Exercises — Conditional expectation

(a) Number. No conditioning at all — a single best guess before any info. (b) Number. We fixed the value $Y=3$, so nothing random is left; it is one specific average. (c) Random variable. Capital $Y$ stays random. It is the function $g(Y)=E[X\mid Y=y]{|}_{y=Y}$, so its value changes as $Y$ changes. (d) Number. The outer $E[\cdot ]$ averages the random variable in (c) over all $Y$; by the Tower Rule it equals $E[X]$.

Rule of thumb: lower-case $y$ (or a fixed event) → number; capital $Y$ still inside → random variable.

Given $B$, the outcomes are $\{2,4,6\}$, each equally likely with conditional probability $\frac{1}{3}$. $E[X\mid B]=2\cdot \frac{1}{3}+4\cdot \frac{1}{3}+6\cdot \frac{1}{3}=\frac{2+4+6}{3}=\frac{12}{3}=4.$ This is just the ordinary mean, but computed using the conditional distribution (only even faces survive).

Let $Y$ be the coin. Condition on it: $E[X\mid Y=H]=3.5,E[X\mid Y=T]=10.$ Tower Rule, weighting by $P(Y=H)=P(Y=T)=\frac{1}{2}$: $E[X]=\frac{1}{2}(3.5)+\frac{1}{2}(10)=\frac{13.5}{2}=\mathrm{6.75.}$

Once $Y$ is known, $Y^2$ is a constant and pulls out: $E[Y^{2}X\mid Y]=Y^{2}E[X\mid Y]=Y^2\cdot 3.5,$ using independence ($X\perp Y$) to write $E[X\mid Y]=E[X]=3.5$. Take expectations (Tower Rule): $E[Y^{2}X]=3.5E[Y^2].$ For a fair die, $E[Y^2]=\frac{1^2+2^2+\cdots +6^2}{6}=\frac{91}{6}$. So $E[Y^{2}X]=3.5\cdot \frac{91}{6}=\frac{7}{2}\cdot \frac{91}{6}=\frac{637}{12}\approx \mathrm{53.083.}$

The mean of a uniform on $[0,y]$ is the midpoint $\frac{0+y}{2}=\frac{y}{2}$, so $E[X\mid Y=y]=\frac{y}{2}⟹E[X\mid Y]=\frac{Y}{2}.$ Tower: $E[X]=E\left[\frac{Y}{2}\right]=\frac{1}{2}E[Y]=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}.$

Condition on the chosen $p$. A geometric variable (first success) with success probability $p$ has mean $\frac{1}{p}$: $E[N\mid p=\frac{1}{3}]=3,E[N\mid p=\frac{1}{2}]=2.$ Tower Rule, equal weights $\frac{1}{2}$: $E[N]=\frac{1}{2}(3)+\frac{1}{2}(2)=\frac{5}{2}=\mathrm{2.5.}$ Conditioning on $p$ turned an ugly mixture into two textbook geometrics.

Given $N=n$, linearity of expectation gives $E[S\mid N=n]=n\cdot 3.5$. Hence $E[S\mid N]=3.5N$. Tower: $E[S]=E[3.5N]=3.5E[N]=3.5\cdot 3.5=\mathrm{12.25.}$ (This is exactly the shape of Wald's identity: $E[S]=E[N]E[X]$ when $N$ is independent of the $X_i$.)

Same midpoint reasoning: $E[X\mid Y=y]=\frac{y}{2}$, so $E[X\mid Y]=\frac{Y}{2}$. Now $E[Y]=\frac{0+2}{2}=1$, so $E[X]=E\left[\frac{Y}{2}\right]=\frac{1}{2}\cdot 1=\frac{1}{2}.$ The red line in the figure is the conditional-mean function $y\mapsto y/2$; averaging its height over the range of $Y$ (weighted by $Y$'s uniform density) lands at $E[X]=\frac{1}{2}$.

$Y$ is known once we condition on it, so pull it out: $E[XY\mid Y]=YE[X\mid Y]=Y\cdot \frac{Y}{2}=\frac{Y^2}{2}.$ Tower: $E[XY]=E\left[\frac{Y^2}{2}\right]=\frac{1}{2}E[Y^2].$ For $Y\sim U(0,1)$, $E[Y^2]={\int }_0^1{y}^{2}dy=\frac{1}{3}$. Hence $E[XY]=\frac{1}{2}\cdot \frac{1}{3}=\frac{1}{6}\approx \mathrm{0.1667.}$

Within-group term. A uniform on $[0,y]$ has variance $\frac{y^2}{12}$, so $\mathrm{Var}(X\mid Y)=\frac{Y^2}{12}$ and $E[\mathrm{Var}(X\mid Y)]=\frac{1}{12}E[Y^2]=\frac{1}{12}\cdot \frac{1}{3}=\frac{1}{36}.$ Between-group term. We have $E[X\mid Y]=\frac{Y}{2}$, so $\mathrm{Var}\left(\frac{Y}{2}\right)=\frac{1}{4}\mathrm{Var}(Y)=\frac{1}{4}\cdot \frac{1}{12}=\frac{1}{48},$ using $\mathrm{Var}(Y)=\frac{(1-0{)}^2}{12}=\frac{1}{12}$ for a uniform on $[0,1]$. Add: $\mathrm{Var}(X)=\frac{1}{36}+\frac{1}{48}.$ Common denominator $144$: $\frac{4}{144}+\frac{3}{144}=\frac{7}{144}\approx \mathrm{0.04861.}$

Mean (Wald / tower). Given $N=n$, $E[S\mid N=n]=n\cdot 10$, so $E[S\mid N]=10N$. Tower: $E[S]=10E[N]=10\cdot 4=40.$ Variance (Eve's Law). Within-group: given $N=n$, the $X_i$ are independent, so $\mathrm{Var}(S\mid N=n)=n\cdot 25$, i.e. $\mathrm{Var}(S\mid N)=25N$. Thus $E[\mathrm{Var}(S\mid N)]=25E[N]=25\cdot 4=100.$ Between-group: $E[S\mid N]=10N$, so $\mathrm{Var}(E[S\mid N])=100\mathrm{Var}(N)=100\cdot 4=400,$ using $\mathrm{Var}(N)=\lambda =4$ for a Poisson. Add: $\mathrm{Var}(S)=100+400=500.$ Sanity/degenerate check. If every customer spent exactly $10$ (so $\mathrm{Var}(X_i)=0$), the within-group term vanishes and $\mathrm{Var}(S)=100\mathrm{Var}(N)=400$ — pure customer-count randomness. Our extra $100$ is precisely the spending randomness. Both effects add, as Eve's Law demands.

If $Y\equiv c$ always, then conditioning on $Y$ tells you nothing you didn't already know — every outcome already has $Y=c$. So the conditional distribution of $X$ equals its unconditional one: $E[X\mid Y]=E[X\mid Y=c]=E[X],$ a constant random variable. Then $E[E[X\mid Y]]=E[E[X]]=E[X],$ since the expectation of a constant is itself. The Tower Rule collapses to the identity $E[X]=E[X]$ — the boundary case where "splitting into groups" makes exactly one group. This mirrors the independence collapse rule: a constant is independent of everything.

Given $E[XY]$ where $Y$ is the conditioning variable — first move? ::: Take out what is known: $E[XY\mid Y]=YE[X\mid Y]$, then apply the tower. Compound sum $S={\sum }_1^N{X}_i$, want the mean? ::: Tower / Wald: $E[S]=E[N]E[X]$. Compound sum, want the variance? ::: Eve's Law: $E[N]\mathrm{Var}(X)+\mathrm{Var}(N)(E[X]{)}^2$. Conditioning variable is a constant — what happens to $E[X\mid Y]$? ::: It equals $E[X]$; the tower becomes a trivial identity. When is "plain average of conditional means" valid? ::: Only when all $P(Y=y)$ are equal.