Question

Let $X$be a Poisson random variable with mean $\lambda$. Show that if $\lambda$is not too small, then

$Var\left(\sqrt{X}\right)\approx .25$

Hint: Use the result of Theoretical Exercise $7.4$to approximate$E\left[\sqrt{X}\right]$.

Short answer

If $\lambda$is not too small, then $Var\left(\sqrt{X}\right)\approx .25$is shown.

Step-by-step solution

Given Information

$\lambda$is not too small.

Explanation

We know,$E\left[X\right]=4,Var\left(X\right)={\sigma }^2$

$E\left[g\right(x\left)\right]\approx f\left(\mu \right)+\frac{g^{\text{'}}\left(\mu \right)}{2}{\sigma }^2$

$E\left[\sqrt{X}\right]\approx \sqrt{\lambda }+\frac{\lambda }{2}\frac{d^2}{d{\lambda }^2}$

$=\sqrt{\lambda }+\frac{\lambda }{2}\left(\frac{-1}{4{\lambda }^{\frac{3}{2}}}\right)$

Solved, $=\sqrt{\lambda }+\frac{\lambda }{2}\left(\frac{-1}{8{\lambda }^{\frac{1}{2}}}\right)$

Explanation

If $E\left[X\right]=\lambda ,Var\left(X\right)=\lambda$

Now, $Var\left(\sqrt{X}\right)=E\left[X\right]-\left(E\right[\sqrt{X}]{)}^2=\lambda -(E\left[\sqrt{X}\right]{)}^2$

$\left(E\right[\sqrt{X}]{)}^2=\lambda +\frac{1}{64\lambda }-\frac{1}{4}$

$Var\left(\sqrt{X}\right)=\lambda -\lambda -\frac{1}{64\lambda }+\frac{1}{4}=\frac{1}{4}+\frac{1}{64\lambda }$

Thus,$Var\left(\sqrt{X}\right)\approx \frac{1}{4}=0.25$.

Final answer

If $\lambda$is not too small, then $Var\left(\sqrt{X}\right)\approx \frac{1}{4}=0.25$is shown.