Compute [X] if X has a density function given by (a) f(x)={14xe−x/2x&gt;00otherwise; (b) f(x)={c(1−x2)−1&lt;x&lt;10otherwise f(x)={5x2x&gt;50x≤5

Step 1To determine E(X) in each of the following cases.We know thatE(X)=∫−∞∞xf(x)dx(a)Given pdf:f(x)={14xe−x/2x&gt;00otherwiseE(x)=∫−∞∞xf(x)dx=∫−∞00dx+∫0∞xd14xe−x2dx=∫0∞x24e−x2dx=14[x2∫e−x2dx−∫(x2)′(∫e−x2dx)dx]0∞=14[x2e−x2−12−∫(2x)e−x2−12dx]0∞=14[−2x2e−x2+∫4xe−x2dx]0∞=14[−2x2e−x2+4xe−x2−12+8∫e−x2dx]0∞=14[−2x2e−x2−8xe−x2−16e−x2]0∞

Step 2 (b)Given pdf: f(x)={c(1−x2)−1&lt;x&lt;10otherwise E(x)=∫−∞∞xf(x)dx=∫−∞−10dx+∫−11xc(1−x2)dx+∫1∞0dx =∫−11xc(1−x2)dx =c∫−11(x−x3)dx =c[x22−x44]−11 =c[12−14−12−(−14)] =c(0) =0 Thus, E(x)=0

Step 3 (c)Given pdf: f(x)={5x2x&gt;50x≤5E(x)=∫−∞∞xf(x)dx=∫−∞50dx+∫5∞x5x2dx=∫5∞5xdx=[ln⁡x]5∞=ln⁡∞−ln⁡5=∞ Thus, E(x)=∞

Step 1To determine E(X) in each of the following cases.We know thatE(X)=∫−∞∞xf(x)dx(a)Given pdf:f(x)={14xe−x/2x&gt;00otherwiseE(x)=∫−∞∞xf(x)dx=∫−∞00dx+∫0∞xd14xe−x2dx=∫0∞x24e−x2dx=14[x2∫e−x2dx−∫(x2)′(∫e−x2dx)dx]0∞=14[x2e−x2−12−∫(2x)e−x2−12dx]0∞=14[−2x2e−x2+∫4xe−x2dx]0∞=14[−2x2e−x2+4xe−x2−12+8∫e−x2dx]0∞=14[−2x2e−x2−8xe−x2−16e−x2]0∞

Step 2 (b)Given pdf: f(x)={c(1−x2)−1&lt;x&lt;10otherwise E(x)=∫−∞∞xf(x)dx=∫−∞−10dx+∫−11xc(1−x2)dx+∫1∞0dx =∫−11xc(1−x2)dx =c∫−11(x−x3)dx =c[x22−x44]−11 =c[12−14−12−(−14)] =c(0) =0 Thus, E(x)=0

Step 3 (c)Given pdf: f(x)={5x2x&gt;50x≤5E(x)=∫−∞∞xf(x)dx=∫−∞50dx+∫5∞x5x2dx=∫5∞5xdx=[ln⁡x]5∞=ln⁡∞−ln⁡5=∞ Thus, E(x)=∞

Expert answers to "Compute [X] if X has a density function..."

High School
Other

fertilizeki

Answered question

2021-12-26

Compute $[X]$ if X has a density function given by

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

(b)
$f (x) = {\begin{cases} c (1 - x^{2}) & - 1 < x < 1 \\ 0 & o t h e r w i s e \end{cases}$
$f (x) = {\begin{cases} \frac{5}{x^{2}} & x > 5 \\ 0 & x \leq 5 \end{cases}$

Answer & Explanation

Pansdorfp6

Beginner2021-12-27Added 27 answers

Step 1
To determine

E (X)

in each of the following cases.
We know that

E (X) = \int_{- \infty}^{\infty} x f (x) d x

(a)Given pdf:

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

E (x) = \int_{- \infty}^{\infty} x f (x) d x = \int_{- \infty}^{0} 0 d x + \int_{0}^{\infty} x d \frac{1}{4} x e^{- \frac{x}{2}} d x

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

= \frac{1}{4} {[x^{2} \int e^{- \frac{x}{2}} d x - \int {(x^{2})}^{'} (\int e^{- \frac{x}{2}} d x) d x]}_{0}^{\infty}

= \frac{1}{4} {[x^{2} \frac{e^{- \frac{x}{2}}}{- \frac{1}{2}} - \int (2 x) \frac{e^{- \frac{x}{2}}}{- \frac{1}{2}} d x]}_{0}^{\infty}

= \frac{1}{4} {[- 2 x^{2} e^{- \frac{x}{2}} + \int 4 x e^{- \frac{x}{2}} d x]}_{0}^{\infty}

= \frac{1}{4} {[- 2 x^{2} e^{- \frac{x}{2}} + 4 x \frac{e^{- \frac{x}{2}}}{- \frac{1}{2}} + 8 \int e^{- \frac{x}{2}} d x]}_{0}^{\infty}

= \frac{1}{4} {[- 2 x^{2} e^{- \frac{x}{2}} - 8 x \frac{e^{- \frac{x}{2}}}{-} 16 e^{- \frac{x}{2}}]}_{0}^{\infty}

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Papilys3q

Beginner2021-12-28Added 34 answers

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

$E (x) = \int_{- \infty}^{\infty} x f (x) d x = \int_{- \infty}^{- 1} 0 d x + \int_{- 1}^{1} x c (1 - x^{2}) d x + \int_{1}^{\infty} 0 d x$
$= \int_{- 1}^{1} x c (1 - x^{2}) d x$
$= c \int_{- 1}^{1} (x - x^{3}) d x$
$= c {[\frac{x^{2}}{2} - \frac{x^{4}}{4}]}_{- 1}^{1}$
$= c [\frac{1}{2} - \frac{1}{4} - \frac{1}{2} - (- \frac{1}{4})]$
$= c (0)$
$= 0$
Thus,
$E (x) = 0$

user_27qwe

Skilled2021-12-30Added 375 answers

Step 3
(c)Given pdf:
$\begin{matrix} f (x) = {\begin{cases} \frac{5}{x^{2}} & x > 5 \\ 0 & x \leq 5 \end{cases} \\ E (x) = \int_{- \infty}^{\infty} x f (x) d x = \int_{- \infty}^{5} 0 d x + \int_{5}^{\infty} x \frac{5}{x^{2}} d x \\ = \int_{5}^{\infty} \frac{5}{x} d x \\ = [\ln x]_{5}^{\infty} \\ = \ln \infty - \ln 5 \\ = \infty \end{matrix}$
Thus,
$E (x) = \infty$

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