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: \(\int |f(X)|^2dx < \infty\)