Solutions - 5160-01

(1) State the PDF of an exponential distribution:

$$f_X(x)=\lambda e^{-\lambda x},x>0$$

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(2) Compute $E[X]$ using the integral formula:

$$\begin{array}{c}E[X]={\int }_0^{\infty }\lambda x{e}^{-\lambda x}dx\\ \\ \gg \gg \underset{b\to \infty }{\lim }{[-x{e}^{-\lambda x}-\frac{e^{-\lambda x}}{\lambda }]|}_0^b\gg \gg \frac{1}{\lambda }\end{array}$$

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(3) Compute $E\left[X^2\right]$ using the integral formula:

$$\begin{array}{c}E\left[X^2\right]={\int }_0^{\infty }\lambda x^2{e}^{-\lambda x}dx\\ \\ \gg \gg \underset{b\to \infty }{\lim }{[-x^2{e}^{-\lambda x}-\frac{2x{e}^{-\lambda x}}{\lambda }-\frac{2e^{-\lambda x}}{{\lambda }^2}]|}_0^b\gg \gg \frac{2}{{\lambda }^2}\end{array}$$

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(4) Compute $\text{Var}[X]$:

$$\begin{array}{c}\text{Var}[X]=E\left[X^2\right]-{\left(E[X]\right)}^2\\ \\ \gg \gg \frac{2}{{\lambda }^2}-\frac{1}{{\lambda }^2}\gg \gg \frac{1}{{\lambda }^2}\end{array}$$