Solutions - 5150-02

(1) Recall the integral formula for variance:

Use the fact that $\text{Var}[X]=E\left[{(X-\mu )}^2\right]$.

$$\text{Var}[X]={\int }_{-\infty }^{\infty }{(x-\mu )}^2{f}_X(x)dx$$

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(2) Compute $\mu =E[X]$:

$$E[X]={\int }_0^{\infty }3x{e}^{-3x}dx=\underset{b\to \infty }{\lim }{[-x{e}^{-3x}-\frac{e^{-3x}}{3}]|}_0^b=\frac{1}{3}$$

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

$$\text{Var}[X]={\int }_0^{\infty }{(x-\frac{1}{3})}^{2}3e^{-3x}dx=\underset{b\to \infty }{\lim }{\int }_0^{b}3x^2{e}^{-3x}-2x{e}^{-3x}+\frac{e^{-3x}}{3}dx$$ $$=\underset{b\to \infty }{\lim }{[-x^2{e}^{-3x}-\frac{e^{-3x}}{9}]|}_0^b=\frac{1}{9}$$