Question

Use the identity of Theoretical Exercise 5.5 to derive E[X2] when X is an exponential random variable with parameter λ.

Short answer

Thus,

${\int }_0^{\mathrm{\infty }}2xP(X>x)dx={\int }_0^{\mathrm{\infty }}2x{e}^{-\lambda x}dx=2{\int }_0^{\mathrm{\infty }}x{e}^{-\lambda x}dx$

$\text{Use partial integration . Let}u=x\text{and}dv=e^{-\lambda x}\text{. Hence, we have that}du=dx\text{and}v=-\frac{e^{-\lambda x}}{\lambda }\text{. Thus}$

$2{\int }_0^{\mathrm{\infty }}x{e}^{-\lambda x}dx=-{2x\frac{e^{-\lambda x}}{\lambda }|}_0^{\mathrm{\infty }}+\frac{2}{\lambda }{\int }_0^{\mathrm{\infty }}{e}^{-\lambda x}dx={\frac{2}{\lambda }\cdot (-\frac{e^{-\lambda x}}{\lambda })|}_0^{\mathrm{\infty }}=\frac{2}{{\lambda }^2}$

Step-by-step solution

Given Information 

Derive E[X2] when X is an exponential random variable with parameter λ.

Explanation

From exercise 5, we have that.

$E\left(X^2\right)={\int }_0^{\mathrm{\infty }}2xP(X>x)dx$

$\text{Now, we are given that}X~\mathrm{Expo}\left(\lambda \right)\text{. Hence, we have that}$

$P(X>x)=1-P(X\le x)=1-(1-e^{-\lambda x})=e^{-\lambda x}$

Thus,

${\int }_0^{\mathrm{\infty }}2xP(X>x)dx={\int }_0^{\mathrm{\infty }}2x{e}^{-\lambda x}dx=2{\int }_0^{\mathrm{\infty }}x{e}^{-\lambda x}dx$

$\text{Use partial integration . Let}u=x\text{and}dv=e^{-\lambda x}\text{. Hence, we have that}du=dx\text{and}v=-\frac{e^{-\lambda x}}{\lambda }\text{. Thus}$

$2{\int }_0^{\mathrm{\infty }}x{e}^{-\lambda x}dx=-{2x\frac{e^{-\lambda x}}{\lambda }|}_0^{\mathrm{\infty }}+\frac{2}{\lambda }{\int }_0^{\mathrm{\infty }}{e}^{-\lambda x}dx={\frac{2}{\lambda }\cdot (-\frac{e^{-\lambda x}}{\lambda })|}_0^{\mathrm{\infty }}=\frac{2}{{\lambda }^2}$

Hence, we have proved that

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

Final Answer 

Thus,

${\int }_0^{\mathrm{\infty }}2xP(X>x)dx={\int }_0^{\mathrm{\infty }}2x{e}^{-\lambda x}dx=2{\int }_0^{\mathrm{\infty }}x{e}^{-\lambda x}dx$

$\text{Use partial integration . Let}u=x\text{and}dv=e^{-\lambda x}\text{. Hence, we have that}du=dx\text{and}v=-\frac{e^{-\lambda x}}{\lambda }\text{. Thus}$

$2{\int }_0^{\mathrm{\infty }}x{e}^{-\lambda x}dx=-{2x\frac{e^{-\lambda x}}{\lambda }|}_0^{\mathrm{\infty }}+\frac{2}{\lambda }{\int }_0^{\mathrm{\infty }}{e}^{-\lambda x}dx={\frac{2}{\lambda }\cdot (-\frac{e^{-\lambda x}}{\lambda })|}_0^{\mathrm{\infty }}=\frac{2}{{\lambda }^2}$