According to textbooks, the mean of a random variable $X\sim\displaystyle\operatorname{Bin}\left(n,\theta\right)$ that follows a binomial distribution is given by $\mathbb{E}\left[X\right]=\theta$.
The mean of a discrete random variable $X$ is given by
$$\mathbb{E}\left[X\right]:=\sum\limits_{x\in\chi}xp\left(X=x\right)$$
where $\chi$ is the state space of $X$.
I am trying to prove using this definition of the mean of the discrete random variable $X\in\displaystyle\operatorname{Bin}\left(n,\theta\right)$, $\mathbb{E}\left[X\right]=\theta$.
In this case, $\chi=[0,n]$ and
$$p\left(X=x\right):={\binom{n}{x}}\theta^{x}\left(1-\theta\right)^{n-x}$$
where
$$\binom{n}{x}:=\frac{n!}{\left(n-x\right)!x!}$$
From the above,
$$\mathbb{E}\left[X\right]=\sum\limits_{x=0}^{n}x{\binom{n}{x}}\theta^{x}\left(1-\theta\right)^{n-x}=\sum\limits_{x=0}^{n}\frac{n!}{(n-x)!(x-1)!}\theta^{x}(1-\theta)^{n-x}$$
I was hoping to "isolate" $\theta$ by using the Taylor series relation:
$$\left(1-\theta\right)^{n-x}=\sum\limits_{i=0}^{\infty}\binom{n-x}{i}\left(-\theta\right)^{i}=\sum\limits_{i=0}^{\infty}\frac{\left(n-x\right)!}{\left(n-x-i\right)!i!}\left(-\theta\right)^{i}$$
which I believe should be convergent given that $\theta\leq1$ based on the fundamental rules of probability. Now I have
$$\mathbb{E}\left[X\right]=\sum\limits_{x=0}^{n}\left(\frac{n!}{(n-x)!(x-1)!}\theta^{x}\left(\sum\limits_{i=0}^{\infty}\frac{\left(n-x\right)!}{\left(n-x-i\right)!i!}\left(-\theta\right)^{i}\right)\right)$$
which i manage to simplify to
$$\mathbb{E}\left[X\right]=\sum\limits_{x=0}^{n}\left(\sum\limits_{i=0}^{\infty}\left(\frac{n!\theta^{x+i}\left(-1\right)^{i}}{\left(x-1\right)!\left(n-x-i\right)!i!}\right)\right)$$
Appreciate any help in trying to explain how this gets to $\mathbb{E}\left[X\right]=\theta$.
Best Answer
Expanding $(1-\theta)^{x}$ makes things too complicated. Instead of that do the following: $ \sum\limits_{x=0}^{n} x \binom {n} {x} \theta^{x}(1-\theta)^{n-x}=\sum\limits_{x=1}^{n} \frac {n!} {(x-1) (n-x)!} \theta^{x}(1-\theta)^{n-x}=n\theta \sum\limits_{x=1}^{n} \frac {(n-1)!} {(x-1) (n-x)!}\theta^{x-1}(1-\theta)^{n-x}$. Put $y=x-1$ to write this as $n\theta \sum\limits_{y=0}^{n-1} \binom {n-1} {y} \theta^{y}(1-\theta)^{(n-1-y)}$. The sum here is nothing but the binomial expansion of $(\theta+(1-\theta))^{n-1}$ so we are left with $n\theta (1)^{n-1}=n\theta$. Hence the mean is $n \theta$.