Conditional expectation: Question bank — Hinglish deep dive

4.9.13 · D5 · Maths › Probability Theory & Statistics › Conditional expectation

Do objects jo tumhe hamesha alag rakhne chahiye:

$E [X ∣ Y = y]$ — ek number, har fixed value $y$ ke liye alag.
$E [X ∣ Y]$ — ek random variable, function $g (y) = E [X ∣ Y = y]$ jo random $Y$ par evaluate hoti hai.

True ya false — justify karo

E [X ∣ Y]

hamesha ek single number hota hai.

False — yeh ek random variable

g (Y)

hai. Sirf tab jab tum

Y = y

fix karo tab yeh number

E [X ∣ Y = y]

ban jaata hai.

E [E [X ∣ Y]] = E [X]

ke liye

X

aur

Y

ka independent hona zaroori hai.

False — Tower Rule kisi bhi

X, Y

ke liye valid hai (finite mean ke saath). Conditional averages ko wapas whole mein average karne ke liye Independence ki kabhi zaroorat nahi hoti.

Agar

X ⊥ Y

toh

E [X ∣ Y] = E [X]

True —

Y

jaanna

X

ke baare mein kuch naya nahi batata, isliye updated guess prior guess

E [X]

ke barabar hoti hai, jo ek constant hai. Dekho Conditional probability.

E [X ∣ Y] = E [X]

ka matlab hai ki

X

aur

Y

independent hain.

False — iska sirf yeh matlab hai ki

Y

ke paas

X

ke baare mein koi linear/mean information nahi hai. Independence isse zyada strong hai; e.g.

Y

phir bhi

X

ka variance badal sakta hai.

Kisi bhi function

g

ke liye

E [g (Y) ∣ Y] = g (Y)

True — ek baar

Y

pata chal jaaye,

g (Y)

ek fixed constant ban jaata hai, aur constant ki expectation woh khud hoti hai.

E [X Y ∣ Y] = Y E [X]

hamesha.

False — bahar nikalte waqt milta hai

E [X Y ∣ Y] = Y E [X ∣ Y]

. Tum

E [X ∣ Y]

ki jagah

E [X]

tabhi rakh sakte ho jab

X ⊥ Y

ho.

Tower Rule conditional means

E [X ∣ Y = y]

ka ek simple average hai.

False — yeh ek weighted average hai,

P (Y = y)

se weighted. Bade groups zyada count karte hain; equal weighting tabhi kaam aati hai jab sab

y

equally likely hon.

E [X ∣ Y]

ke values

X

ki range se bahar ja sakte hain.

False — har

E [X ∣ Y = y]

possible

X

values ka average hai, isliye yeh

X

ki min–max range ke andar hi rehta hai.

Var (E [X ∣ Y]) \leq Var (X)

True — conditioning fluctuation ka ek hissa average kar deta hai, isliye "explained" variance total se zyada nahi ho sakti. Yeh hai Conditional variance and Eve's law.

Agar

E [X ∣ Y] = c

(ek constant) hai, toh

c = E [X]

True — Tower apply karo:

E [X] = E [c] = c

, isliye constant unconditional mean hi honi chahiye.

Error dhundho

E [X ∣ Y = y] = \sum_{x} x P (X = x, Y = y)

P (Y = y)

se divide karna bhool gaye — tumhe conditional distribution

P (X = x ∣ Y = y) = P (X = x, Y = y) / P (Y = y)

use karni chahiye, joint nahi.

"Tower se,

E [X] = \frac{1}{n} \sum_{y} E [X ∣ Y = y]

Y

n

values ke liye."

Galat weights — sahi sum hai

\sum_{y} E [X ∣ Y = y] P (Y = y)

1/ n

se average karna silently assume karta hai ki har

Y = y

equally likely hai.

E [X + Z ∣ Y] = E [X ∣ Y] + E [Z]

Doosra term

E [Z]

nahi,

E [Z ∣ Y]

hona chahiye. Conditional expectation usi conditioning ko rakhte hue linear hai — tum ek term se

Y

nahi hata sakte.

"Kyunki

E [X ∣ Y]

random hai,

E [E [X ∣ Y]]

bhi random hai."

Bahari

E [\cdot]

Y

par average karta hai, jo constant

E [X]

produce karta hai.

E

apply karne se random variable ek fixed number ban jaata hai.

E [X^{2} ∣ Y] = (E [X ∣ Y])^{2}

Generally false — yeh Jensen's inequality ka area hai;

E [X^{2} ∣ Y] \geq (E [X ∣ Y])^{2}

with a gap equal to conditional variance

Var (X ∣ Y)

X ∣ Y = y \sim U (0, y)

ke liye,

E [X ∣ Y] = 1/2

hai."

U (0, y)

ka mean

y /2

hai, isliye

E [X ∣ Y] = Y /2

— ek random variable, constant

1/2

nahi. Sirf

E [\cdot]

lene ke baad humein

E [X] = 1/4

milta hai.

P (A ∣ Y)

aur

E [1_{A} ∣ Y]

alag-alag cheezein hain."

Ye dono same object hain — kisi event ki conditional probability uske indicator

1_{A}

ki conditional expectation hoti hai. Dekho Law of total probability.

Why questions

E [X ∣ Y]

ko sirf numbers ki table ki jagah

Y

ka function kyun define kiya jaata hai?

Har answer "

E [X ∣ Y = y]

sabhi

y

ke liye" ko ek function

g (Y)

mein pack karne se hum random

Y

ko wapas feed kar sakte hain, taaki hum uski expectation, variance le sakein, aur usse Martingales ke andar use kar sakein.

Inner marginalising sum

\sum_{y} P (X = x, Y = y)

P (X = x)

mein collapse kyun ho jaata hai?

Joint ko sab

Y

values par sum karna

X = x

hone ke har possible tarike ko account karta hai, jo exactly marginal

P (X = x)

hai. Dekho Joint and marginal distributions.

Hum "jo known hai usse bahar nikal sakte hain" —

E [g (Y) X ∣ Y] = g (Y) E [X ∣ Y]

— kyun?

Y

par conditioning karte waqt

Y

ko fixed maana jaata hai, isliye

g (Y)

ek constant multiplier ban jaata hai, aur constants kisi bhi expectation se Expectation and its linearity ke through factor out ho jaate hain.

Tower ko

P (Y = y)

se weight kyun karna chahiye aur sab groups ko equal kyun nahi maana ja sakta?

Kyunki zyada probable

y

E [X]

mein zyada sample mass contribute karte hain; weights ignore karne par rare groups zyada count ho jaate hain aur overall mean distort ho jaata hai.

Tower Rule "split karke phir recombine karne se mean nahi badalta" intuitively kyun sach hai?

Har group ke andar average karna aur phir group size se blend karna har original outcome ko exactly ek baar uske sahi weight ke saath touch karta hai, isliye yeh global average reproduce karta hi hai.

Wald's identity conditional expectation par kyun rely karta hai?

Yeh sum

S_{N} = \sum_{i = 1}^{N} X_{i}

ko random count

N

par condition karta hai,

E [S_{N} ∣ N] = N E [X]

compute karta hai, phir Tower apply karta hai — dekho Wald's identity.

E [X ∣ X] = X

kyun hai?

Jab

X

khud hi woh cheez hai jis par tum condition kar rahe ho, toh woh poori tarah known hai, isliye

X

ke liye best guess

X

khud hai aur koi averaging baaki nahi hai.

Edge cases

Agar

P (Y = y) = 0

ho toh

E [X ∣ Y = y]

kya hai?

Ratio formula se undefined — tum probability-zero event par condition nahi kar sakte discrete definition mein; continuous case ise densities ke zariye handle karta hai.

Jab

Y

ek constant

c

ho (degenerate, koi information nahi), toh

E [X ∣ Y]

kya hoga?

Yeh

E [X]

ke barabar hoga — kisi aisi cheez par condition karna jo kabhi vary nahi karti koi naya information nahi deta, isliye guess prior mean par hi rehti hai.

Agar

X

ka finite mean nahi hai (e.g. Cauchy), toh kya

E [X ∣ Y]

defined hai?

Nahi — conditional expectation finite-mean requirement inherit karta hai; bina well-defined

E [X]

ke averages exist nahi karte.

Jab

Y = X

ho (perfect information), toh

E [X ∣ Y]

kya hai?

Yeh

X

khud hai —

Y = X

jaanna

X

ko exactly pin down kar deta hai, average karne ke liye kuch bacha hi nahi.

Continuous

X, Y

ke liye

P (X = x ∣ Y = y)

ki jagah, jo

0/0

hai, kya aata hai?

Conditional density

f_{X ∣ Y} (x ∣ y) = f_{X, Y} (x, y) / f_{Y} (y)

, aur

E [X ∣ Y = y] = \int x f_{X ∣ Y} (x ∣ y) d x

Agar

Y

sirf ek possible value probability 1 se leta hai, toh

Var (E [X ∣ Y])

kya hoga?

Zero —

E [X ∣ Y]

tab constant

E [X]

hai, aur constants ka koi variance nahi hota, jo is intuition se match karta hai ki

Y

ne kuch bhi explain nahi kiya.

Jab

X

constant

a

ho toh Tower ka kya hoga?

E [X ∣ Y] = a

har

y

ke liye, aur

E [a] = a = E [X]

— rule trivially hold karta hai, kyunki koi bhi group ka average

a

se alag nahi ho sakta.

Connections

Conditional expectation (parent)
Expectation and its linearity
Conditional probability
Law of total probability
Conditional variance and Eve's law
Joint and marginal distributions
Wald's identity
Martingales