Question

Verify that$Var\left(X\right)=\frac{\alpha }{{\lambda }^2}$when$X$is a gamma random variable with parameters $\alpha$and $\lambda$

Short answer

To get the Variance, find the generic expression for $E\left(X^k\right)$and use it.

Step-by-step solution

Determine the positive integral. 

Let's find the first and second moments of $X$if $X$$~$Gamma $(\alpha ,\lambda )$

For any positive integer$k$.

we have that

localid="1649619459546" $$\begin{gathered} E\left(X^k\right)={\int }_0^{\infty }{x}^k·\frac{{\lambda }^{\alpha }}{\Gamma \left(\alpha \right)}{x}^{\alpha -1}{e}^{-\lambda x}dx \\ =\frac{{\lambda }^{\alpha }}{\Gamma \left(\alpha \right)}{\int }_0^{\infty }{x}^{k+\alpha -1}{e}^{-\lambda x}dx \end{gathered}$$

Expression the integral.

in order to evaluate the integral, make the substitution$s=\lambda x$

Hence, the expression above is equal to localid="1649619493912" $\frac{{\lambda }^{\alpha }}{\Gamma \left(\alpha \right)}{\int }_0^{\infty }{x}^{k+\alpha -1}{e}^{-\lambda x}dx=\frac{{\lambda }^{\alpha }}{\Gamma \left(\alpha \right)}{\int }_0^{\infty }{\left(\frac{s}{\lambda }\right)}^{k+\alpha -1}{e}^{-s}\frac{ds}{\lambda }$

localid="1649619476293" $$\begin{gathered} =\frac{1}{{\lambda }^k·\Gamma \left(\alpha \right)}{\int }_0^{\infty }{s}^{k+\alpha -1}{e}^{-s}ds \\ =\frac{\Gamma (k+\alpha )}{{\lambda }^k·\Gamma \left(\alpha \right)} \end{gathered}$$

So, we have obtained that,

$E\left(X^k\right)=\frac{\Gamma (k+\alpha )}{{\lambda }^k·\Gamma \left(\alpha \right)}$

Implies the value.

The implies that

$$\begin{gathered} E\left(X\right)=\frac{\Gamma (\alpha +1)}{\lambda \times \Gamma \left(\alpha \right)} \\ =\frac{\alpha \times \Gamma \left(\alpha \right)}{\lambda \times \Gamma \left(\alpha \right)} \\ =\frac{\alpha }{\lambda } \end{gathered}$$

and

$$\begin{gathered} E\left(X^2\right)=\frac{\Gamma (\alpha +2)}{{\lambda }^2\times \Gamma \left(\alpha \right)} \\ =\frac{(\alpha +1)\alpha \Gamma \left(\alpha \right)}{{\lambda }^2\times \Gamma \left(\alpha \right)} \\ =\frac{(\alpha +1)\alpha }{{\lambda }^2} \end{gathered}$$

Finally, we have that

$$\begin{gathered} Var\left(X\right)=E\left(X^2\right)-E(X{)}^2 \\ =\frac{(\alpha +1)\alpha }{{\lambda }^2}-\frac{{\alpha }^2}{{\lambda }^2} \\ =\frac{\alpha }{{\lambda }^2} \end{gathered}$$

Which has to be proved.