Find the expected value of $\sqrt{K}$ where $K$ is a random variable according to Poisson distribution with parameter $\lambda$.

I don't know how to calculate the following sum:

$E[\sqrt{K}]= e^{-\lambda} \sum_{k=0}^{\infty} \sqrt{k} \frac{\lambda^k}{k!} $

Based on Wiki (https://en.wikipedia.org/wiki/Poisson_distribution) I know that should be approximately $\sqrt{\lambda}$.

  • 1
    $\begingroup$ Are you sure that has a closed form solution...? $\endgroup$ Commented Nov 19, 2015 at 7:03
  • $\begingroup$ @BenjaminLindqvist No. $\endgroup$ Commented Nov 19, 2015 at 7:03
  • $\begingroup$ I'd guess it does not, sadly... $\endgroup$ Commented Nov 19, 2015 at 7:06
  • $\begingroup$ @BenjaminLindqvist So, how I may find an approximation of it? $\endgroup$ Commented Nov 19, 2015 at 7:07
  • $\begingroup$ W|A doesn't know either. For small $\lambda$ you can approximate by $\lambda$. For large $\lambda$ you can approximate by $\sqrt{\lambda}$ $\endgroup$
    – A.S.
    Commented Nov 19, 2015 at 7:14

2 Answers 2


In general, for smooth $g(X)$ you can do a Taylor expansion around the mean $\mu=E(X)$:

$$g(X)=g(\mu) + g'(\mu)(X-\mu)+ \frac{g^{''}(\mu)}{2!}(X-\mu)^2+ \frac{g^{'''}(\mu)}{3!}(X-\mu)^3+\cdots$$


$$E[g(X)]=g(\mu) + \frac{g^{''}(\mu)}{2!}m_2+ \frac{g^{'''}(\mu)}{3!} m_3+\cdots $$

where $m_i$ is the $i$-th centered moment. In our case $m_2=m_3 =\lambda$, so:

$$E[g(X)]=\sqrt{\lambda} - \frac{\lambda^{-1/2}}{8} + \frac{ \lambda^{-3/2}}{16} +\cdots $$

This approximation is useful only if $\lambda \gg 1$

  • $\begingroup$ Would you mind explaining why the approximation is only useful for $\lambda \gg 1$? $\endgroup$
    – Bergson
    Commented Mar 19, 2019 at 1:46
  • 1
    $\begingroup$ @ThomasBladt Only for $\lambda \gg 1$ the terms of the sum decrease quickly. $\endgroup$
    – leonbloy
    Commented Mar 19, 2019 at 1:54

Around $\lambda=0$ you have already written the series expansion so $$E(\sqrt{ P_\lambda})=(1-\lambda+O(\lambda^2))(\lambda+\frac {\sqrt 2}2\lambda^2+O(\lambda^3))=\lambda-(1-\frac {\sqrt 2}2)\lambda^2+O(\lambda^3)$$ Around $\lambda=\infty$, $\epsilon_\lambda=\frac {P_\lambda-\lambda}\lambda$ is tightly concentrated (approximately normal) around $0$ with mean $0$ and variance $\frac 1 \lambda$, so you can expand $\sqrt {P_\lambda}=\sqrt \lambda(1+\frac 1 2\epsilon_\lambda-\frac 1 8\epsilon_\lambda^2+O(\epsilon_\lambda^3))$ so that $$E(\sqrt {P_\lambda})=\sqrt \lambda(1-\frac 1 8 \lambda^{-1}+o(\lambda^{-{3/2}})).$$

  • $\begingroup$ For the series around $\lambda = 0$ the coefficient of $\lambda^n$ is $\displaystyle\sum_{k=0}^n \dfrac{(-1)^{n-k} \sqrt{k}}{k!(n-k)!}$. $\endgroup$ Commented Nov 19, 2015 at 18:41
  • $\begingroup$ Thanks, are you sure about the third term? Is the third term on the order of $\lambda^{-1}$? Your answer says this. $\endgroup$ Commented Nov 9, 2017 at 23:59

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