Conditional expectation: Worked examples — Hinglish deep dive

4.9.13 · D3 · Maths › Probability Theory & Statistics › Conditional expectation

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Recall Notation ka matlab (parent se)

$E [X]$ ::: $X$ ke liye aapka best single-number guess koi bhi info milne se pehle — ek plain number. $E [X ∣ Y = y]$ ::: aapka guess $Y$ ki specific value $y$ jaanne ke baad — phir bhi ek plain number, har $y$ ke liye ek. $E [X ∣ Y]$ (capital $Y$ ) ::: $Y$ ko unknown chhod do → guess khud ek random variable ban jaata hai, function $g (Y) = E [X ∣ Y = y]$ jo random $Y$ ko feed kiya gaya ho. Tower Rule ::: $E [E [X ∣ Y]] = E [X]$ — group-averages ka average lao, group sizes se weighted.

Scenario matrix

Har conditional-expectation problem in cells mein se kisi ek mein aati hai. Neeche ke examples cell ke label ke saath hain.

Cell	Case class	Tricky kyon hai	Example
A	Discrete, ek event $B$ par condition karo	$P (B)$ se renormalise karna padta hai	Ex 1
B	Discrete, ek variable $Y$ par condition karo; Tower use karo	$P (Y = y)$ se weight karna padta hai	Ex 2
C	Continuous, $E [X ∣ Y] = g (Y)$ phir integrate karo	inner density $f_{X ∣ Y}$	Ex 3
D	Degenerate: $X ⊥ Y$	$Y$ jaanna kuch nahi badalta	Ex 4
E	$g (Y) X$ ke saath take-out-what-is-known	rules ko fuse mat karo	Ex 5
F	Real-world word problem (random count)	Wald-style sum	Ex 6
G	Exam twist: tower se hidden mean recover karo	tower ko ulta chalao	Ex 7
H	Limiting/edge: $P (B) \to 0$ , ya $Y$ boundary ke paas	check karo ki sane rehta hai	Ex 8

Cell A — ek event par condition karo

Worked example Ex 1 (Cell A). Ek fair die roll karo. Given ki outcome

kam se kam 3 hai, $E [X ∣ B]$ nikalo. Yahan $X$ = die face, $B = {X \geq 3}$ .

Forecast: pehle guess karo — unconditional mean $3.5$ hai. ${1, 2}$ kaatne se average upar jayega. $3.5$ se upar? Neeche? Bet lagao.

$P (B)$ nikalo. Favourable faces hain ${3, 4, 5, 6}$ , isliye $P (B) = 4/6 = 2/3$ . Yeh step kyon? Cell-A problems ek event par condition karti hain, aur definition $E [X ∣ B] = \sum_{x} x P (X = x, B) / P (B)$ $P (B)$ se divide karti hai — hume pehle yeh jaanna zaroori hai.
Renormalise karo. Har surviving face $x \in {3, 4, 5, 6}$ ke liye, $P (X = x ∣ B) = \frac{1/6}{2/3} = \frac{1}{4}$ . Kyon? $B$ par conditioning sirf yahi outcomes rakhti hai aur unki probabilities stretch karti hai taaki woh 1 sum karein.
Average nikalo. $E [X ∣ B] = \frac{1}{4} (3 + 4 + 5 + 6) = \frac{1}{4} \cdot 18 = 4.5$ . Kyon? Yeh renormalised distribution ke under sirf ordinary expectation hai.

Verify: $4.5 > 3.5$ ✓ (do sabse chhote faces hatane se mean badhna chahiye). Jo chaar faces rakhe hain woh $4.5$ ke baare mein symmetric hain, isliye unka equal-weight average bilkul middle mein baithta hai — sanity confirm hua.

Cell B — ek variable par condition karo, phir Tower

Worked example Ex 2 (Cell B). Do fair dice roll karo;

$X$ = sum, $Y$ = pehla die. Tower Rule ke zariye $E [X]$ nikalo, weights ke baare mein careful rehte hue.

Forecast: do dice, har ek ka mean $3.5$ , toh shayad aap pehle se $E [X] = 7$ smell kar rahe hain. Achha — lekin aao usse tower ke zariye derive karein taaki weighting ki aadat pakki ho jaye.

Conditional mean. $E [X ∣ Y = y] = y + E [second die] = y + 3.5$ . Yeh step kyon? Doosra die $Y$ se independent hai; conditioning sirf pehle term ko fix karti hai, doosre ka mean untouched rehta hai.
$P (Y = y)$ se weight karo. Har $y \in {1, \dots, 6}$ ke liye $P (Y = y) = 1/6$ hai: $E [X] = \sum_{y = 1}^{6} (y + 3.5) \cdot \frac{1}{6} .$ Kyon? Tower Rule hai $E [X] = \sum_{y} E [X ∣ Y = y] P (Y = y)$ — group means ka weighted average, plain nahi.
Compute karo. $\sum_{y = 1}^{6} y = 21$ , isliye $E [X] = \frac{1}{6} (21) + 3.5 = 3.5 + 3.5 = 7$ . Yeh step kyon? Sum distribute karke, $\frac{1}{6} \sum_{y} (y + 3.5) = \frac{1}{6} \sum_{y} y + \frac{1}{6} \cdot 6 \cdot 3.5$ ; constant $3.5$ ko chhe equal weights par average karo toh woh $3.5$ hi rehta hai, aur $\frac{1}{6} \cdot 21 = 3.5$ , toh dono halves $7$ mein add hote hain.

Verify: $E [X] = E [die_{1}] + E [die_{2}] = 3.5 + 3.5 = 7$ linearity se match karta hai (Expectation and its linearity) ✓.

Cell B ke andar Twist — even pehle die par condition karo (yeh Cell A renormalisation reuse karta hai):

Pehle die ko renormalise karo. ${1, \dots, 6}$ mein even faces hain ${2, 4, 6}$ , isliye $P (Y even) = 3/6 = 1/2$ , aur har even $y$ ke liye, $P (Y = y ∣ Y even) = \frac{1/6}{1/2} = \frac{1}{3}$ . Yeh step kyon? Bilkul Cell A ki tarah, event ${Y even}$ par conditioning sirf ${2, 4, 6}$ rakhti hai aur unki probabilities stretch karti hai taaki 1 sum ho.
$y$ ka weighted average. $E [Y ∣ Y even] = \frac{1}{3} (2 + 4 + 6) = \frac{1}{3} \cdot 12 = 4$ . Yeh step kyon? Teen even faces ab $1/3$ probability ke saath equally likely hain, $E [Y ∣ Y even] = \sum y P (y ∣ even)$ unka equal-weight average hai — ${2, 4, 6}$ ka midpoint $4$ .
Doosra die add karo. $E [X ∣ Y even] = 4 + 3.5 = 7.5$ (parent Example 1 se match karta hai) ✓. Yeh step kyon? Linearity se $E [X ∣ Y even] = E [Y ∣ Y even] + E [second die]$ ; doosra die independent hai isliye uska mean $3.5$ rehta hai, jo $4 + 3.5$ deta hai.

Cell C — continuous

Worked example Ex 3 (Cell C).

$Y \sim U (0, 1)$ aur, given $Y = y$ , $X \sim U (0, y)$ . $E [X]$ aur $E [X^{2}]$ nikalo.

Forecast: $X$ $[0, y]$ ke andar rehta hai aur $Y < 1$ hai, isliye $X$ chhoti values ki taraf squeeze hota hai. Guess karo $E [X]$ $0.5$ se kaafi neeche hoga.

Inner mean. $[0, y]$ par uniform ke liye, $E [X ∣ Y = y] = \frac{0 + y}{2} = \frac{y}{2}$ . Yeh step kyon? Ek uniform ka mean uska midpoint hota hai. Yeh function $g (y) = y /2$ deta hai, isliye random variable $E [X ∣ Y] = Y /2$ milta hai.
Mean ke liye Tower. $E [X] = E [Y /2] = \frac{1}{2} E [Y] = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} .$ Kyon? Joint density par bade double integral ki jagah do aasaan one-liners aa jaate hain.
Tower se second moment. $U (0, y)$ ke liye, $E [X^{2} ∣ Y = y] = y^{2} /3$ . Kyon? $\int_{0}^{y} x^{2} \cdot \frac{1}{y} d x = \frac{y ^{2}}{3}$ — uniform ka variance-plus-mean-squared, lekin hum seedha integrate karte hain. Phir $E [X^{2}] = E [\frac{Y ^{2}}{3}] = \frac{1}{3} E [Y^{2}] = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} .$ ( $E [Y^{2}] = \int_{0}^{1} y^{2} d y = 1/3$ use kiya.)

Verify: $E [X] = 1/4 = 0.25 < 0.5$ ✓ (chhoti values ki taraf squeeze, jaise forecast kiya tha). Ek doosre sanity check ke liye hum variance $Var (X)$ use karte hain, jo mean se squared distance ka average hota hai aur $Var (X) = E [X^{2}] - E [X]^{2}$ ke roop mein compute kiya ja sakta hai. Check ke liye kyon? Kyunki variance squares ka average hai, yeh kabhi bhi negative nahi ho sakta — toh agar humare $E [X^{2}]$ aur $E [X]$ ek-doosre se inconsistent hote, yeh negative aata aur error flag karta. Yahan $Var (X) = \frac{1}{9} - \frac{1}{16} = \frac{7}{144} > 0$ ✓ (safely positive — tools toot nahi gaye). Yeh continuous case Joint and marginal distributions aur Conditional variance and Eve's law se connect hota hai.

Cell D — degenerate: independence

Worked example Ex 4 (Cell D).

$X$ = 3 fair-coin flips mein heads ki sankhya; $Y$ = ek bilkul alag fair die. $E [X ∣ Y]$ aur $E [X ∣ Y = 6]$ nikalo.

Forecast: die ka coins se koi lena-dena nahi. Kya $Y$ jaanna $X$ ke baare mein aapka guess badalta hai? Padhne se pehle predict karo.

Independence check karo. Coin flips aur die physically unrelated hain, isliye $X ⊥ Y$ . Yeh step kyon? Independence-collapse rule $E [X ∣ Y] = E [X]$ tabhi fire hoti hai jab $X$ aur $Y$ independent hon — hume isse justify karna hoga, assume nahi kar sakte. Dekho Conditional probability.
Collapse karo. $E [X ∣ Y] = E [X]$ , ek constant random variable. Kyon? $Y$ jaanna $X$ ke baare mein zero new information deta hai, isliye updated guess original ke barabar rehta hai.
Value. $X \sim Binomial (3, \frac{1}{2})$ , isliye $E [X] = 3 \cdot \frac{1}{2} = 1.5$ . Isliye $E [X ∣ Y = 6] = 1.5$ bhi — har $y$ ke liye same. Yeh step kyon? Binomial $(n, p)$ $n$ independent trials mein successes count karta hai, aur linearity se uska mean $n p$ hota hai; yahan $n = 3, p = \frac{1}{2}$ se $n p = 1.5$ milta hai. Kyunki collapse ne $E [X ∣ Y]$ ko constant bana diya, koi bhi $y$ (including $6$ ) plug in karo toh wahi $1.5$ return hota hai.

Verify: Tower phir bhi hold karni chahiye: $E [E [X ∣ Y]] = E [1.5] = 1.5 = E [X]$ ✓. Constant ki expectation khud wahi hoti hai, isliye tower trivially close ho jaati hai — degenerate case consistent hai.

Cell E — take out what is known (rules ko fuse mat karo)

Worked example Ex 5 (Cell E).

$Y \sim U (0, 1)$ , aur given $Y = y$ , $X \sim U (0, y)$ (Ex 3 jaisa hi setup). $E [X Y]$ nikalo.

Forecast: yeh likhna tempting hai $E [X Y ∣ Y] = Y E [X]$ . Yeh classic trap hai. Dekho.

Known factor bahar nikalo. Given $Y$ , value $Y$ ek constant hai, isliye $E [X Y ∣ Y] = Y E [X ∣ Y] .$ Yeh step kyon? "Take out what is known": jab $Y$ fix ho jaata hai, $Y$ ek constant hai aur inner expectation chhod deta hai. Yeh hamesha valid hai.
$E [X ∣ Y]$ ko $E [X]$ se replace MAT karo. Yahan $E [X ∣ Y] = Y /2$ hai (Ex 3 se), jo $Y$ par depend karta hai — $X$ aur $Y$ independent nahi hain. Kyon? $E [X]$ swap karne ke liye independence chahiye (Cell D); dono rules fuse karne se galat answer $Y \cdot \frac{1}{4}$ milta hai. Inhe alag rakho.
Assemble karo aur outer expectation lo. $E [X Y ∣ Y] = Y \cdot \frac{Y}{2} = \frac{Y ^{2}}{2}, E [X Y] = E [\frac{Y ^{2}}{2}] = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} .$ Yeh step kyon? $E [X Y ∣ Y]$ ek random variable hai ( $Y$ ka function); Tower Rule kehti hai uski outer expectation $E [E [X Y ∣ Y]]$ $E [X Y]$ ke barabar hai. Constant $\frac{1}{2}$ bahar nikaalo aur Ex 3 se $E [Y^{2}] = \frac{1}{3}$ use karo, kaam ho gaya.

Verify: Galat raasta contrast karo: $Y E [X] = Y \cdot \frac{1}{4} \Rightarrow E [X Y] = ? \frac{1}{4} E [Y] = \frac{1}{8}$ . Kyunki $\frac{1}{6} \neq = \frac{1}{8}$ , rules fuse karna genuinely break karta hai — humaara careful answer $\frac{1}{6}$ sahi hai ✓.

Cell F — real-world word problem (random count)

Worked example Ex 6 (Cell F). Ek call centre ko ek ghante mein

$N \sim Poisson (λ = 10)$ calls aate hain. Har call independently $L$ minutes chalta hai jahan $E [L] = 4$ hai. Maano $S = \sum_{i = 1}^{N} L_{i}$ total minutes hai. $E [S]$ nikalo.

Forecast: roughly "10 calls times 4 minutes" ⇒ 40 minutes. Aao prove karein ki tower yeh exact bana deti hai bhale hi $N$ random ho.

Count $N$ par condition karo. Given $N = n$ , $S$ $n$ fixed-mean pieces ka sum hai: $E [S ∣ N = n] = \sum_{i = 1}^{n} E [L_{i}] = n \cdot 4.$ Yeh step kyon? $N$ fix hone par sum mein terms ki sankhya jaani-pehchani hai, isliye expectation ki linearity seedha apply hoti hai (Expectation and its linearity).
Random variable ke roop mein package karo. $E [S ∣ N] = 4 N$ . Kyon? Har value $n$ map hoti hai $4 n$ par, isliye function $g (N) = 4 N$ hi conditional expectation hai.
Tower. $E [S] = E [4 N] = 4 E [N] = 4 \cdot 10 = 40$ minutes. Kyon? Conditional means ka average, $P (N = n)$ se weighted — exactly Tower Rule — $4 E [N]$ par collapse ho jaata hai.

Verify: Yeh Wald's identity hai: $E [S] = E [N] E [L] = 10 \cdot 4 = 40$ ✓. Units: (calls) $\times$ (minutes/call) = minutes ✓.

Cell G — exam twist: tower ko ulta chalao

Worked example Ex 7 (Cell G). Ek exam ke do versions hain. Ek student version

$A$ probability $0.3$ se aur version $B$ probability $0.7$ se paata hai. Overall mean score jaana hua hai $E [X] = 68$ . Version- $B$ waale average $E [X ∣ B] = 72$ karte hain. Version- $A$ mean $E [X ∣ A]$ nikalo.

Forecast: overall 68 hai, aur B (majority) 72 average karta hai (overall se upar). Toh A ko mean neeche kheenchna chahiye — guess karo A ka mean 68 se neeche hoga.

Tower ko weighted sum ke roop mein likho. $E [X] = P (A) E [X ∣ A] + P (B) E [X ∣ B] .$ Yeh step kyon? Kaun sa version hai par conditioning ek two-value partition hai; tower kehti hai overall mean dono group means ka crowd-weighted blend hai (Law of total probability).
Jaane hue values plug karo. $68 = 0.3 E [X ∣ A] + 0.7 \cdot 72 = 0.3 E [X ∣ A] + 50.4.$ Kyon? Hume chaar mein se teen quantities pata hain; tower ek unknown mein ek equation deti hai.
Solve karo. $0.3 E [X ∣ A] = 68 - 50.4 = 17.6 \Rightarrow E [X ∣ A] = \frac{17.6}{0.3} = \frac{176}{3} = 58. \overline{6} \approx 58.67.$ Yeh step kyon? Yeh ek single linear equation hai $68 = 0.3 m + 50.4$ unknown $m = E [X ∣ A]$ mein; jaana hua $50.4$ subtract karne se $0.3 m$ term isolate ho jaata hai, aur coefficient $0.3$ se divide karne par $m$ mil jaata hai.

Verify: $0.3 \cdot \frac{176}{3} + 0.7 \cdot 72 = 17.6 + 50.4 = 68$ ✓, aur $\frac{176}{3} < 68$ ✓ (A mean neeche kheenchta hai, jaise forecast kiya tha). Yahan trap yeh hai ki weights bhool jaana — ek naive $\frac{68 + 72}{2}$ reasoning bilkul bekaar hogi.

Cell H — limiting / edge behaviour

Worked example Ex 8 (Cell H). Do edge checks taaki reader koi unseen case na hit kare.

(H1) Ek shrinking event. Ek fair die ke liye, $B_{k} = {X \geq k}$ par condition karo aur $k \to 6$ jaane do.

Forecast: jaise event sirf face 6 tak narrow hoti hai, conditional mean upar slide karke 6 ho jaana chahiye.

$E [X ∣ X \geq k] = \frac{k + ( k + 1 ) + \dots + 6}{7 - k}$ = kept faces ka midpoint $= \frac{k + 6}{2}$ . Yeh step kyon? Kept faces ${k, \dots, 6}$ equally spaced aur equally likely hain, isliye unka mean midpoint hota hai.
$k = 6$ par: $E [X ∣ X \geq 6] = \frac{6 + 6}{2} = 6$ , aur $P (B_{6}) = 1/6 > 0$ toh formula abhi bhi legal hai. Kyon? Conditional expectation sirf $P (B) > 0$ chahti hai; yahan yeh singleton ${6}$ tak bilkul hold karta hai.

Verify: $k = 1$ se $\frac{1 + 6}{2} = 3.5 = E [X]$ ✓ (certain event par conditioning kuch nahi badalta); $k = 6$ se $6$ ✓. $3.5$ se $6$ tak monotone increase jaise event shrink hoti hai — exactly jaise forecast kiya tha. Forbidden case hai $P (B) = 0$ (jaise "given $X = 7$ "), jahan $E [X ∣ B]$ undefined hai, zero nahi.

(H2) Ek boundary continuous value. Ex 3 mein, $Y \to 0^{+}$ jaane do. Tab $X ∣ Y = y \sim U (0, y)$ point $0$ ki taraf collapse hota hai.

Boundary ke paas conditional mean padho. Ex 3 se, $E [X ∣ Y = y] = y /2$ , isliye $y \to 0^{+}$ jaane par $E [X ∣ Y = y] = y /2 \to 0$ milta hai. Yeh step kyon? Edge case ko usi formula se handle kiya jaata hai jo hum pehle se derive kar chuke hain — hum bas dekh rahe hain ki boundary par kya hota hai na ki koi naya rule invent kar rahe hain.
Collapse ki sanity. Shrinking interval $[0, y]$ par uniform ka saara mass $0$ ki taraf slide karta hai, isliye uska mean bhi $0$ jaana chahiye — aur khaas baat yeh ki formula $y /2$ kabhi kisi cheez se divide nahi karta jo vanish ho, isliye kuch blow up nahi hota. Yeh step kyon? Hume confirm karna hai ki limit finite aur sensible hai: degenerate inputs exactly woh jagah hai jahan careless formulas explode karte hain, aur yahan woh nahi karte ✓.

Figure — Left panel (Cells A & H): blue faded bars fair die ki original $P = \frac{1}{6}$ hain; ${X \geq 3}$ par conditioning faces 1 aur 2 cross out karti hai (yellow X's) aur pink bars survivors ko $P = \frac{1}{4}$ par renormalised dikhaate hain, jinka balance point (yellow dashed line) $E [X ∣ B] = 4.5$ par baithta hai. Right panel (Cells B & G): do bars jinki heights group means hain ( $E [X ∣ A] \approx 58.7$ pink, $E [X ∣ B] = 72$ blue) aur jinki widths crowd weights hain ( $0.3$ aur $0.7$ ); yellow dashed line width-weighted blend $E [X] = 68$ hai — tower literally "heights ka average lo, widths se weighted" hai.

Upar ka chalkboard do workhorse pictures dikhaata hai: left, Cell A/H — conditioning as outcomes kaatna aur renormalise karna; right, Cell B/G — tower as weighted bars jinki blended height overall mean hai.

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

Scenario Map

Neeche ka flowchart matrix table ka restatement nahi hai — yeh ek decision procedure hai: ise top-to-bottom padho taaki pata chale kisi fresh problem ko kaun si technique chahiye ( $P (B)$ se divide karo? $P (Y = y)$ se weight karo? integrate karo? collapse karo?), phir labelled example par jump karo. Table classify karta hai; yeh diagram route karta hai.