Expected value, variance, standard deviation — properties: Question bank

4.9.4 · D5 · Maths › Probability Theory & Statistics › Expected value, variance, standard deviation — properties

Definition Is page par use hone wale symbols

== $X, Y$ == random variables hain — woh numbers jo ek chance experiment se nikalte hain.
== $E [X]$ expected value (mean) hai==, $X$ ka probability-weighted average; hum ise $μ$ (mu) se abbreviate karte hain. Toh $μ = E [X]$ distribution ka balance point hai.
== $Var (X) = E [(X - μ)^{2}]$ variance hai==, mean se average squared distance.
== $SD (X) = Var (X)$ standard deviation hai==, abbreviated $σ$ (sigma); yeh spread $X$ ke apne units mein measured hoti hai, isliye $σ^{2} = Var (X)$ .
== $Cov (X, Y) = E [(X - E [X]) (Y - E [Y])] = E [X Y] - E [X] E [Y]$ covariance hai==, jo measure karta hai ki $X$ aur $Y$ saath mein kaise move karte hain (dekho Covariance and Correlation).
== $X_{1}, \dots, X_{n}$ $X$ ke $n$ independent copies hain (same distribution, unlinked outcomes), aur overbar $\overset{ˉ}{X} = \frac{1}{n} (X_{1} + \dots + X_{n})$ sample mean hai==, unka average. Yahan $n$ sample size hai.
== $p$ ek probability hai== ( $0$ aur $1$ ke beech ka number), neeche Bernoulli $0/1$ variable ke liye use hota hai.

Is page par har symbol inhi mein se ek hai.

Teen pictures jo dimaag mein rakhni hain

Traps se pehle, teen quick visuals — is page ke zyaadatar concept errors dissolve ho jaate hain jab aap inhe dekhte hain.

Figure 1 — Squaring cancellation ko kyun rokta hai. Deviations $X - μ$ left (negative) aur right (positive) point karte hain; averaged raw, woh exactly $0$ par cancel ho jaate hain. Squaring dono sides ko upar fold kar deta hai taaki kuch cancel na ho.

Figure — Expected value, variance, standard deviation — properties

Figure 2 — Variances "in quadrature" kyun add hote hain. Do independent spreads ek right triangle ki do legs ki tarah combine hote hain: total SD hypotenuse $σ_{X}^{2} + σ_{Y}^{2}$ hota hai, jo hamesha $σ_{X} + σ_{Y}$ se chhota hota hai.

Figure 3 — $E [X^{2}] \geq (E [X])^{2}$ kyun (Jensen). Convex cup $y = x^{2}$ ke liye, curve heights ka average curve ke upar hota hai average input par. Woh vertical gap exactly variance hai.

True ya false — justify karo

E [X + Y] = E [X] + E [Y]

ke liye

X

aur

Y

ka independent hona zaroori hai.

False — linearity of expectation hamesha hoti hai, chahe variables dependent hon, kyunki expectation sirf ek weighted sum hai aur sums unconditionally split ho jaate hain.

Var (X + Y) = Var (X) + Var (Y)

ke liye independence zaroori hai.

Iske liye sirf

Cov (X, Y) = 0

(uncorrelatedness) chahiye; independence sufficient hai par zaroori se thoda zyaada strong hai. Dekho Covariance and Correlation.

Variance negative ho sakta hai agar

X

negative values leta hai.

False — variance squared deviations ka average hai, aur squares kabhi negative nahi hote, isliye

Var (X) \geq 0

regardless of the sign of

X

Var (X) = 0

ek genuine random variable ke liye possible hai.

True, lekin sirf tab jab

X

ek constant ho (ek "degenerate" distribution): zero spread ka matlab hai ki har outcome probability one ke saath mean

μ

ke barabar hai.

E [X^{2}] \geq (E [X])^{2}

har random variable ke liye.

True — unka difference exactly

Var (X) \geq 0

hai, jo convex function

x^{2}

par apply ki gayi Jensen's inequality bhi hai.

Standard deviation

X

ke same units mein measured hoti hai.

True — yehi poora reason hai ki hum square root lete hain; variance units-squared mein rehta hai, aur

hume original scale par wapas laata hai.

SD (a X + b) = a SD (X)

False — sahi factor

∣ a ∣

hai, absolute value, kyunki standard deviation ek distance hai aur

a < 0

hone par bhi

\geq 0

rehni chahiye.

Agar

Y = - X

hai toh

Var (Y) = Var (X)

True — scale rule ko

a = - 1

ke saath use karne par

(- 1)^{2} Var (X) = Var (X)

milta hai; number line ko flip karna cluster ko reflect karta hai par scatter nahi badalta.

E [X Y] = E [X] E [Y]

sabhi random variables ke liye.

False — yeh factorisation sirf tab hoti hai jab

X

aur

Y

uncorrelated hon (khaaske, independent); warna gap exactly

Cov (X, Y)

hai.

Mean

E [X]

zaroori ek aisi value honi chahiye jo

X

actually le sakta hai.

False — ek fair die mein

E [X] = 3.5

hota hai, jo kabhi roll nahi hota; mean ek balance point hai, koi attainable outcome nahi.

Error dhundho

Var (2 X) = 2 Var (X)

kyunki variance linear hai."

Error yeh hai ki variance ko expectation ki tarah treat kiya gaya; variance squared deviations use karta hai, isliye constant squared hota hai:

Var (2 X) = 4 Var (X)

Var (X + X) = Var (X) + Var (X) = 2 Var (X)

Sum rule ke liye independence chahiye, lekin

X

apne aap se perfectly correlated hai; sahi tarika:

X + X = 2 X

deta hai

4 Var (X)

E [1/ X] = 1/ E [X]

kyunki expectation functions se pass through ho jaati hai."

Sirf linear functions pass through hote hain;

1/ x

nonlinear hai, isliye generally

E [1/ X] \neq = 1/ E [X]

(Jensen gap ki direction bhi fix karta hai).

SD (X + Y) = SD (X) + SD (Y)

independent

X, Y

ke liye."

Variances add hote hain, SDs nahi:

Var (X + Y) = Var (X) + Var (Y)

, isliye SDs in quadrature add hote hain,

SD = σ_{X}^{2} + σ_{Y}^{2}

, jo

σ_{X} + σ_{Y}

se chhota hai (dekho Figure 2).

Var (X - Y) = Var (X) - Var (Y)

Andar subtraction se variances subtract nahi hote;

Var (X - Y) = Var (X) + Var (Y) - 2 Cov (X, Y)

, aur independent variables ke liye yeh

Var (X) + Var (Y)

tak add ho jaata hai.

"Ek Bernoulli variable ke liye,

E [X^{2}] = (E [X])^{2} = p^{2}

0/1

variable ke liye

X^{2} = X

hota hai, isliye

E [X^{2}] = p

hai, na ki

p^{2}

; variance

p - p^{2} = p (1 - p)

exactly yahi gap hai. Dekho Bernoulli and Binomial Distributions.

"Constant

b

add karne se mean shift hota hai aur variance bhi increase hota hai."

b

add karna poori distribution ko slide karta hai lekin spread untouched rehti hai, isliye

Var (X + b) = Var (X)

— sirf mean

μ

move karta hai.

Cov (X, X) = 1

kyunki ek variable apne aap se perfectly correlated hai."

Perfect correlation

1

hoti hai, lekin

X

ka apne aap se covariance

Var (X)

hai, jo koi bhi nonnegative number ho sakta hai — tum covariance ko correlation ke saath confuse kar rahe ho.

Why questions

Hum deviations ko square kyun karte hain instead of simply

X - μ

average karne ke (yaad raho

μ = E [X]

Kyunki

E [X - μ] = 0

hamesha hota hai (positive aur negative deviations mean ki definition se cancel ho jaate hain), isliye squaring zaroori hai taaki cancellation ruke aur spread ka ek real measure mile — exactly woh fold jo Figure 1 mein dikhaya gaya hai.

Square kyun lein, absolute value

∣ X - μ ∣

kyun nahi?

Squaring clean, differentiable algebra deta hai (ek computational formula, independence ke under additivity) jo

∣ \cdot ∣

mein nahi hota, aur yeh large deviations ko zyaada heavily penalise karta hai.

Var (a X + b)

mein

+ b

kyun gaayab ho jaata hai?

Kyunki naya mean bhi

b

se shift hota hai, isliye

(a X + b) - (a μ + b)

mein dono

+ b

terms cancel ho jaate hain, sirf

a

se scaling bachti hai.

Independence covariance ko zero kyun banata hai?

Independence force karta hai

E [X Y] = E [X] E [Y]

, aur covariance exactly

E [X Y] - E [X] E [Y]

define hoti hai, isliye yeh

0

ho jaata hai.

E [X^{2}]

kabhi

(E [X])^{2}

se chhota kyun nahi hota?

Unka difference hi variance hai, squares ka average, jo negative nahi ho sakta; equality sirf tab hoti hai jab

X

constant ho (Figure 3 mein convex-cup ka gap).

Kai independent copies ka average lene se variance kyun shrink hota hai?

Sample mean

\overset{ˉ}{X} = \frac{1}{n} (X_{1} + \dots + X_{n})

ke liye, sum ke variances add hote hain lekin

1/ n

factor average mein square hota hai, deta hai

Var (\overset{ˉ}{X}) = σ^{2} / n \to 0

— yahi mechanism Law of Large Numbers ke peeche hai.

Bernoulli variance

p = 0.5

par largest kyun hota hai?

p (1 - p)

ek downward parabola hai jo

p = 0.5

par peak karti hai; ek fair coin maximally uncertain hota hai, jabki

p

near

0

1

hona nearly deterministic hota hai aur isliye barely spread karta hai.

Edge cases

Var (X)

kya hota hai jab

X

ek single value probability

1

ke saath leta hai?

Zero — mean

μ

se koi deviation nahi hota, isliye spread exactly

0

hai (degenerate distribution).

Var (0 \cdot X)

kya hai?

Zero, kyunki

0 \cdot X

constant

0

hai; scale rule deta hai

0^{2} Var (X) = 0

, jo ek aise distribution se match karta hai jisme koi spread nahi.

Agar

X

aur

Y

uncorrelated hain lekin dependent hain, toh kya

Var (X + Y) = Var (X) + Var (Y)

phir bhi hold karta hai?

Haan — sum rule sirf

Cov (X, Y) = 0

maangta hai, jo uncorrelatedness deta hai, chahe independence (ek stronger property) fail ho.

Kya kisi distribution ka finite mean ho sakta hai lekin infinite (undefined) variance?

Haan — heavy-tailed distributions ka

E [X]

well-defined ho sakta hai jabki

E [X^{2}]

diverge karta hai, variance ko infinite banata hai; aise cases Central Limit Theorem ki usual guarantees tod dete hain.

a \to 0

ke saath

SD (a X + b)

ka kya hota hai?

Yeh

0

ki taraf tend karta hai: distribution constant

b

ki taraf squash ho jaati hai, isliye saari spread disappear ho jaati hai — consistent with

∣ a ∣ σ \to 0

, jahan

σ = SD (X)

Agar do variables perfectly negatively correlated hain aur equal standard deviation

σ

hai, toh

Var (X + Y)

kya hai?

Zero — equal variance

σ^{2}

aur

Cov (X, Y) = - σ^{2}

ke saath, sum rule deta hai

σ^{2} + σ^{2} - 2 σ^{2} = 0

; fluctuations exactly cancel ho jaate hain, jaise ek perfect hedge.