Chapter 5: Q.5.6 (page 212)

5.6. Compute $E [X]$ if $X$ has a density function given by
$(a) f (x) = \{\begin{cases} \frac{1}{4} x e^{\frac{- x}{2}} & x > 0 \\ 0 & o t h e r w i s e \end{cases}$ ;
$(b) f (x) = \{\begin{cases} c (1 - x^{2}) & - 1 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$ ;
$(c) f (x) = \{\begin{cases} \frac{5}{x^{2}} & x > 5 \\ 0 & x \leq 5 \end{cases}$ .

Short Answer

Expert verified

Thus the answer of given question is

$(a), E (X) = 4$

$(b), E (X) = 0$

$(c), E (X) = \infty$

Step by step solution

Given Information.

$f (x) = \{\begin{cases} \frac{1}{4} x e^{\frac{- x}{2}} & x > 0 \\ 0 & o t h e r w i s e \end{cases}$

Explanation.

$E (x) = \int_{- \infty}^{\infty} x . \frac{1}{4} x e^{\frac{- x}{2}} d x$

$= \frac{1}{4} \int_{0}^{\infty} x^{9} e^{\frac{- x}{2}} d x$

$\frac{x}{2} = 4, x = 2 y$

$E (X) = \int_{0}^{\infty} 2 . y^{2} e^{- y} d y = 2 \int_{0}^{\infty} y^{3 - 1} e^{- 1} d y$

$= 2 \times 2 = 4$

Explanation.

$f (x) = \{\begin{cases} c (1 - x^{2}) & - 1 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$

$\int_{- \infty}^{\infty} (1 - x^{2}) d x = 1, \int_{- \infty}^{\infty} c (1 - x^{2}) d x = 1$

$c . \frac{4}{3} = 1 \Rightarrow c = \frac{3}{4}$

$E (X) = \int_{- \infty}^{\infty} x . \frac{3}{4} (1 - x^{2}) d x = \frac{3}{4} \int_{- 1}^{1} (x - \frac{3}{4}) d x$

$= \frac{3}{4} {[\frac{x^{2}}{2} - \frac{x^{4}}{4}]}^{1}_{- 1}$

$= \frac{3}{4} \times 0 = 0$

Explanation.

$E (X) = \int_{- \infty}^{\infty} \frac{5}{x^{0}} d x = \int_{}^{} \frac{5}{0 x} d x = 5$

$= 5 {|\frac{u}{x^{1}}|}_{5}^{\infty}$

$= \infty$

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5.6. ComputeE[X]if Xhas a density function given by(a)f(x)=14xe-x2x>00otherwise;(b)f(x)=c(1−x2)-1<x<10otherwise;(c)f(x)=5x2x>50x≤5.

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