Expected value and standard deviation of $\sqrt{X}$

Let X be a poisson-distributed stochastic variable, where $$E(X) = m$$ $$V(X) = m.$$ Let $Y=\sqrt{X}.$

Calculate (approximatively) $$E(Y),$$ $$V(Y).$$

Now, the answer is $$E(Y) \approx \sqrt{m},$$ $$V(Y) \approx \left( \frac{1}{2\sqrt{m}} \right)^2$$

I've looked in my textbook but I don't know where this is coming from. Why is this the case? What (simple) theorem can I use to verify that it's true?.

• – Math1000 Nov 1 '17 at 19:56
• What is the source of this exercise? – Did Nov 2 '17 at 11:31

Using the law of the unconscious statistician, $$E(\sqrt X) = \sum_{k=0}^\infty \sqrt k P(X=k)=\sum_{k=0}^\infty \sqrt k e^{-m}\frac{m^k}{k!}=e^{-m}\sum_{k=0}^\infty \sqrt k \frac{m^k}{k!}$$
An asymptotic expansion of $\sum_{k=0}^\infty \sqrt k \frac{m^k}{k!}$ as $m\to \infty$ is needed. From this answer, $$\sum_{n=1}^{\infty}\frac{\sqrt{n}\,z^n}{n!}= \sqrt{z}\ e^z\left[1-\frac 1{8\,z}-\frac 7{128\,z^2}-\frac {75}{1024\,z^3}+O\left(\frac 1{z^4}\right)\right]$$
Hence $$e^{-m}\sum_{k=0}^\infty \sqrt k \frac{m^k}{k!} = \sqrt m - \frac{1}{8 \sqrt m} + o\left(\frac{1}{\sqrt m} \right)$$
Hence $E(\sqrt X)\sim \sqrt m$ and $$Var(\sqrt X)=E(X)-E(\sqrt X)^2=m-(m-\frac 14+o(1)) \sim \frac 14$$