Q.4.19 - A First Course in Probability

Question

Show that $X$is a Poisson random variable with parameter $\lambda$, then

$E\left[X^n\right]=\lambda E\left[(X+1{)}^{n-1}\right]$

Now use this result to compute $E\left[X^3\right]$.

Step-by-Step Solution

Answer

$E\left(X^3\right)={\lambda }^3+3{\lambda }^2+\lambda$

Step 1 : Given information

Given in the question that$X$ is a Poisson random variable with parameter $\lambda$, then $E\left[Xn\right] = \lambda E\left[\right(X + 1) n-1]$. We need to find $E\left[X3\right]$

Step 2 : Explanation

Using the theorem about the mean of function of random variable, 

we have that

$E\left(X^n\right)=\sum _{k=0}^{\infty }{k}^n\frac{{\lambda }^k}{k!}{e}^{-\lambda }=\sum _{k=1}^{\infty }{k}^n\frac{{\lambda }^k}{k!}{e}^{-\lambda }=\lambda \sum _{k=1}^{\infty }{k}^n\frac{{\lambda }^{k-1}}{k!}{e}^{-\lambda }$

$=\lambda \sum _{k=1}^{\infty }{k}^{n-1}\frac{{\lambda }^{k-1}}{(k-1)!}{e}^{-\lambda }=\lambda \sum _{k=0}^{\infty }(k+1{)}^{n-1}\frac{{\lambda }^k}{k!}{e}^{-\lambda }=\lambda E\left((X+1{)}^{n-1}\right)$

which had to be proved. Using that, we have that

$E\left(X^3\right)=\lambda E\left((X+1{)}^2\right)=\lambda E\left(X^2+2X+1\right)=\lambda \left(E\left(X^2\right)+2E\left(X\right)+1\right)$

Now, use that $E\left(X^2\right)=Var\left(X\right)+E{X}^2=\lambda +{\lambda }^2$ 

and that$EX=\lambda$,

so we have that the expression above is equal to

$=\lambda \left(\lambda +{\lambda }^2+2\lambda +1\right)={\lambda }^3+3{\lambda }^2+\lambda$

Step 3 : Final answer

$E\left(X^3\right)={\lambda }^3+3{\lambda }^2+\lambda$