Find $\mathbb{E}[X]$ given a discrete $2$ variable probability function

Solution 1:

First we need $c$.

$$\sum_{n=0}^{\infty}\sum_{m=0}^{146}c e^{-m} \frac{ m^n}{ n!} \binom{146}{m} =1$$.

So $$\sum_{m=0}^{146}\sum_{n=0}^{\infty}c e^{-m} \frac{ m^n}{ n!} \binom{146}{m}=1$$.

So $$\sum_{m=0}^{146}c\binom{146}{m}=1\implies c\cdot 2^{146}=1\implies c=\frac{1}{2^{146}}$$.

So $$\mathbb{E}[X]=\frac{1}{2^{146}}\sum_{n=0}^{\infty}\sum_{m=0}^{146}ne^{-m}\frac{m^{n}}{n!}\binom{146}{m}$$

$$=\frac{1}{2^{146}}\sum_{m=0}^{146}\sum_{n=1}^{\infty}me^{-m}\frac{m^{n-1}}{(n-1)!}\binom{146}{m}$$

$$=\frac{1}{2^{146}}\sum_{m=0}^{146}m\binom{146}{m}=\sum_{m=1}^{146}\frac{1}{2^{146}}\frac{m\cdot 146!}{m!(146-m)!}$$

$$=\frac{146}{2^{146}}\sum_{m=1}^{146}\frac{145!}{(m-1)!(145-(m-1))!}=\frac{146}{2^{146}}\sum_{m=0}^{145}\binom{145}{m}=\frac{146}{2^{146}}\cdot 2^{145}$$

$$=\frac{146}{2}=73$$

The change of order of summation is justified by Fubini as the summands are positive.