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Find expected value of $\frac{1}{1+\xi}$ if $\xi$ has Poisson distribution.

My try:

$$E\left[ \frac{1}{1+\xi}\right]=\sum_{k=0}^{\infty}{x}_{k}{p}_{k}=\sum_{k=0}^{\infty}\frac{1}{k+1}\frac{\lambda^ke^{-\lambda}}{k!}$$

Is it right?

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    $\begingroup$ Yes, but it can be reduced. $\endgroup$
    – Paul
    Commented Jun 8, 2016 at 15:13

5 Answers 5

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Observe that $$ \sum_{k=0}^\infty\frac{1}{k+1}\frac{\lambda^ke^{-\lambda}}{k!}=\frac{1}{\lambda}\sum_{k=0}^\infty\frac{\lambda^{k+1}e^{-\lambda}}{(k+1)!}=\frac{1}{\lambda}\sum_{k={\color{red}1}}^\infty\frac{\lambda^ke^{-\lambda}}{k!}=\frac{1}{\lambda}\left(1-e^{-\lambda}\right). $$

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I wanted to point out a probabilistic interpretation and manipulation.

$\xi$ is the number of events in a Poisson process of rate $\lambda$ in the time interval $[0,1]$. These events split the interval into $\xi+1$ subintervals which are identically distributed. The expected length of each subinterval is $E [ \frac{1}{1+\xi}]$. But this is also equal to the expectation of the length first subinterval: the expectation of the minimum between $T$, the time of the first event of the Poisson process, and the time $1$. Recall that $T$ is exponential with parameter $\lambda$. Summarizing:

$$E[ \frac{1}{1+\zeta} ] = E[T\wedge 1].$$

To compute the RHS, observe that

$$ E [ T] = E [ T \wedge 1] + E [ (T-1), T>1] = E [ T\wedge 1]+ E[ T-1 | T>1] P(T>1)$$

By the memoryless property of exponential random variables, $T-1$ conditioned on $T>1$ is also exponential $\lambda$, so we obtain

$$ E[T] = E[T\wedge 1] + E[T]P(T>1) \Rightarrow E[T\wedge 1] = E[T] (1-P(T>t) )=\frac{1}{\lambda} (1-e^{-\lambda}).$$

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By the law of the unconscious statistician, $$ \mathbb{E}\left[\frac{1}{1+\xi}\right] = \sum_{k=0}^\infty \frac{p_k}{1+k} = e^{-\lambda} \sum_{k=0}^\infty \frac{\lambda^k}{k!(1+k)} = \frac{1}{\lambda e^\lambda} \sum_{k=0}^\infty \frac{\lambda^{k+1}}{(1+k)!} = \frac{e^\lambda - 1}{\lambda e^\lambda} $$

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Right, and: $$ \sum_{k\geq 0}\frac{\lambda^k e^{-\lambda}}{(k+1)k!} = \frac{e^{-\lambda}}{\lambda}\sum_{k\geq 0}\frac{\lambda^{k+1}}{(k+1)!}=\frac{e^{-\lambda}}{\lambda}\left(e^{\lambda}-1\right)=\color{red}{\frac{1-e^{-\lambda}}{\lambda}}.$$

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To start where you left off, $$\sum_{k=0}^{\infty}\frac{1}{k+1}\frac{\lambda^ke^{-\lambda}}{k!}$$ $$=\sum_{k=0}^{\infty}\frac{\lambda^ke^{-\lambda}}{(k+1)!}$$ $$=\frac{e}{\lambda}\sum_{k=0}^{\infty}\frac{\lambda^{k+1}e^{-(\lambda+1)}}{(k+1)!}$$ $$=\frac{e}{\lambda}(\sum_{k=0}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}-1)$$

Can you take it from there?

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