esa mediana 4 bilan javob beradi. MAE bo‘yicha mediana yaxshiroq (mediana — ≈ 2,33, o‘rtacha — ≈ 2,67), MSE bo‘yicha — o‘rtacha (o‘rtacha — ≈ 8,67, mediana — ≈ 9,67). Bu umumiy qoida: barcha o‘zgarmaslar orasida MAE ni mediana, MSE ni esa o‘rtacha minimallashtiradi.

Misol 15★★☆

Ishorali o‘rtacha xato aldaydi

$y = (10;\ 10;\ 10;\ 10)$,   $\hat y = (15;\ 5;\ 15;\ 5)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 10-15\rvert = \lvert -5\rvert = 5\\\lvert e_{2}\rvert &= \lvert 10-5\rvert = \lvert 5\rvert = 5\\\lvert e_{3}\rvert &= \lvert 10-15\rvert = \lvert -5\rvert = 5\\\lvert e_{4}\rvert &= \lvert 10-5\rvert = \lvert 5\rvert = 5\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{4}\lvert e_i\rvert = 5 + 5 + 5 + 5 = 20$$

3. MAE — yig‘indini $n = 4$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{4}\cdot 20 = \frac{20}{4} = 5$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (10-15)^2 = (-5)^2 = 25\\e_{2}^2 &= (10-5)^2 = 5^2 = 25\\e_{3}^2 &= (10-15)^2 = (-5)^2 = 25\\e_{4}^2 &= (10-5)^2 = 5^2 = 25\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{4}e_i^2 = 25 + 25 + 25 + 25 = 100$$

6. MSE — yig‘indini $n = 4$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{4}\cdot 100 = \frac{100}{4} = 25$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
11015−5525
21055525
31015−5525
41055525
Yig‘indi20100
O‘rtacha (÷ 4)525
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(15-10)}{4} = 0{,}25\\g_{2} &= \frac{\operatorname{sign}(5-10)}{4} = -0{,}25\\g_{3} &= \frac{\operatorname{sign}(15-10)}{4} = 0{,}25\\g_{4} &= \frac{\operatorname{sign}(5-10)}{4} = -0{,}25\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(15-10)}{4} = 2{,}5\\g_{2} &= \frac{2\cdot(5-10)}{4} = -2{,}5\\g_{3} &= \frac{2\cdot(15-10)}{4} = 2{,}5\\g_{4} &= \frac{2\cdot(5-10)}{4} = -2{,}5\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} = \begin{pmatrix}0{,}25\\-0{,}25\\0{,}25\\-0{,}25\end{pmatrix}\\[8pt]\nabla\mathrm{MSE} = \begin{pmatrix}2{,}5\\-2{,}5\\2{,}5\\-2{,}5\end{pmatrix}\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,25 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Javob: $\mathrm{MAE} = 5$,   $\mathrm{MSE} = 25$

Misol nimani ko‘rsatadi

−5, +5, −5, +5 xatolar yig‘indisi 0 — «o‘rtacha model adashmaydi»mi? Yo‘q: har bir bashorat 5 ga xato. Shuning uchun metrikalarda modul yoki kvadrat olinadi: MAE = 5, MSE = 25.

Hayotdagidek

O‘lchov birliklari, baholar, yetkazib berish vaqti, «teskari» masala va asosiy amaliy savol — qaysi modelni tanlash.

Misol 16★★☆

Mingliklardagi narxlar: o‘lchov birliklari

$y = (100;\ 150;\ 200)$,   $\hat y = (110;\ 140;\ 230)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 100-110\rvert = \lvert -10\rvert = 10\\\lvert e_{2}\rvert &= \lvert 150-140\rvert = \lvert 10\rvert = 10\\\lvert e_{3}\rvert &= \lvert 200-230\rvert = \lvert -30\rvert = 30\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 10 + 10 + 30 = 50$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 50 = \frac{50}{3} \approx 16{,}67$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (100-110)^2 = (-10)^2 = 100\\e_{2}^2 &= (150-140)^2 = 10^2 = 100\\e_{3}^2 &= (200-230)^2 = (-30)^2 = 900\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 100 + 100 + 900 = 1100$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 1100 = \frac{1100}{3} \approx 366{,}67$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
1100110−1010100
21501401010100
3200230−3030900
Yig‘indi501100
O‘rtacha (÷ 3)≈ 16,67≈ 366,67
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(110-100)}{3} \approx 0{,}33\\g_{2} &= \frac{\operatorname{sign}(140-150)}{3} \approx -0{,}33\\g_{3} &= \frac{\operatorname{sign}(230-200)}{3} \approx 0{,}33\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(110-100)}{3} \approx 6{,}67\\g_{2} &= \frac{2\cdot(140-150)}{3} \approx -6{,}67\\g_{3} &= \frac{2\cdot(230-200)}{3} = 20\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (0{,}33;\ -0{,}33;\ 0{,}33)\\[4pt]\nabla\mathrm{MSE} \approx (6{,}67;\ -6{,}67;\ 20)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE 3-obyektni eng kuchli tortadi: uning signali modul bo‘yicha 20 — unda eng katta xato (30).

Javob: $\mathrm{MAE} \approx 16{,}67$,   $\mathrm{MSE} \approx 366{,}67$

Misol nimani ko‘rsatadi

Narxlar ming dollarda. MAE ≈ 16,67 ham mingda: «o‘rtacha 16,7 mingga adashamiz». MSE ≈ 366,67 esa «mingning kvadrati»da o‘lchanadi — buni buyurtmachiga tushuntirish qiyin. Shuning uchun yonida ko‘pincha RMSE = √MSE ≈ 19,15 ming hisoblanadi.

Misol 17★★☆

Olti o‘quvchining baholari

$y = (5;\ 4;\ 3;\ 5;\ 4;\ 2)$,   $\hat y = (4;\ 4;\ 4;\ 4;\ 4;\ 4)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 5-4\rvert = \lvert 1\rvert = 1\\\lvert e_{2}\rvert &= \lvert 4-4\rvert = \lvert 0\rvert = 0\\\lvert e_{3}\rvert &= \lvert 3-4\rvert = \lvert -1\rvert = 1\\\lvert e_{4}\rvert &= \lvert 5-4\rvert = \lvert 1\rvert = 1\\\lvert e_{5}\rvert &= \lvert 4-4\rvert = \lvert 0\rvert = 0\\\lvert e_{6}\rvert &= \lvert 2-4\rvert = \lvert -2\rvert = 2\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{6}\lvert e_i\rvert = 1 + 0 + 1 + 1 + 0 + 2 = 5$$

3. MAE — yig‘indini $n = 6$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{6}\cdot 5 = \frac{5}{6} \approx 0{,}83$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (5-4)^2 = 1^2 = 1\\e_{2}^2 &= (4-4)^2 = 0^2 = 0\\e_{3}^2 &= (3-4)^2 = (-1)^2 = 1\\e_{4}^2 &= (5-4)^2 = 1^2 = 1\\e_{5}^2 &= (4-4)^2 = 0^2 = 0\\e_{6}^2 &= (2-4)^2 = (-2)^2 = 4\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{6}e_i^2 = 1 + 0 + 1 + 1 + 0 + 4 = 7$$

6. MSE — yig‘indini $n = 6$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{6}\cdot 7 = \frac{7}{6} \approx 1{,}17$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
154111
244000
334−111
454111
544000
624−224
Yig‘indi57
O‘rtacha (÷ 6)≈ 0,83≈ 1,17
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(4-5)}{6} \approx -0{,}17\\g_{2} &= \frac{\operatorname{sign}(4-4)}{6} = 0\\g_{3} &= \frac{\operatorname{sign}(4-3)}{6} \approx 0{,}17\\g_{4} &= \frac{\operatorname{sign}(4-5)}{6} \approx -0{,}17\\g_{5} &= \frac{\operatorname{sign}(4-4)}{6} = 0\\g_{6} &= \frac{\operatorname{sign}(4-2)}{6} \approx 0{,}17\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(4-5)}{6} \approx -0{,}33\\g_{2} &= \frac{2\cdot(4-4)}{6} = 0\\g_{3} &= \frac{2\cdot(4-3)}{6} \approx 0{,}33\\g_{4} &= \frac{2\cdot(4-5)}{6} \approx -0{,}33\\g_{5} &= \frac{2\cdot(4-4)}{6} = 0\\g_{6} &= \frac{2\cdot(4-2)}{6} \approx 0{,}67\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx \begin{pmatrix}-0{,}17\\0\\0{,}17\\-0{,}17\\0\\0{,}17\end{pmatrix}\\[8pt]\nabla\mathrm{MSE} \approx \begin{pmatrix}-0{,}33\\0\\0{,}33\\-0{,}33\\0\\0{,}67\end{pmatrix}\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,17 signal beradi: faqat xatoning yo‘nalishi muhim. MSE 6-obyektni eng kuchli tortadi: uning signali modul bo‘yicha 0,67 — unda eng katta xato (2).

Javob: $\mathrm{MAE} \approx 0{,}83$,   $\mathrm{MSE} \approx 1{,}17$

Misol nimani ko‘rsatadi

Model hammaga «4» qo‘yadi. Uchtasida 1 ga, bittasida 2 ga adashadi: MAE = 5 / 6 ≈ 0,83 ball. MSE da 2 xato 7 dan 4 ni beradi, shuning uchun MSE ≈ 1,17 MAE dan sezilarli katta.

Misol 18★★☆

Yetkazib berish vaqti daqiqalarda

$y = (30;\ 45;\ 60)$,   $\hat y = (35;\ 45;\ 50)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 30-35\rvert = \lvert -5\rvert = 5\\\lvert e_{2}\rvert &= \lvert 45-45\rvert = \lvert 0\rvert = 0\\\lvert e_{3}\rvert &= \lvert 60-50\rvert = \lvert 10\rvert = 10\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 5 + 0 + 10 = 15$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 15 = \frac{15}{3} = 5$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (30-35)^2 = (-5)^2 = 25\\e_{2}^2 &= (45-45)^2 = 0^2 = 0\\e_{3}^2 &= (60-50)^2 = 10^2 = 100\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 25 + 0 + 100 = 125$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 125 = \frac{125}{3} \approx 41{,}67$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
13035−5525
24545000
360501010100
Yig‘indi15125
O‘rtacha (÷ 3)5≈ 41,67
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(35-30)}{3} \approx 0{,}33\\g_{2} &= \frac{\operatorname{sign}(45-45)}{3} = 0\\g_{3} &= \frac{\operatorname{sign}(50-60)}{3} \approx -0{,}33\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(35-30)}{3} \approx 3{,}33\\g_{2} &= \frac{2\cdot(45-45)}{3} = 0\\g_{3} &= \frac{2\cdot(50-60)}{3} \approx -6{,}67\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (0{,}33;\ 0;\ -0{,}33)\\[4pt]\nabla\mathrm{MSE} \approx (3{,}33;\ 0;\ -6{,}67)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE 3-obyektni eng kuchli tortadi: uning signali modul bo‘yicha 6,67 — unda eng katta xato (10).

Javob: $\mathrm{MAE} = 5$,   $\mathrm{MSE} \approx 41{,}67$

Misol nimani ko‘rsatadi

MAE = 5 daqiqa — mijoz uchun tushunarli javob: «o‘rtacha kuryer 5 daqiqa kechikadi yoki erta keladi». MSE ≈ 41,67 «daqiqa kvadrati» modelni o‘qitish uchun qulay, lekin tushuntirish uchun emas.

Misol 19★★★

Teskari masala: bashoratni toping

Ma’lum: $y = (4;\ 6)$, birinchi bashorat $\hat y_1 = 5$, $\mathrm{MAE} = 1$. Ikkinchi bashorat $\hat y_2$ va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

1. $n = 2$ uchun MAE ni yozamiz va ma’lumlarni qo‘yamiz:

$$\frac{1}{2}\bigl(|4-5| + |6-\hat y_2|\bigr) = 1$$

2. Ikkala tomonni 2 ga ko‘paytiramiz va ma’lum modulni o‘tkazamiz:

$$\begin{aligned}1 + |6-\hat y_2| &= 2\\|6-\hat y_2| &= 1\end{aligned}$$

3. Modul 1 ga teng, demak uning ichidagi ifoda $1$ yoki $-1$:

$$\begin{aligned}6-\hat y_2 = 1 &\Rightarrow \hat y_2 = 5\\6-\hat y_2 = -1 &\Rightarrow \hat y_2 = 7\end{aligned}$$

4. Ikkala javob uchun MSE ni tekshiramiz — u bir xil:

$$\begin{aligned}\hat y_2 = 5:\quad \mathrm{MSE} = \frac{1 + 1}{2} = 1\\\hat y_2 = 7:\quad \mathrm{MSE} = \frac{1 + 1}{2} = 1\end{aligned}$$

Javob: $\hat y_2 = 5$ yoki $\hat y_2 = 7$; MSE = 1.

Misol nimani ko‘rsatadi

Teskari masala ham xuddi shu retsept bilan yechiladi: formulani yozish, ma’lumni qo‘yish va noma’lumni topish. Modul ikkita javob beradi — bashorat yuqoriga ham, pastga ham adashishi mumkin edi.

Misol 20★★★

Qaysi model yaxshiroq?

$y = (10;\ 20;\ 30)$. Model A: $\hat y = (12;\ 18;\ 33)$; Model B: $\hat y = (10;\ 20;\ 36)$. MAE va MSE ni toping.

Avval o‘zingiz yechib ko‘ring, keyin yechimni oching.

Yechim va javob

Model A

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 10-12\rvert = \lvert -2\rvert = 2\\\lvert e_{2}\rvert &= \lvert 20-18\rvert = \lvert 2\rvert = 2\\\lvert e_{3}\rvert &= \lvert 30-33\rvert = \lvert -3\rvert = 3\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 2 + 2 + 3 = 7$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 7 = \frac{7}{3} \approx 2{,}33$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (10-12)^2 = (-2)^2 = 4\\e_{2}^2 &= (20-18)^2 = 2^2 = 4\\e_{3}^2 &= (30-33)^2 = (-3)^2 = 9\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 4 + 4 + 9 = 17$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 17 = \frac{17}{3} \approx 5{,}67$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
11012−224
22018224
33033−339
Yig‘indi717
O‘rtacha (÷ 3)≈ 2,33≈ 5,67
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(12-10)}{3} \approx 0{,}33\\g_{2} &= \frac{\operatorname{sign}(18-20)}{3} \approx -0{,}33\\g_{3} &= \frac{\operatorname{sign}(33-30)}{3} \approx 0{,}33\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(12-10)}{3} \approx 1{,}33\\g_{2} &= \frac{2\cdot(18-20)}{3} \approx -1{,}33\\g_{3} &= \frac{2\cdot(33-30)}{3} = 2\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (0{,}33;\ -0{,}33;\ 0{,}33)\\[4pt]\nabla\mathrm{MSE} \approx (1{,}33;\ -1{,}33;\ 2)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE 3-obyektni eng kuchli tortadi: uning signali modul bo‘yicha 2 — unda eng katta xato (3).

Model B

1. Har bir obyektning xatosi $e_i = y_i - \hat y_i$ va uning moduli:

$$\begin{aligned}\lvert e_{1}\rvert &= \lvert 10-10\rvert = \lvert 0\rvert = 0\\\lvert e_{2}\rvert &= \lvert 20-20\rvert = \lvert 0\rvert = 0\\\lvert e_{3}\rvert &= \lvert 30-36\rvert = \lvert -6\rvert = 6\end{aligned}$$

2. Modullar yig‘indisi:

$$\sum_{i=1}^{3}\lvert e_i\rvert = 0 + 0 + 6 = 6$$

3. MAE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MAE} = \frac{1}{3}\cdot 6 = \frac{6}{3} = 2$$

4. Har bir xatoning kvadrati:

$$\begin{aligned}e_{1}^2 &= (10-10)^2 = 0^2 = 0\\e_{2}^2 &= (20-20)^2 = 0^2 = 0\\e_{3}^2 &= (30-36)^2 = (-6)^2 = 36\end{aligned}$$

5. Kvadratlar yig‘indisi:

$$\sum_{i=1}^{3}e_i^2 = 0 + 0 + 36 = 36$$

6. MSE — yig‘indini $n = 3$ ga bo‘lamiz:

$$\mathrm{MSE} = \frac{1}{3}\cdot 36 = \frac{36}{3} = 12$$
$i$$y_i$$\hat y_i$$e_i$$|e_i|$$e_i^2$
11010000
22020000
33036−6636
Yig‘indi636
O‘rtacha (÷ 3)212
Teskari tarqatish uchun hosilalar

7. MAE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{1}{n}\operatorname{sign}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{\operatorname{sign}(10-10)}{3} = 0\\g_{2} &= \frac{\operatorname{sign}(20-20)}{3} = 0\\g_{3} &= \frac{\operatorname{sign}(36-30)}{3} \approx 0{,}33\end{aligned}$$

8. MSE ning har bir bashorat bo‘yicha hosilasi: $g_i = \frac{2}{n}(\hat y_i - y_i)$:

$$\begin{aligned}g_{1} &= \frac{2\cdot(10-10)}{3} = 0\\g_{2} &= \frac{2\cdot(20-20)}{3} = 0\\g_{3} &= \frac{2\cdot(36-30)}{3} = 4\end{aligned}$$

Bashoratlar bo‘yicha gradiyent vektorlari ($\nabla = \nabla_{\hat y}$):

$$\begin{gathered}\nabla\mathrm{MAE} \approx (0;\ 0;\ 0{,}33)\\[4pt]\nabla\mathrm{MSE} = (0;\ 0;\ 4)\end{gathered}$$

MAE har bir adashishga bir xil kuchli $\pm\tfrac{1}{n}$ = ±0,33 signal beradi: faqat xatoning yo‘nalishi muhim. MSE da noldan farqli barcha signallar modul bo‘yicha bir xil: xatolar bir xil kattalikda.

Solishtirish

MAEMSE
Model A≈ 2,33≈ 5,67
Model B212

Javob: Model A: $\mathrm{MAE} \approx 2{,}33$,   $\mathrm{MSE} \approx 5{,}67$; Model B: $\mathrm{MAE} = 2$,   $\mathrm{MSE} = 12$

Misol nimani ko‘rsatadi

MAE bo‘yicha B yaxshiroq (B — 2, A — ≈ 2,33): u uchta nuqtadan ikkitasida aniq. MSE bo‘yicha A yaxshiroq (A — ≈ 5,67, B — 12): unda katta adashish yo‘q. Javob masalaga bog‘liq: kamdan-kam katta xatolar xavfli bo‘lsa (masalan, ko‘prikka tushadigan yukni bashorat qilish), MSE bo‘yicha tanlang; odatiy xato muhim bo‘lsa — MAE bo‘yicha.

Bashoratdan vaznlarga: zanjir qoidasi

Aktivatsiya funksiyasiz eng oddiy neyronni olamiz: $\hat y_i = w\,x_i + b$. Kirishlar $x = (1;\ 2;\ 3)$, to‘g‘ri javoblar $y = (3;\ 5;\ 8)$, joriy vaznlar $w = 2$, $b = 0$ — demak, bashoratlar $\hat y = (2;\ 4;\ 6)$. Bashoratlar bo‘yicha gradiyent qanday qilib vaznlar bo‘yicha gradiyentga aylanadi?

Har bir bashorat $w$ ga $x_i$ ko‘paytuvchi orqali, $b$ ga esa $1$ ko‘paytuvchi bilan bog‘liq: $\frac{\partial \hat y_i}{\partial w} = x_i$, $\frac{\partial \hat y_i}{\partial b} = 1$. Zanjir qoidasi bo‘yicha:

$$\begin{aligned}\frac{\partial L}{\partial w} &= \sum_{i=1}^{n} \frac{\partial L}{\partial \hat y_i}\cdot\frac{\partial \hat y_i}{\partial w} = \sum_{i=1}^{n} g_i\,x_i\\ \frac{\partial L}{\partial b} &= \sum_{i=1}^{n} g_i\end{aligned}$$

MSE bo‘yicha

Bashoratlar bo‘yicha gradiyent:

$$\nabla_{\hat y}\mathrm{MSE} \approx (-0{,}67;\ -0{,}67;\ -1{,}33)$$

Vaznlar bo‘yicha gradiyent:

$$\begin{aligned}\frac{\partial\,\mathrm{MSE}}{\partial w} &= (-\tfrac{2}{3})\cdot 1 + (-\tfrac{2}{3})\cdot 2\\ &\quad + (-\tfrac{4}{3})\cdot 3 = -6\\[6pt]\frac{\partial\,\mathrm{MSE}}{\partial b} &= (-\tfrac{2}{3}) + (-\tfrac{2}{3}) + (-\tfrac{4}{3})\\ &= -\frac{8}{3} \approx -2{,}67\end{aligned}$$

$\eta = 0{,}1$ bilan tushish qadami:

$$\begin{aligned}w &\leftarrow 2 - 0{,}1\cdot (-6) = 2{,}6\\b &\leftarrow 0 - 0{,}1\cdot (-\tfrac{8}{3}) = \tfrac{4}{15} \approx 0{,}27\end{aligned}$$

Qadamdan keyingi yo‘qotish: MSE 2 → 0,08.

MAE bo‘yicha

Bashoratlar bo‘yicha gradiyent:

$$\nabla_{\hat y}\mathrm{MAE} \approx (-0{,}33;\ -0{,}33;\ -0{,}33)$$

Vaznlar bo‘yicha gradiyent:

$$\begin{aligned}\frac{\partial\,\mathrm{MAE}}{\partial w} &= (-\tfrac{1}{3})\cdot 1 + (-\tfrac{1}{3})\cdot 2\\ &\quad + (-\tfrac{1}{3})\cdot 3 = -2\\[6pt]\frac{\partial\,\mathrm{MAE}}{\partial b} &= (-\tfrac{1}{3}) + (-\tfrac{1}{3}) + (-\tfrac{1}{3})\\ &= -1\end{aligned}$$

$\eta = 0{,}1$ bilan tushish qadami:

$$\begin{aligned}w &\leftarrow 2 - 0{,}1\cdot (-2) = 2{,}2\\b &\leftarrow 0 - 0{,}1\cdot (-1) = 0{,}1\end{aligned}$$

Qadamdan keyingi yo‘qotish: MAE ≈ 1,33 → ≈ 0,83.

MSE da $w$ bo‘yicha hosila −6 ga, MAE da — −2 ga teng: uchinchi obyekt 2 ga adashadi va MSE balandroq «qichqiradi» — qadam kattaroq chiqadi. Bir xil xato signali yo‘qotishdan vaznlarga shunday qaytadi; chuqur tarmoqda zanjir shunchaki uzunroq: signal har bir qatlamning hosilasiga ko‘paytiriladi.

O‘z misolingizni hisoblang

To‘g‘ri javoblar $y$ va bashoratlar $\hat y$ ni kiriting — kalkulyator yechimni xuddi shu qadamlar bo‘yicha — teskari tarqatish uchun hosilalar bilan birga — yozib beradi. Sonlarni nuqtali vergul yoki bo‘sh joy bilan ajrating; kasr qismini vergul yoki nuqta bilan yozing.

Qo‘yish:

Tez-tez uchraydigan xatolar

  • Modulni unutish. −2 va +2 xatolar nolga «qisqardi» — lekin model ikkala marta ham adashgan (5- va 15-misollar).
  • Minusni noto‘g‘ri kvadratga oshirish. $(-3)^2 = 9$, $-9$ emas: istalgan sonning kvadrati manfiy emas (8-misol).
  • Noto‘g‘ri $n$ ga bo‘lish. Obyektlar soniga bo‘lamiz, noldan farqli xatolar soniga emas (7-misol).
  • Avval o‘rtacha olib, keyin kvadratga oshirish. MSE — kvadratlarning o‘rtachasi, o‘rtachaning kvadrati emas: −5 va +5 xatolar uchun MSE = 25, «o‘rtacha xatoning kvadrati» esa 0 bo‘lardi.
  • MAE va MSE ni «kattaligi bo‘yicha» solishtirish. Ularning birliklari har xil: xatolar 1 dan kichik bo‘lsa, MSE MAE dan kichik, katta bo‘lsa — katta (6- va 3-misollar).
  • Ayirish tartibini almashtirish. Qo‘rqinchli emas: $|y - \hat y| = |\hat y - y|$ va $(y - \hat y)^2 = (\hat y - y)^2$. Lekin ishorali xato odatda $y_i - \hat y_i$ deb olinadi.
  • Hosila ishorasini adashtirish. $\frac{\partial}{\partial \hat y_i}(y_i - \hat y_i)^2 = 2(y_i - \hat y_i)\cdot(-1) = 2(\hat y_i - y_i)$: ichki hosila minus beradi. Noto‘g‘ri ishora bilan tushish yuqoriga ketadi va yo‘qotish o‘sadi.
  • $\tfrac{2}{n}$ yoki $\tfrac{1}{n}$ ko‘paytuvchini unutish. Yo‘nalish o‘zgarmaydi, lekin qadam $n$ (yoki $2n$) baravar katta chiqadi — u bilan birga mos o‘rganish tezligi ham o‘zgaradi.
  • MAE hosilasini nolda izlash. U yerda hosila yo‘q: modul sinadi. Subgradiyent olinadi — odatda $0$, PyTorch dagi kabi.

Xulosa

MAEMSE
Formula$\tfrac{1}{n}\sum |e_i|$$\tfrac{1}{n}\sum e_i^2$
Birliklar$y$ bilan bir xil$y$ birligining kvadrati
Katta xatolarmutanosib hisobga oladikuchaytiradi: xato ×2 → hissa ×4
Chetga chiquvchilarbardoshlisezgir
Eng yaxshi o‘zgarmasmedianao‘rtacha
Qachon tanlash kerakodatiy xato muhim, chetga chiquvchilar borkatta adashishlar ayniqsa xavfli
$\hat y_i$ bo‘yicha hosila$-\tfrac{1}{n}\operatorname{sign} e_i$$-\tfrac{2}{n}\,e_i$
Signal kuchibir xil: $\pm\tfrac{1}{n}$xatoga mutanosib
  • Retsept bitta: xatolar → modullar yoki kvadratlar → yig‘indi → $n$ ga bo‘lish.
  • Modul va kvadrat ishorani olib tashlaydi, shuning uchun yuqoriga va pastga adashishlar bir-birini yo‘qqa chiqarmaydi.
  • MSE katta xatolarni «jazolaydi», MAE — yo‘q; bitta model bir metrika bo‘yicha yaxshiroq, boshqasi bo‘yicha yomonroq bo‘lishi mumkin.
  • Teskari tarqatish uchun yo‘qotishning bashorat bo‘yicha hosilasi kerak: MSE da $\tfrac{2}{n}(\hat y_i - y_i)$, MAE da $\tfrac{1}{n}\operatorname{sign}(\hat y_i - y_i)$. Keyin u zanjir qoidasi bo‘yicha qatlamlar hosilalariga ko‘paytiriladi.
  • To‘liq nazariya — qo‘llanmaning MAE, MSE va RMSE sahifalarida.

O‘zingizni tekshiring

1. $y = (2;\ 4)$, $\hat y = (3;\ 1)$. MAE nimaga teng?
2. Xuddi shu sonlar uchun MSE nimaga teng?
3. Barcha xatolar uch baravar oshirildi. MSE necha baravar oshadi?
4. $(1;\ 2;\ 10)$ ma’lumotlar uchun MAE ni qaysi bitta $c$ soni minimallashtiradi?
5. Model xatolari: $0{,}1;\ 0{,}2;\ 0{,}3$. Qaysi biri katta?
6. $y = (2;\ 4)$, $\hat y = (3;\ 1)$. Bashoratlar bo‘yicha MSE gradiyenti nimaga teng?
7. Xuddi shu sonlar uchun MAE gradiyenti nimaga teng?
8. Bitta obyektning xatosi 1 dan 10 gacha oshdi. Uning MAE gradiyenti komponenti qanday o‘zgaradi?

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