Q. 7.10 - A First Course in Probability

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 $E\left[\sqrt{X}\right]$.

Step-by-Step Solution

Answer

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

Step 1: Given Information

$\lambda$ is not too small.

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

Step 3: 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, $Var\left(\sqrt{X}\right)\approx \frac{1}{4}=0.25$.

Step 4: Final answer

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