Why is the expected value of $E(MX|M=X) \neq E(X^2)$

Question

If $X$ and $Y$ are independent exponential random variables with different parameters and $ M = min\{X,Y\}$, why would $E(MX|M=X) \neq E(X^2)$?

By intuition it seems like these terms would be equal ($E(MX|M=X) = E(XX) = E(X^2)$).

Answer 1

The main reason is because $M \le X$, sometimes with strict inequality, and so $E[M] < E[X]$ and $E[M^2] < E[X^2]$. You do have $E(M \mid M=X) = E(X \mid M=X)$, but that is not enough.

What is true is $E(MX \mid M=X) = E(X^2 \mid M=X) = E(M^2 \mid M=X)$. With $X$ and $Y$ being independently exponentially distributed, this is equal to the unconditional $E(M^2)$ but not equal to $E(X^2)$.

If the rates are $\lambda_X$ and $\lambda_Y$, so means and standard deviations are $\frac1{\lambda_X}$ and $\frac1{\lambda_Y}$, then $M$ is exponentially distributed with mean and standard deviation $\frac1{\lambda_X +\lambda_Y}$. You get $E(MX \mid M=X) = E(M^2)= \frac2{(\lambda_X +\lambda_Y)^2}$ while $E(X^2)= \frac2{\lambda_X^2}$.

Answer 2

Perhaps a discrete analogy will be helpful. Let $A$ and $B$ be Bernoulli random variables with parameters $p$ and $q$ respectively, and $C = \text{min}(A,B)$. Note that $C=A$ unless $A=1$ and $B=0$. Then

$$ E[CA | C=A] = \frac{pq}{1-p(1-q)}$$ while $$ E[A^2 ] = p$$