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I know the answer must be $\frac{1}{2}$, because I have done the numerical simulation and as the parameter of the poisson distribution increases, the quantity gets closer to 0.5. Can we show this analytically?

Edit: For whoever comes across this in the future, you can solve this by expanding $X \log X$ into its Taylor series around $\lambda$. After some algebra, you can show that $E[X \log X] - \lambda \log \lambda = \frac{1}{2} + \frac{1}{12\lambda} + \frac{1}{12 \lambda^2} + O(\frac{1}{\lambda^3})$

Using the equation above we now have: $\lim_{\lambda \rightarrow \infty} \big[ E[X \log X] - \lambda \log \lambda \big] = \frac{1}{2}$

Eaman
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  • Welcome to MSE. Here's how to ask a good question. Follow these guidelines to get help in this forum. – jjagmath Jul 06 '22 at 14:26
  • How did you get this power series in $\lambda^{-1}$ ? It doesn't seem obvious to me at all... – Maximilian Janisch Jul 07 '22 at 23:04
  • @MaximilianJanisch did you try the Taylor series expansion of XlogX around $\lambda$? – Eaman Jul 09 '22 at 11:41
  • @Eaman I don’t know what that means, $X\log(X)$ is a function of $\omega\in\Omega$, not of $\lambda$. If you have a reference solving the original question I would be really interested in how it’s done! – Maximilian Janisch Jul 10 '22 at 13:11
  • @MaximilianJanisch Taylor series expansion of $X \log(X)$ around $\lambda$ is $\lambda \log(\lambda) + (log(\lambda) + 1)(X - \lambda) + \frac{1}{2\lambda}(X - \lambda)^2 - \frac{1}{6\lambda^2}(X-\lambda)^3 + ... $

    Plug this in the expectation. After some algebra, you get the answer. I don't have a reference other than my notes.

     – Eaman Jul 19 '22 at 00:23

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