Chapter 3 Random Variables - 3.6 The Variance - Questions - Page 158: 11

See explanation

If $Y$ has an exponential distribution so that $f_Y(y) = \lambda e^{\large -\lambda y}$, then $\mu_Y = E(Y) = \dfrac{1}{\lambda}$ by Exercise 3.5.11 (p. 146). Now, $\begin{align*} E(Y^2) &= \int_{\mathbb{R}} y^2 \cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y^2)\ ] \\ &= \int_0^\infty y^2\cdot \lambda e^{\large -\lambda y}\ dy \qquad \left[\ \text{since}\ f_Y(y)= \lambda e^{\large -\lambda y},\ y \ge 0\ \right] \\ &= \lambda \int_0^\infty y^2e^{\large -\lambda y}\ dy \\ & \\ & \qquad \text{Use twice the reduction formula:} \\ & \qquad\qquad \int y^ne^{ay} = \frac{y^n}{a} - \frac{n}{a}\int y^{n-1}e^{ay}\ dy,\ a \ne 0 \\ & \qquad\qquad \text{Thus,} \\ & \qquad\qquad \lambda \underbrace{\int y^2\cdot e^{\large -\lambda y}\ dy}_{\text{use reduction formula}} \\ & \qquad\qquad = \lambda\left[ \frac{y^2e^{\large -\lambda y}}{-\lambda} - \frac{2}{-\lambda}\underbrace{\int ye^{\large -\lambda y}\ dy}_{\text{use reduction formula}} \right] \\ & \qquad\qquad = \lambda\left[-\frac{y^2e^{\large -\lambda y}}{\lambda} - \frac{2}{-\lambda}\left(\frac{ye^{\large -\lambda y}}{-\lambda} - \frac{1}{-\lambda}\int e^{\large -\lambda y}\ dy \right)\right] \\ & \qquad\qquad = \lambda\left[-\frac{y^2e^{\large -\lambda y}}{\lambda} - \frac{2}{-\lambda}\left(\frac{ye^{\large -\lambda y}}{-\lambda} - \frac{1}{-\lambda}\cdot\frac{e^{\large -\lambda y}}{-\lambda} \right)\right] \\ & \qquad\qquad = \lambda\left[-\frac{y^2e^{\large -\lambda y}}{\lambda} - \frac{2ye^{\large -\lambda y}}{\lambda^2} - \frac{2e^{\large -\lambda y}}{\lambda^3}\right] \\ & \qquad\qquad = -y^2e^{\large -\lambda y} - \frac{2ye^{\large -\lambda y}}{\lambda} - \frac{2e^{\large -\lambda y}}{\lambda^2} \\ & \\ &= \lim_{t\ \to \infty} \left( \lambda \int_0^t y^2e^{\large -\lambda y}\ dy \right) \\ &= \lim_{t\ \to \infty} \left( -y^2e^{\large -\lambda y} - \frac{2ye^{\large -\lambda y}}{\lambda} - \frac{2e^{\large -\lambda y}}{\lambda^2}\ \right\vert_0^t \\ &= \lim_{t\ \to \infty} \Biggl[\left(-\frac{t^2}{e^{\large t\lambda}} - \frac{2t}{\lambda e^{\large t\lambda}} - \frac{2\lambda}{\lambda^2e^{\large t\lambda}}\right) \\ & \qquad\qquad\qquad - \left(-(0^2)e^{\large -\lambda\cdot 0} - \frac{2(0)e^{\large -\lambda\cdot 0}}{\lambda} - \frac{2e^{\large -\lambda\cdot 0}}{\lambda^2}\right)\Biggr] \\ &= \left(\underbrace{-\frac{\infty}{\infty} - \frac{\infty}{\infty}}_{\text{use Hospital's Rule}} - \frac{2\lambda}{\infty}\right) - \left(-(0^2)e^{\large -\lambda\cdot 0} - \frac{2(0)e^{\large -\lambda\cdot 0}}{\lambda} - \frac{2e^{\large -\lambda\cdot 0}}{\lambda^2}\right) \\ &= (-0 -0 -0) - \left(-0 -0 - \frac{2}{\lambda^2} \right) \\ E(Y^2)\ &= \frac{2}{\lambda^2} \end{align*}$ By Theorem 3.6.1, since $\mu_Y = \dfrac{1}{\lambda}$ exists and $E(Y^2) = \dfrac{2}{\lambda^2}$ is finite, $\begin{align*} \text{Var}(Y) &= E(Y^2) - \mu_Y^2 \\ &= \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 \\ &= \frac{2}{\lambda^2} - \frac{1}{\lambda^2} \\ \color{blue}{\text{Var}(Y)}\ &\color{blue}{= \frac{1}{\lambda^2}} \end{align*}$