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i have problems with calulating

$$E(e ^X)$$

X has a binomialdistributon with parameters n,p. E is the expectation.

My approach $$ E(e^X)= \sum_{k=0}^n e^k \binom{n}{k}p^k (1-p)^{n-k} = ... ?$$

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3 Answers

Hint Note that $e^k \cdot p^k = (e \cdot p)^k$. Apply the Binomial Theorem...

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Binomial theorem... do you think of this: $(ep+1-p)^n$ ? –  user1324258 Feb 3 '13 at 18:00
    
@user1324258 Exactly! –  saz Feb 3 '13 at 18:02
    
Ohh its very simple thank you! –  user1324258 Feb 3 '13 at 18:02
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$$ E(e^X)= \sum_{k=0}^n e^{k} \binom{n}{k}p^k (1-p)^{n-k} = \sum_{k=0}^n \binom{n}{k}(ep)^k (1-p)^{n-k} =(1-p+pe)^{n}$$

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hint:$$E(e^X)= \sum_{k=0}^n e^k \binom{n}{k}p^k (1-p)^{n-k} $$ $$\sum_{k=0}^n \binom{n}{k}({ep})^k (1-p)^{n-k}=(ep+1-p)^n$$

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