-1
$\begingroup$

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?

$\endgroup$
1
  • 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

3
$\begingroup$

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 $$ \mathbb E[2^X] = (1+p)^n. $$

$\endgroup$
2
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.