Expected value $E[e^X]$ when $X$ has a binomial distribution

Question

I was given $X\sim B(3,0.7)$.

I don't get why $E [e^X] = 10.6887$.

Any help is appreciated.

Answer 1

\begin{align}\mathbb{E} [\exp(X)]&= \sum_{x=0}^n P(X=x) \exp(x) \\&= \sum_{x=0}^n \binom{n}{x}p^x(1-p)^{n-x} e^x \\&= \sum_{x=0}^n \binom{n}{x} (pe)^x(1-p)^{n-x} \\&= (pe+1-p)^n \\&=(p(e-1)+1)^n \end{align}

Let $p=0.7$ and $n=3$ and you should be able to obtain your result.

Remark: $\mathbb{E}[\exp(Xt)]$ is known as the moment generating function.

Answer 2

Hint: In general, you can use the following formula:$$E(f(X))=\sum_{x=0}^3\Bbb P(X=x)f(x).$$ Applied to your example, since $f(X)=e^X$, this gives: $$E(e^X)=\sum_{x=0}^3\Bbb P(X=x)e^x.$$ Do you think you can finish it from here?