As the title states I want to calculate $$ \mathbb{E}[2^X], \quad 2^X\sim \text{Binom}. $$ I know the Law of the unconscious statistician but this leads nowhere because I cannot simplify $$ 2^k\binom{n}{k}. $$ Any hints?

  • 1
    $\begingroup$ You mean $\sum_k 2^k \binom nk$? But look at the binomial expansion of $3^n=(1+2)^n$. $\endgroup$
    – lulu
    Nov 24 at 12:23

2 Answers 2


If $X\sim\mathsf{Binom}(n,p)$ then we may write $X=\sum_{i=1}^n X_i$ where the $X_i$ are independent $\mathsf{Ber}(p)$ random variables. For any real $s$ it is clear that $$ \mathbb E[s^{X_1}] = 1-p +ps = 1+p(s-1), $$ and hence $$ \mathbb E[s^X] = \mathbb E \left[s^{\sum_{i=1}^n X_i}\right]=\prod_{i=1}^n\mathbb E[s^{X_i}] = \mathbb E[s^{X_1}]^n = (1+p(s-1))^n. $$ So in this case, where $s=2$, we simply have $$ \mathbb E[2^X] = (1+p)^n. $$

  • $\begingroup$ I dont get your first equation, you say it is clear but why? $\endgroup$ 2 days ago
  • 1
    $\begingroup$ @calculatormathematical Simplify $\Bbb E[s^{X_1}]=P(X_1=0)s^0+P(X_1=1)s^1$. $\endgroup$
    – J.G.
    2 days ago

We have $$ \mathbb E[2^X]=\sum_{k=0}^{n}2^k{n \choose k} p^k(1-p)^{n-k}=\sum_{k=0}^{n}{n \choose k} (2p)^k(1-p)^{n-k}=((1-p)+2p)^n=(1+p)^n. $$


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