$$P(X = m, Y = n) = \frac{e^{-7}4^m3^{n-m}}{m!(n-m)!}$$
$m \in 0, 1, 2, ..., n$
$n \in N$

P otherwise zero.

Find $E(X)$.

It can be shown that $$P(X = m) = \frac{e^{-4}4^m}{m!}$$

Then to my understanding $E(X)$ = $\sum_{m=0}^n mP(X = m)$

$$E(X) = \sum_{m=0}^n \frac{me^{-4}4^m}{m!}$$

$$4e^{-4}\sum_{m=1}^n \frac{4^{m-1}}{(m-1)!}$$

$$4e^{-4}\sum_{m=0}^{n-1} \frac{4^m}{m!}$$

Is this the answer? Or do you let n approach infinity, which would give $E(X) = 4$. This seems more elegant, but I'm not sure why you'd allow n to approach infinity, as you should sum over m, not n.

  • $\begingroup$ Rewrite the set of $(m,n)$ with positive probability as $m\ge0, n\ge m$ and rethink your argument from the "Then to my understanding". $\endgroup$ – kimchi lover Apr 26 at 0:59
  • $\begingroup$ Do you mean use a double summation? $\endgroup$ – Vahan Apr 26 at 2:14
  • $\begingroup$ Your expectation is a double sum. You have incorrectly identified which terms to sum over. That's all. $\endgroup$ – kimchi lover Apr 26 at 2:17
  • $\begingroup$ But I think they are correct. $P(X=m)$ was found by summing over $n$. Then the expected value of X sums over $m$, as $X = m$. $\endgroup$ – Vahan Apr 26 at 3:50

You cannot have $n$ as the argmax for the series when you have "summed out" that term. After all, $n$ should not appear in the final answer. Look to the support again.

$$\{(m,n)\in\Bbb N^2:1\leq n\leq \infty, 1\leq m\leq n\}=\{(m,n)\in\Bbb N^2:1\leq m\leq \infty, m\leq n\leq\infty\}$$

So, indeed the solution is::

$$\begin{align}\mathsf E(X)&=\sum_{(m,n):1\leq m\leq n}~m~\mathsf P(X=m,Y=n)\\[1ex]&=\sum_{m:1\leq m}~m~\sum_{n:m\leq n}~\frac{e^{-7}4^m3^{n-m}}{m!(n-m)!}\\[1ex]&=\sum_{m=1}^\infty \dfrac{e^{-4}4^m}{(m-1)!}\sum_{n=m}^\infty \frac{e^{-3}3^{n-m}}{(n-m)!}\\[1ex]&=4\sum_{j=0}^\infty \dfrac{e^{-4}4^j}{j!}\sum_{k=0}^\infty \frac{e^{-3}3^{k}}{k!}\\[1ex]&= 4\sum_{j=0}^\infty \dfrac{e^{-4}4^j}{j!}\\[2ex]&=4\end{align}$$

  • $\begingroup$ Was my second solution incorrect, and I just arrived at the answer by luck? $\endgroup$ – Vahan Apr 26 at 4:49
  • $\begingroup$ Your solution was correct except for the argmax needing to be $\infty$ rather than $n$. I just performed the calculation for $\mathsf P(X=m)$ inline so it was clear how the support was separated into the double summation. $\endgroup$ – Graham Kemp Apr 26 at 4:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.