Expected value, variance, standard deviation — properties: Worked examples

4.9.4 · D3 · Maths › Probability Theory & Statistics › Expected value, variance, standard deviation — properties

Yeh page expected value, variance aur standard deviation ki properties ke har case ko drill karne ke liye hai. Kuch bhi solve karne se pehle hum saare case classes ka ek map banate hain taaki aap dekh sako ki kuch bhi choot nahi raha. Phir har worked example ko us map cell ke saath tag kiya gaya hai jo woh cover karta hai.

Agar koi symbol yahan aajnabi lage, toh woh parent note mein banaya gaya tha — lekin hum har ek ko use karte waqt zero se re-anchor karte hain.

The scenario matrix

Ise padhein jaise: "problem se problem mein kya badal sakta hai?" Har row ek alag tarah ki situation hai. Neeche ke humare examples is tarah chuné gaye hain ki har row ko kam se kam ek baar hit kiya jaaye.

#	Case class	Usmein tricky kya hai	Covered by
A	Positive scale, shift ( $a X + b$ , $a > 0$ )	shift vanish ho jaata hai, scale square hota hai	Ex 1
B	Negative scale ( $a < 0$ )	SD mein $∣ a ∣$ use hota hai; sign traps	Ex 2
C	Zero scale / degenerate constant ( $a = 0$ )	variance $0$ ho jaata hai	Ex 3
D	Independent variables ka sum	covariance term khatam ho jaata hai	Ex 4
E	Dependent variables ka sum (incl. $X + X$ )	$2 Cov$ rakhna zaroori hai	Ex 5
F	Limiting behaviour (Bernoulli spread vs $p$ )	spread sabse bada/zero kahan?	Ex 6
G	Real-world word problem (units, money)	words ko symbols mein translate karo	Ex 7
H	Exam twist — $n$ i.i.d. ka average (shrinking spread)	bade theorems se connect hota hai	Ex 8

Hum teen tools par lean karenge. Yeh rahe sab ek jagah, har ek ke saath woh sawaal jo woh answer karta hai:

Yahan $E [X]$ ka matlab hai $X$ ki average value (balance point); $Var (X)$ us balance point se average squared distance hai; $SD (X) = Var (X)$ hume ordinary units mein wapas laata hai; aur $Cov (X, Y) = E [X Y] - E [X] E [Y]$ measure karta hai ki do variables saath-saath kaise move karte hain.

Humaara running character hai fair die: $X \in {1, 2, 3, 4, 5, 6}$ , har ek probability $\frac{1}{6}$ ke saath. Parent note se hum pehle se jaante hain uske do summaries:

$E [X] = 3.5, Var (X) = \frac{35}{12} \approx 2.9167, SD (X) \approx 1.708.$

Neeche ki picture pehle teen examples ke liye stretch-and-slide ka poora idea ek mass distribution par dikhati hai — pehle teen examples ke liye ise apne dimag mein rakho.

Figure — Expected value, variance, standard deviation — properties

Example 1 — Cell A: positive scale + shift

Worked example Die pips ko score mein convert karo

Ek game mein har die roll ek score $Y = 10 X + 5$ mein badal jaata hai. $E [Y]$ , $Var (Y)$ , $SD (Y)$ nikalo.

Forecast: padhne se pehle andaza lagao — kya " $+ 5$ " spread badalta hai? Kya " $10$ " variance ko $10$ se multiply karta hai ya kisi aur cheez se?

Step 1. $E [Y] = 10 E [X] + 5 = 10 (3.5) + 5 = 40$ . Yeh step kyun? Mean linear hai — ek stretch centre ko scale karta hai, ek slide use move karta hai. Dono bachte hain.

Step 2. $Var (Y) = 1 0^{2} Var (X) = 100 \cdot \frac{35}{12} = \frac{3500}{12} \approx 291.67$ . Yeh step kyun? $+ 5$ poore mass ko rigidly slide karta hai — centre-se-distances unchanged rehti hain, isliye yeh variance se vanish ho jaata hai. $10$ har distance ko $10$ se stretch karta hai; squared distances ko $1 0^{2}$ se.

Step 3. $SD (Y) = ∣ 10 ∣ \cdot 1.708 = 17.08$ . Yeh step kyun? SD ek distance hai, isliye yeh raw factor (absolute value) se scale hoti hai, uske square se nahi.

Verify: $291.67 \approx 17.08$ ✓, Step 3 se match karta hai. Units check: agar pips $10$ se scale hon, toh "score units" mein spread $10$ se scale hogi — consistent hai.

Example 2 — Cell B: negative scale (the sign trap)

Worked example Die ko flip karo

Maano $Z = - 2 X + 1$ . $E [Z]$ , $Var (Z)$ , $SD (Z)$ nikalo.

Forecast: $- 2$ distribution ko left-to-right flip karta hai. Kya variance negative ho jaayega? Kya SD $- 2 \cdot 1.708$ banega?

Step 1. $E [Z] = - 2 (3.5) + 1 = - 6$ . Yeh step kyun? Mean ki linearity $a$ ke kisi bhi sign ke liye hold karti hai; flip centre ko negative side par le jaata hai.

Step 2. $Var (Z) = (- 2)^{2} Var (X) = 4 \cdot \frac{35}{12} = \frac{140}{12} = \frac{35}{3} \approx 11.667$ . Yeh step kyun? $a^{2} = (- 2)^{2} = 4$ — squaring sign ko erase kar deta hai. Variance kabhi negative nahi ho sakta; ek mirror flip yeh nahi badalta ki points kitne scattered hain.

Step 3. $SD (Z) = ∣ - 2 ∣ \cdot 1.708 = 2 \cdot 1.708 = 3.416$ . Yeh step kyun? SD mein $∣ a ∣$ use hota hai. $- 2 \cdot 1.708$ likhne se ek negative "distance" milta, jo impossible hai.

Verify: $35/3 \approx 3.416$ ✓. Sanity: darts ke ek cloud ko ek point ke baare mein reflect karne se cloud utna hi wide rehta hai.

Example 3 — Cell C: zero scale / degenerate constant

Example 4 — Cell D: independent variables ka sum

Worked example Do independent dice

Do independent fair dice $X_{1}, X_{2}$ roll karo aur $S = X_{1} + X_{2}$ maano. $E [S]$ , $Var (S)$ , $SD (S)$ nikalo.

Forecast: mean zaroor double hokar $7$ ho jaayega. Kya variance bhi bas double ho jaata hai?

Step 1. $E [S] = E [X_{1}] + E [X_{2}] = 3.5 + 3.5 = 7$ . Yeh step kyun? Expectation ki linearity hamesha hold karti hai, independent ho ya na ho.

Step 2. Kyunki dice independent hain, $Cov (X_{1}, X_{2}) = 0$ , isliye $Var (S) = Var (X_{1}) + Var (X_{2}) = \frac{35}{12} + \frac{35}{12} = \frac{70}{12} = \frac{35}{6} \approx 5.833.$ Yeh step kyun? Ek die doosre ke baare mein kuch nahi batata, isliye koi "saath-saath move karna" wala term add nahi hota. Dekho Covariance and Correlation ki independence $Cov = 0$ kyun force karti hai.

Step 3. $SD (S) = 35/6 \approx 2.415$ . Yeh step kyun? Root pip units mein wapas laata hai.

Verify: $2 \times 2.9167 = 5.833$ ✓. Note karo $SD (S) \approx 2.415$ kam hai $2 \times 1.708 = 3.416$ se: independent spreads quadrature mein add hoti hain (Pythagoras jaisi), linearly nahi. Yeh Law of Large Numbers ka beej hai.

Example 5 — Cell E: dependent sum, including $X + X$

Worked example The self-sum trap

Ek die $X$ ke liye, $Var (X + X)$ ko "independent guess" $2 Var (X)$ se compare karo.

Forecast: $X + X$ aur $2 X$ ek hi cheez hain. Kya sum rule phir bhi $2 Var (X)$ deta hai?

Step 1. $X + X = 2 X$ , isliye scale rule se $Var (2 X) = 2^{2} Var (X) = 4 \cdot \frac{35}{12} = \frac{35}{3} \approx 11.667$ . Yeh step kyun? Yeh $2$ se stretch karna hai, cell B/A machinery — koi independence assumed nahi.

Step 2. Ab sum rule se $Y = X$ ke saath: $Cov (X, X) = Var (X)$ (ek variable apne aap se perfectly correlated hota hai). Isliye $Var (X + X) = Var (X) + Var (X) + 2 Var (X) = 4 Var (X) .$ Yeh step kyun? $Cov (X, X) = E [X \cdot X] - E [X] E [X] = E [X^{2}] - μ^{2} = Var (X)$ — yahi toh variance ki definition hai.

Step 3. Dono routes agree karte hain: $4 \cdot \frac{35}{12} = \frac{35}{3} \approx 11.667$ , nahi $2 \cdot \frac{35}{12} \approx 5.833$ . Yeh step kyun? "Bas variances add karo" wala naive answer exactly $2 Cov$ term ki wajah se galat hai.

Verify: Step 1 se $35/3 \approx 11.667$ , $4 \times 2.9167$ ke barabar ✓. Dependent answer Ex 4 ke independent answer se double hai — dependence bahut zyada matter karta hai.

Example 6 — Cell F: limiting behaviour (spread sabse bada kahan?)

Worked example Bernoulli spread jab

$p$ vary karta hai Ek indicator $X$ , probability $p$ ke saath $1$ hai, warna $0$ . Uska variance hai $Var (X) = p (1 - p)$ . Kis $p$ par spread sabse bada hai? Limits $p \to 0$ aur $p \to 1$ par kya hota hai?

Forecast: ek coin $p = \frac{1}{2}$ par "sabse uncertain" hota hai. Kya aap expect karte ho ki variance wahan peak kare aur ends par vanish ho jaaye?

Step 1. $v (p) = p (1 - p) = p - p^{2}$ likho. Yeh step kyun? Hum "spread vs $p$ " ko ek variable ki plain function mein badal dete hain taaki uski top find kar sakein.

Step 2. Top wahan hai jahan slope zero hai: $v^{'} (p) = 1 - 2 p = 0 \Rightarrow p = \frac{1}{2}$ , jo deta hai $v (\frac{1}{2}) = \frac{1}{4}$ . Derivative kyun? $v^{'} (p)$ answer deta hai "kya spread abhi bhi badh raha hai jab main $p$ ko nudge karta hun?" Yeh exactly peak par zero cross karta hai.

Step 3. Limits: $v (0) = 0$ aur $v (1) = 0$ . Yeh step kyun? $p = 0$ par (kabhi nahi hota) ya $p = 1$ par (hamesha hota hai) outcome ek constant hai — yeh wahi Ex 3 wala degenerate zero-variance case hai.

Verify: $v (0.5) = 0.25$ ; $v (0.1) = 0.09$ ; $v (0) = v (1) = 0$ ✓. Curve ek ulta parabola hai jo fair coin par peak karta hai — dekho Bernoulli and Binomial Distributions.

Example 7 — Cell G: real-world word problem (money & units)

Worked example Daily profit

Ek stall ki daily demand $D$ (units sold) mein $E [D] = 40$ aur $SD (D) = 6$ hai. Har unit se $3 profit hota hai lekin ek fixed $20 daily rent hai. Profit hai $P = 3 D - 20$ . Mean profit aur uski standard deviation nikalo.

Forecast: kya $20 rent affect karti hai ki profit din-o-din kitna variable hai?

Step 1. Words ko symbols mein translate karo: $P = 3 D - 20$ , isliye $a = 3$ (dollars per unit), $b = - 20$ (fixed rent). Yeh step kyun? Har "per-unit price" ek scale $a$ hai; har "fixed cost" ek shift $b$ hai.

Step 2. $E [P] = 3 E [D] - 20 = 3 (40) - 20 = 100$ dollars. Yeh step kyun? Mean ki linearity.

Step 3. $Var (D) = SD (D)^{2} = 6^{2} = 36$ , isliye $Var (P) = 3^{2} \cdot 36 = 324$ , aur $SD (P) = ∣ 3 ∣ \cdot 6 = 18$ dollars. Yeh step kyun? Fixed rent profit ko neeche slide karti hai lekin din-o-din scatter nahi badlati, isliye $b = - 20$ drop out ho jaata hai. $3/unit scatter ko $3$ se stretch karta hai.

Verify: $324 = 18$ ✓. Units: SD of $P$ dollars mein hai ( $=3\ \tfrac{\$ }{\text{unit}}\times 6\ \text{units} $) ✓. Mean \$ 100, typical swing $18.

Example 8 — Cell H: exam twist ( $n$ i.i.d. ka average)

Worked example Averaging randomness ko kyun kabu karta hai

$n$ independent fair dice lo aur sample mean $\overset{ˉ}{X}_{n} = \frac{X _{1} + \dots + X _{n}}{n}$ banao. $E [\overset{ˉ}{X}_{n}]$ aur $Var (\overset{ˉ}{X}_{n})$ nikalo. $n \to \infty$ par kya hota hai?

Forecast: average abhi bhi " $3.5$ ke aas-paas" hai. Kya uska spread $n$ badhne par same rehta hai, badh jaata hai, ya shrink hota hai?

Step 1. $E [\overset{ˉ}{X}_{n}] = \frac{1}{n} \sum_{i = 1}^{n} E [X_{i}] = \frac{1}{n} \cdot n \cdot 3.5 = 3.5$ . Yeh step kyun? Linearity: same-mean variables ke average ka mean woh common mean hota hai, kisi bhi $n$ ke liye.

Step 2. Independence variances ko add karne deta hai: $Var (\sum_{i} X_{i}) = n Var (X)$ . Phir $\frac{1}{n}$ ek scale hai, squared: $Var (\overset{ˉ}{X}_{n}) = \frac{1}{n ^{2}} \cdot n Var (X) = \frac{Var ( X )}{n} = \frac{35/12}{n} .$ Yeh step kyun? Sum rule (cell D) phir scale rule (cell A) $a = \frac{1}{n}$ ke saath, isliye $a^{2} = \frac{1}{n ^{2}}$ .

Step 3. Jab $n \to \infty$ , $Var (\overset{ˉ}{X}_{n}) = \frac{35/12}{n} \to 0$ , aur $SD (\overset{ˉ}{X}_{n}) = \frac{1.708}{n}$ $1/ n$ ki tarah shrink hota hai. Yeh step kyun? $n$ se divide karna spread ko zero par drive kar deta hai — average $3.5$ par concentrate ho jaata hai.

Verify (numeric): $n = 100$ ke liye, $Var (\overset{ˉ}{X}_{100}) = \frac{35/12}{100} \approx 0.029167$ aur $SD \approx 0.1708$ ✓. Mean $3.5$ par pinned hai lekin swing ek die se $10 \times$ chhota hai. Yahi mechanism hai Law of Large Numbers ke peeche aur Central Limit Theorem mein $1/ n$ scale ka.

Recall

Recall Poore matrix ke ek-line answers

$Var$ mein shift $b$ ? ::: Hamesha vanish ho jaata hai — spread shift-invariant hai. $Var$ mein scale $a$ ? ::: $a^{2}$ se multiply hota hai; SD mein $∣ a ∣$ se. Ek constant ka $Var$ ? ::: Exactly $0$ (degenerate). Independent $X, Y$ ke liye $Var (X + Y)$ ? ::: $Var (X) + Var (Y)$ . $Var (X + X)$ ? ::: $4 Var (X)$ , $2 Var (X)$ nahi — $Cov (X, X) = Var (X)$ . Bernoulli spread sabse bada kahan? ::: $p = \frac{1}{2}$ par, value $\frac{1}{4}$ ; $p = 0, 1$ par zero. $n$ i.i.d. ke average ka $Var$ ? ::: $Var (X) / n \to 0$ ; SD $1/ n$ ki tarah shrink hoti hai.