# 2 2013-Spring

## 2.2 Problem 1

Given: $$E[X|Y]=X\ and \ E[Y|X]=X$$

To Show:

### 2.2.1 Part (a): $$P(X=Y)=1$$

$E[Y]=E[E[Y|X]]=E[X]=\mu_x$

Thus, $$\mu_y=\mu_x$$

Also, \begin{align} Cov(X,Y) &=E[XY]-E[X]E[Y]\\ &= E[E[XY|X]]-\mu_x^2\\ &= E[XE[Y|X]]-\mu_x^2\\ &= E[X^2]-\mu_x^2\\ &= \sigma_x^2 \end{align}

Repeating the above with $$E[XY]= E[E[XY|Y]]$$ would give $$Cov(X,Y)=\sigma_y^2$$ and hence $$Cov(X,Y)=\sigma_x^2=\sigma_y^2=Var(X)$$ which implies $$X=Y$$ or $$P(X=Y)=1$$

Note, we implicitly used the requirement of the variance being finite[This is what is implied by the function being squared integrable: $$\int |f(X)|^2dx < \infty$$