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

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.

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  • $\begingroup$ 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
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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::

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

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  • $\begingroup$ 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

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