Exercises — Conditional expectation

(a) Number. Koi bhi conditioning nahi — kisi bhi info ke bina ek single best guess. (b) Number. Humne value $Y=3$ fix kar di, toh kuch bhi random nahi bacha; ye ek specific average hai. (c) Random variable. Capital $Y$ random rehta hai. Ye function $g(Y)=E[X\mid Y=y]{|}_{y=Y}$ hai, toh iski value $Y$ ke saath badlati hai. (d) Number. Outer $E[\cdot ]$ (c) ke random variable ko saare $Y$ ke upar average karta hai; Tower Rule se ye $E[X]$ ke barabar hai.

Rule of thumb: lower-case $y$ (ya koi fixed event) → number; capital $Y$ andar ho → random variable.

$B$ given hone par, outcomes $\{2,4,6\}$ hain, har ek ka conditional probability $\frac{1}{3}$ hai. $E[X\mid B]=2\cdot \frac{1}{3}+4\cdot \frac{1}{3}+6\cdot \frac{1}{3}=\frac{2+4+6}{3}=\frac{12}{3}=4.$ Ye sirf ordinary mean hai, lekin conditional distribution use karke compute kiya gaya (sirf even faces bachte hain).

$Y$ ko coin maano. Uske upar condition karo: $E[X\mid Y=H]=3.5,E[X\mid Y=T]=10.$ Tower Rule, $P(Y=H)=P(Y=T)=\frac{1}{2}$ se weight karke: $E[X]=\frac{1}{2}(3.5)+\frac{1}{2}(10)=\frac{13.5}{2}=\mathrm{6.75.}$

Jab $Y$ known hai, $Y^2$ ek constant hai aur bahar aa jaata hai: $E[Y^{2}X\mid Y]=Y^{2}E[X\mid Y]=Y^2\cdot 3.5,$ independence ($X\perp Y$) use karke $E[X\mid Y]=E[X]=3.5$ likhte hain. Expectations lo (Tower Rule): $E[Y^{2}X]=3.5E[Y^2].$ Fair die ke liye, $E[Y^2]=\frac{1^2+2^2+\cdots +6^2}{6}=\frac{91}{6}$. Toh $E[Y^{2}X]=3.5\cdot \frac{91}{6}=\frac{7}{2}\cdot \frac{91}{6}=\frac{637}{12}\approx \mathrm{53.083.}$

$[0,y]$ par uniform ka mean midpoint $\frac{0+y}{2}=\frac{y}{2}$ hota hai, toh $E[X\mid Y=y]=\frac{y}{2}⟹E[X\mid Y]=\frac{Y}{2}.$ Tower: $E[X]=E\left[\frac{Y}{2}\right]=\frac{1}{2}E[Y]=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}.$

Chosen $p$ par condition karo. Success probability $p$ wale geometric variable (first success) ka mean $\frac{1}{p}$ hota hai: $E[N\mid p=\frac{1}{3}]=3,E[N\mid p=\frac{1}{2}]=2.$ Tower Rule, equal weights $\frac{1}{2}$: $E[N]=\frac{1}{2}(3)+\frac{1}{2}(2)=\frac{5}{2}=\mathrm{2.5.}$ $p$ par conditioning ne ek ugly mixture ko do textbook geometrics mein badal diya.

$N=n$ given hone par, expectation ki linearity se $E[S\mid N=n]=n\cdot 3.5$ milta hai. Isliye $E[S\mid N]=3.5N$. Tower: $E[S]=E[3.5N]=3.5E[N]=3.5\cdot 3.5=\mathrm{12.25.}$ (Ye bilkul Wald's identity ki shape hai: $E[S]=E[N]E[X]$ jab $N$ $X_i$'s se independent ho.)

Same midpoint reasoning: $E[X\mid Y=y]=\frac{y}{2}$, toh $E[X\mid Y]=\frac{Y}{2}$. Ab $E[Y]=\frac{0+2}{2}=1$, toh $E[X]=E\left[\frac{Y}{2}\right]=\frac{1}{2}\cdot 1=\frac{1}{2}.$ Figure mein red line conditional-mean function $y\mapsto y/2$ hai; $Y$ ki range par iska height average karo ($Y$ ki uniform density se weighted) toh $E[X]=\frac{1}{2}$ milta hai.

$Y$ known hota hai jab hum uske upar condition karte hain, toh use bahar nikalo: $E[XY\mid Y]=YE[X\mid Y]=Y\cdot \frac{Y}{2}=\frac{Y^2}{2}.$ Tower: $E[XY]=E\left[\frac{Y^2}{2}\right]=\frac{1}{2}E[Y^2].$ $Y\sim U(0,1)$ ke liye, $E[Y^2]={\int }_0^1{y}^{2}dy=\frac{1}{3}$. Isliye $E[XY]=\frac{1}{2}\cdot \frac{1}{3}=\frac{1}{6}\approx \mathrm{0.1667.}$

Within-group term. $[0,y]$ par uniform ka variance $\frac{y^2}{12}$ hota hai, toh $\mathrm{Var}(X\mid Y)=\frac{Y^2}{12}$ aur $E[\mathrm{Var}(X\mid Y)]=\frac{1}{12}E[Y^2]=\frac{1}{12}\cdot \frac{1}{3}=\frac{1}{36}.$ Between-group term. Hamare paas $E[X\mid Y]=\frac{Y}{2}$ hai, toh $\mathrm{Var}\left(\frac{Y}{2}\right)=\frac{1}{4}\mathrm{Var}(Y)=\frac{1}{4}\cdot \frac{1}{12}=\frac{1}{48},$ $[0,1]$ par uniform ke liye $\mathrm{Var}(Y)=\frac{(1-0{)}^2}{12}=\frac{1}{12}$ use karke. Add karo: $\mathrm{Var}(X)=\frac{1}{36}+\frac{1}{48}.$ Common denominator $144$: $\frac{4}{144}+\frac{3}{144}=\frac{7}{144}\approx \mathrm{0.04861.}$

Mean (Wald / tower). $N=n$ given hone par, $E[S\mid N=n]=n\cdot 10$, toh $E[S\mid N]=10N$. Tower: $E[S]=10E[N]=10\cdot 4=40.$ Variance (Eve's Law). Within-group: $N=n$ given hone par, $X_i$ independent hain, toh $\mathrm{Var}(S\mid N=n)=n\cdot 25$, yaani $\mathrm{Var}(S\mid N)=25N$. Isliye $E[\mathrm{Var}(S\mid N)]=25E[N]=25\cdot 4=100.$ Between-group: $E[S\mid N]=10N$, toh $\mathrm{Var}(E[S\mid N])=100\mathrm{Var}(N)=100\cdot 4=400,$ Poisson ke liye $\mathrm{Var}(N)=\lambda =4$ use karke. Add karo: $\mathrm{Var}(S)=100+400=500.$ Sanity/degenerate check. Agar har customer exactly $10$ kharchta (toh $\mathrm{Var}(X_i)=0$), within-group term vanish ho jaata aur $\mathrm{Var}(S)=100\mathrm{Var}(N)=400$ — pure customer-count randomness. Humara extra $100$ precisely spending randomness hai. Dono effects add hote hain, jaisa Eve's Law demand karta hai.

Agar $Y\equiv c$ hamesha hai, toh $Y$ par condition karne se tumhe kuch naya nahi pata — har outcome mein already $Y=c$ hai. Toh $X$ ki conditional distribution uski unconditional distribution ke barabar hai: $E[X\mid Y]=E[X\mid Y=c]=E[X],$ ek constant random variable. Phir $E[E[X\mid Y]]=E[E[X]]=E[X],$ kyunki ek constant ki expectation woh khud hoti hai. Tower Rule $E[X]=E[X]$ identity mein collapse ho jaata hai — boundary case jahan "groups mein split karna" exactly ek group banata hai. Ye independence collapse rule ko mirror karta hai: ek constant har cheez se independent hota hai.

$E[XY]$ given hai jahan $Y$ conditioning variable hai — pehla move kya? ::: Take out what is known: $E[XY\mid Y]=YE[X\mid Y]$, phir tower apply karo. Compound sum $S={\sum }_1^N{X}_i$, mean chahiye? ::: Tower / Wald: $E[S]=E[N]E[X]$. Compound sum, variance chahiye? ::: Eve's Law: $E[N]\mathrm{Var}(X)+\mathrm{Var}(N)(E[X]{)}^2$. Conditioning variable ek constant hai — $E[X\mid Y]$ ka kya hota hai? ::: Ye $E[X]$ ke barabar ho jaata hai; tower ek trivial identity ban jaati hai. "Plain average of conditional means" kab valid hai? ::: Sirf jab saare $P(Y=y)$ equal hon.