Question

The second moment of a Poisson-distributed random variable is 2. The mean of the random variable is .

• A. 1

Answer 1

The correct option is A 1
${\text{n}}^{th}$ moment of a Random variable is given as

$E\left[x^n\right]={\int }_{-\infty }^{\infty }{x}^n{f}_x\left(x\right)dx$

Now, $x\to x-\overline{x};\overline{x}=$ mean value of Random Variable x

$\Rightarrow E\left[\right[x-\overline{x}{]}^n]={\int }_{-\infty }^{\infty }(x-\overline{x}{)}^n{f}_x(x)dx$

Now, if $n=2$ [second moment]

$\Rightarrow E\left[\right(x-\overline{x}{)}^2]={\int }_{-\infty }^{\infty }(x-\overline{x}{)}^2{f}_x(x)dx={\sigma }_x^2$

(Variance) [as per definition]

Hence, Variance of a random variable:

${\sigma }_x^2=E\left[x^2\right]-E^2\left[x\right]$ ...(i)

Let mean value of poisson-distributed random variable x is $\lambda$ and we know that Mean = Variance.

In case of Poisson distribution,

So $\lambda =2-{\lambda }^2$ (using (i))

$\Rightarrow {\lambda }^2+\lambda -2=0$

$\Rightarrow \lambda =1,-2$

$\because \lambda \ne -2$

$\therefore \lambda =1$