# STT 861 Theory of Prob and STT I Lecture Note - 11

2017-11-15

Proof of the "Tower" property; discrete conditional distribution; continuous conditional distribution expectation and variance and their examples; linear predictor and mean squared error.

# Lecture 11 - Nov 15 2017

Reinterpretation of last part of item (c) in proof of Theorem 5.2.1 in textbook.

Recall : let $X$, $Y$ be 2 random variables. Just to make things simple, assume $X$ and $Y$ are discrete, with PMF’s $P_x$ and $P_y$ and joint PMF $P_{X,Y}$.

Generally, for $h$ a function $\mathbb{R}\rightarrow\mathbb{R}$,

$E(h(X)Y) = E(h(X)g(X))$

where $g(x)=E(Y\vert X=x)$.

We want to think of that result in the following way:

• The meaning of the notation $g(X)$ is the value of the function ( g(x) ) where $x$ is replaced by the random variable $x$.

• However, we reinterpret $g(x)$ as:

$g(X)=E(Y\vert X)$

(think of this as a definition.)

Now we reinterpret the formula above like this:

\begin{align*} E(h(x)|Y) &= E(E(h(X)Y|X)) (\star) \\ &= E(h(X)E(Y|X)) \end{align*}

The first line means “An expectation can always be written as the expectation of a conditional expectation”. It is known as the “tower” property of conditional expectation.

The second line means: “when conditioning by $X$, $X$ can be considered as known (non-random) and any factor depending on $X$ can be pulled out of the conditional expectation”.

Proof the Star ($\star$):

\begin{align*} RHS &= E(h(X)g(X)) \\ &= E(h(X)E(Y|X)) \\ &= \sum_{X} h(x)E(Y|X=x)P_X(x) \\ &= \sum_xh(x)\sum_y f \frac{P_{X,Y}(x,y)}{P_X(x)}\\ &= \sum_x h(x) \sum_y yP_{X,Y}(x,y) \\ &= \sum_x\sum_yh(x)yP_{X,Y}(x,y)\\ &= E(h(X)Y) \end{align*}

(RHS = right hand side) This is the proof of star.

Next, we use star ($\star$) to compute the unconditional variance $Var(Y)$ by conditional by $X$.

\begin{align*} Var(Y) &= E((Y-E(Y))^2) \\ &= E((Y-g(X)+g(X)-E(Y))^2) \\ &\triangleq E(A^2+2AB+B^2) \end{align*}

where $A=Y-g(X)$, $B=g(X)-E(Y)$.

We will first compute

\begin{align*} E(AB) &= E((Y-g(X))(g(X)-E(Y)))\\ &=E(E((Y-g(X))(g(X)-E(Y))|X)) \\ & = E(E(Y-g(X)|X)(g(X)-E(Y))) \\ &= E((g(X)-g(X))g(X)-E(Y)) = 0 \end{align*}

We have proved

$Var(Y) = E(A^2)+E(B^2)$

Now we interpret

\begin{align*} E(A^2) &= E((Y-g(X))^2) \\ &= E(E((Y-g(X))^2|X)) \\ &= E(Var(Y|X)) \end{align*}

$Var(Y\vert X)$ is also known as $v(X)$.

So we see $E(A^2)=E(v(X))$ (expectation of conditional variance).

Finally,

\begin{align*} E(B^2) &= E((g(X)-E(Y))^2)\\ &= E((g(X)-E(E(Y|X)))^2) \\ &= E((g(X)-E(g(X)))^2) \\ &= Var(g(X)) \end{align*}

this is the variance of the conditional expectation.

### Homework Problem 5.2.5 Part b

$X$ takes value ${0,1,2}$ with probability ${0.3, 0.4, 0.3}$, $\varepsilon=\pm1$ with probability 0.5 and 0.5. $Y=5-X^2+\varepsilon$.

Q: Find $\rho=\rho(X,Y)$

$\rho(X,Y) = \frac{cov(X,Y)}{\sqrt{Var(X)Var(Y)}}$

\begin{align*} E((X-\mu_X)-(Y-\mu_Y)) &= E(XY)-\mu_X\mu_Y\\ &=E(X(5-X^2+\varepsilon))\\ &= E(5X-X^3+X\varepsilon) \\ &= 5E(X) -E(X^3) =-E(X\varepsilon)\\ &=5\mu_X - E(X^3) \end{align*}

### Example 1

$N$ people come into a store in a given day, customer spends $X_i$ dollars. Let $T$ be the total \$ of sales for the day.

$T= X_1+X_2+\cdots + X_N=\sum_{i=1}^{N}X_i$

A: Find $E(T)$ and $Var(T)$.

Assume:

• $N$ is independent of all the $X_i$’s
• $X_i$s are i.i.d.

Let’s compute

$E(T) = E(E(\sum X_i\vert N))$

\begin{align*} E(T) &= E(\sum E(X_i|N))\\ &= E(NE(X_i))\\ &=E(X_1)E(N) \end{align*}

We know $T$ is related to $N$. Therefore we must compute conditional variance

• Conditional variance is $v(n) = Var (T\vert N=n)$
• Conditional expectation is $g(n) = E(T\vert N=n)$

$Var(T\vert N=n)=Var(\sum X_i\vert N=n) = Var(\sum X_i) = nVar(X_1)$

We have just proved that $v(n)=nVar(X_1)$.

Next,

$E(T\vert N=n) = E(\sum X_i\vert N=n) = E(\sum X_i) = nE(X_i)$

We proved here $g(n)=nE(X_1)$, now finally go back to the original formula,

$Var(T) = E(v(N)+ Var(g(N)) = E(NVar(X_1))+Var(N(E(X_1))) = Var(X_1)E(N)+ E(X_1)^2Var(N)$

where $T=\sum X_i$, where $X_i$ are i.i.d and $N$ independent of $X_i$’s.

Exercise: Prove the following (using similar method of proof as for the $Var(T)$ formula).

$Cov(N,T)=E(X_1)Var(N)$

and therefore,

$\rho (N,T)=\frac{1}{\sqrt{1+\theta}}$

where $\theta=Var(X)/(E(X_1)Var(N))$.

Also, for $N\sim Poi(\lambda)$ and $X\sim Ber(\theta)$, compute $Var(T)$ and $\rho(N,T)$.

### Example 2

let $X\sim Geom(p)$, $D\sim NegBin(p,r=X)$, therefore, $D$ is a certain $T$, where the $N$ is the $X$ above and each $X_i$ is $\sim Geom(p)$, i.i.d.

Let $Y=X+D$, find $E(Y)F$.

$E(Y)+E(X)+E(D)=\frac{1}{p} +E(X_1)E(X)=\frac{1}{p}+\frac{1}{p^2}$

$Var(Y) = Var(X+D)=Var(X)+Var(D)+ 2cov(X,D)$

## Continuous Case

### Example 5.3.2

$X\sim \Gamma (\alpha,1)$, $Y\sim \Gamma(\beta,1)$.

Let $V=X+Y$, therefore, $V\sim Gamma(\alpha+\beta,1)$.

Let $V=\frac{X}{X+Y}$, this is called Beta random variable $\sim B(\alpha+\beta)$.

Let’s now try to prove that $U$ and $V$ are independent.

A: Let $g(u)=E(X\vert U=u)$. It turns out (Wikipedia) $E(V)=\frac{\alpha}{\beta}$.

Therefore $E(UV\vert U=u)=uE(V\vert U=u)=uE(V)=u \frac{\alpha}{\alpha+\beta}$.

This gives us an example where the function $g$ is linear as a function of $u$ because $E(X|U=u)=E(UV|U=u)=u \frac{\alpha}{\alpha+\beta}$.

This situation where $X$ is linear given $U$ is pretty exceptional.

We call $g(x)=E(Y|X=x)$ the predictor of $Y$ given $X$. But what is the linear predictor?

## Linear Predictor and Mean Squared Error

We would like to predict $Y$ using a linear function of $X$.

Let $aX+b$ be the linear predictor. Consider the error in replacing $Y$ by $aX+b$.

We can choose $a$ and $b$ such that $E(Y-aX-b)=0$.

More systematically, let’s consider what statistic cases might called the mean square error (MSE)

$E((Y-(aX+b))^2)$

we want to minimize MSE over all possible choices of the 2 values $a$ and $b$. It turns out that $a=Corr(X,Y) \frac{\sigma_X}{\sigma_Y}$ and best $b=E(Y)-aE(X)$.

Note: this is the closely allied to the question of linear regression. It turns out the MSE fir that pair of $(a,b)$ is

$1-\rho^2Var(Y)$

This says: the uncertainty level on $Y$ is $Var(Y)$. The proposition of that variance which is explained by $X$ is the variance of $aX+b$ is

$Var(aX)=a^2Var(X) =\rho\frac{\sigma_Y^2}{\sigma_X^2}\sigma_X^2$

and what is not explained by $X$ is the MSE $(1-\rho^2)Var(Y)$.

Summary: with $(a,b)$ as above and $\sigma_X^2=Var(X)$, $\sigma_Y^2$. we see that the amount of variance of $Y$ explained by $X$ is $Var(aX)=\rho^2\sigma_Y^2$ The MSE $=(1-\rho^2)\rho_Y^2$ is the variance of $Y$unexplained by $X$.