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Let X be a poisson distributed variable with parameter $\lambda$, what is $E[2^{-x}]$?

I know it should equal to

$\sum_{i=0}^\infty p(\frac{1}{2^x}=i)i$

but how do I calculate this sum?

If I isolate X in $1/2^x$, I get $p(x=log_{\frac{1}{2}}i)$ and then

$E[2^{-x}]=\sum_{i=0}^\infty i * \frac{e^{-\lambda}\lambda^{log_{1/2}i}}{(log_{1/2}i)!} = e^{-\lambda}\sum_{i=0}^\infty i * \frac{\lambda^{log_{1/2}i}}{(log_{1/2}i)!}$

But because of the 'i' in the sum I don't know how to solve this either.

Thanks in advance, and sorry if there were English mistakes.

1 Answers 1

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Note for any given $\lambda>0$, using the property of mpf of Poisson distribution, we have $$\sum_{i=0}^\infty \frac{\lambda^i}{i!}e^{-\lambda}=1.$$

Thus $$E[2^{-X}]=\sum_{i=0}^\infty 2^{-i}\frac{\lambda^i}{i!}e^{-\lambda}=e^{-\lambda/2}\sum_{i=0}^\infty \frac{(\lambda/2)^i}{i!}e^{-\lambda/2}=e^{-\lambda/2}.$$

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    Thank you, but how did you reach the first sum? is it correct to just replace the i with the function of X, and if so, why? Also, why did the e^lambda get out of the sum with e^-lambda/2? – 2017-02-04
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    Nevermind, I just realized that you split the e^-lambda to two and only ejected 1 of them. Still wondering about the first question though. – 2017-02-04
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    @hrsidkpi For the first question, by definition, $E[2^{-X}]=\sum_{i=0}^\infty 2^{-i}P(X=i)$. – 2017-02-04
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    thank you, the identity I was looking for was E[g(x)]. – 2017-02-06