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Question

Let $X$ be Poisson with parameter $\lambda$. Compute the mean of $(1 + X)^{-1}$. (Introduction to Probability Theory, Hoel, pp. 104)

Answer Key

The answer key shows that the mean is $\lambda^{-1}(1 - e^{-\lambda})$

My Solution

$$E(1 + X)^{-1} = \sum\limits_{j = 1}^\infty (1 + j)^{-1} \dfrac{\lambda^{(1 + j)^{-1}}e^{-\lambda}}{(1 + j)^{-1}!}$$ $$= e^{-\lambda} \sum\limits_{j = 1}^\infty \dfrac{\lambda^{(1 + j)^{-1}}}{((1 + j)^{-1} - 1)!}$$ $$= \lambda^{-1} e^{-\lambda} \sum\limits_{j = 1}^\infty \dfrac{\lambda^{(1 + j)^{-1} - 1}}{((1 + j)^{-1} - 1)!}$$ $$= \lambda^{-1} e^{-\lambda}(e^{\lambda} - 1)$$ $$= \lambda^{-1}(1 - e^{-\lambda})$$

EDIT: I found why I got the answer incorrect. Thanks for those who answer the question! :)

StubbornAtom
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NasuSama
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3 Answers3

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Your solution is not quite correct: $$ \mathsf{E}\left(\frac{1}{1+X}\right) = \sum_{k=0}^\infty \frac{1}{1+k} \Pr\left(X=k\right) = \sum_{k=0}^\infty \frac{1}{1+k} \frac{\lambda^k \mathrm{e}^{-\lambda}}{k!} = \frac{\mathrm{e}^{-\lambda}}{\lambda} \sum_{k=0}^\infty \frac{\lambda^{k+1} }{(k+1)!} = \frac{\mathrm{e}^{-\lambda}}{\lambda} \left( \sum_{k=1}^\infty \frac{\lambda^{k} }{k!} \right) = \frac{\mathrm{e}^{-\lambda}}{\lambda} \left( \sum_{k=0}^\infty \frac{\lambda^{k} }{k!} -1 \right) = \frac{\mathrm{e}^{-\lambda}}{\lambda} \left(\mathrm{e}^\lambda -1 \right) $$

Sasha
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  • I edit my work in case you didn't know. I changed the answer because I figured out why already. Anyway, thanks for the answer! – NasuSama Sep 18 '13 at 16:11
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We do the calculation. The expectation is $$\sum_{n=0}^\infty \frac{1}{1+n}e^{-\lambda}\frac{\lambda^n}{n!}.$$ This is $$e^{-\lambda} \sum_{n=0}^\infty \frac{\lambda^n}{(n+1)!}.$$ Divide, multiply by $\lambda$. We get $$\frac{e^{-\lambda}}{\lambda}\sum_{n=0}^\infty \frac{\lambda^{n+1}}{(n+1)!}.$$ The inner sum above is equal to $e^{\lambda}-1$, so the whole thing simplifies to $$\frac{1}{\lambda}(1-e^{-\lambda}).$$

obataku
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André Nicolas
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1

Your formula for expectation has a problem, you should not plug in $1+X$ inverse in the probability as you are summing over the sample space, you should be plugging in $X$ only.

Evan
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