A nonnegative random variable X has distribution function
$F(x)=1-\frac{3}{5}\mathrm{e}^{-5x}-\frac{2}{5}\mathrm{e}^{-2x},\quad x\geq0.$
(i) Verify that F is a proper distribution function.

(ii) Find E(X).

Question

A nonnegative random variable X has distribution function

[latex]F(x)=1-\frac{3}{5}\mathrm{e}^{-5x}-\frac{2}{5}\mathrm{e}^{-2x},\quad x\geq0.[/latex]

(i) Verify that F is a proper distribution function.

(ii) Find E(X).

Accepted Answer

(i) We need to check that all conditions for a distribution function are satisfied. First, it is obvious that F(x) ≥ 0 for all x (in particular, F(x) = 0 for x < 0, since X is nonnegative). Next, we have

[latex]\operatorname*{lim}_{x\rightarrow-\infty}F(x)=0,\quad\operatorname*{lim}_{x\rightarrow\infty}F(x)=1[/latex]

and finally, we note that F is right continuous (in fact, it is both left and right continuous for any real x). It is also straightforward to check that F is a nondecreasing function. Thus, it is a proper distribution function of a random variable.

(ii) We use the second formula in Proposition 6.7, because X takes only nonnegative values. This gives

[latex]{ E}(X)=\int_{0}^{\infty}(1-F(x))\mathrm{d}x=\int_{0}^{\infty}\left({\frac{3}{5}}\mathrm{e}^{-5x}+{\frac{2}{5}}\mathrm{e}^{-2x}\right)\mathrm{d}x={\frac{3}{5}}\int_{0}^{\infty}\mathrm{e}^{-5 x}\mathrm{d}x+{\frac{2}{5}}\int_{0}^{\infty}\mathrm{e}^{-2x}\mathrm{d}x={\frac{3}{5}}\left[{\frac{\mathrm{e}^{-5x}}{-5}}\right]_{0}^{\infty}+{\frac{2}{5}}\left[{\frac{\mathrm{e}^{-2x}}{-2}}\right]_{0}^{\infty}={\frac{3}{5}}\cdot{\frac{1}{5}}+{\frac{2}{5}}\cdot{\frac{1}{2}}=\frac{8}{25}.[/latex]

Question 6.13: A nonnegative random variable X has distribution function F(......

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