Show that $var(\sqrt{x})\approx.25$

Let $X$ be a Poisson random variable with mean $\lambda$. Show that $var(\sqrt{x})\approx.25$

Here is my attempt:

In a previous part of the problem, I showed $E[g(x)]\approx g(\mu)+\frac{g''(\mu)\sigma^2}{2}$, where $\mu$ is the mean and $\sigma^2$ is the variance of the random variable.

Let $g(x)=\sqrt{x}$. Then $E[\sqrt{x}]\approx \sqrt{\mu}-\frac{\sigma^2}{8{\mu}^{3/2}}$

Thus, $var(\sqrt{x})=E[(\sqrt{x})^2]-(E[\sqrt{x}])^2\approx E[x]-(\sqrt{\mu}-\frac{\sigma^2}{8{\mu}^{3/2}})^2=\lambda-\mu+\frac{\sigma^2}{4{\mu}}-\frac{\sigma^4}{64{\mu}^3}$.

I was hoping something would cancel out to $.25$. Did I do something wrong or is there something I missed that allows me to simplify further?

• what is $\sigma$ ? $\mu$? – Canardini Dec 6 '16 at 2:20
• @Canardini $\sigma^2$ is the variance and $\mu$ is the mean of the random variable I used to find $E[g(x)]\approx g(\mu)+\frac{g''(\mu)\sigma^2}{2}$ – Silvia Rossi Dec 6 '16 at 2:27
• @Canardini I was not given a distribution for that random variable – Silvia Rossi Dec 6 '16 at 2:30
• If the r.v. is $X$ then you should replace $x$'s by $X$'s in the formulation. – A.G. Dec 6 '16 at 2:51

You know the distribution of X which is Poisson of parameter $\lambda$, therefoire
$var(X)=\sigma^2=\lambda$
$E(X)=\mu=\lambda$
you get that $var(\sqrt(X))=\frac{1}{4}-\frac{1}{64\lambda}$
Remember that $\lambda=\mu=\sigma^2$ for a Poisson variable, so $\operatorname{Var}(\sqrt x)\approx \frac 1 4 - \frac {1}{64\lambda}$. So if $\lambda$ is large enough, we can say that this is about $\frac 1 4$.