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Statistics / Mathematical Statistics with Applications 7 / Chapter 5 / Problem 77E

Mathematical Statistics with Applications | 7th Edition | ISBN: 9780495110811 | Authors: Dennis Wackerly; William Mendenhall; Richard L. Scheaffer

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In Exercise 5.9, we determined that ReferenceLet Y1 and Y2

Chapter 5, Problem 77E

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QUESTION:

In Exercise $5.9$ , we determined that

\(f\left(y_{1}, y_{2}\right)=\left\{\begin{array}{ll}

6\left(1-y_{2}\right), & 0 \leq y_{1} \leq y_{2} \leq 1, \\

0, & \text { elsewhere }

\end{array}\right.

\)

is a valid joint probability density function. Find
a \(E\left(Y_{1}\right)\) and \(E\left(Y_{2}\right)\).
b \(V\left(Y_{1}\right)\) and \(V\left(Y_{2}\right)\).
c \(E\left(Y_{1}-3 Y_{2}\right)\).

Equation Transcription:

$f({y}_{1},{y}_{2})={\{}_{0,\ \ \ \ \ elsewhere}^{6(1-{y}_{2}),\ 0\leq{}{y}_{1}\leq{}{y}_{2}\leq{}1,}$

$E({Y}_{1})$

$E({Y}_{2})$

$V({Y}_{1})$

$V({Y}_{2})$

$E({Y}_{1}-3{Y}_{2})$

Text Transcription:

f(y_1,y_2)={_0, elsewhere ^6(1-y_2), 0</=y_1</=y_2</=1,

E(Y_1)

E(Y_2)

V(Y_1)

V(Y_2)

E(Y_1-3Y_2)

Questions & Answers

QUESTION:

In Exercise $5.9$ , we determined that

\(f\left(y_{1}, y_{2}\right)=\left\{\begin{array}{ll}

6\left(1-y_{2}\right), & 0 \leq y_{1} \leq y_{2} \leq 1, \\

0, & \text { elsewhere }

\end{array}\right.

\)

is a valid joint probability density function. Find
a \(E\left(Y_{1}\right)\) and \(E\left(Y_{2}\right)\).
b \(V\left(Y_{1}\right)\) and \(V\left(Y_{2}\right)\).
c \(E\left(Y_{1}-3 Y_{2}\right)\).

Equation Transcription:

$f({y}_{1},{y}_{2})={\{}_{0,\ \ \ \ \ elsewhere}^{6(1-{y}_{2}),\ 0\leq{}{y}_{1}\leq{}{y}_{2}\leq{}1,}$

$E({Y}_{1})$

$E({Y}_{2})$

$V({Y}_{1})$

$V({Y}_{2})$

$E({Y}_{1}-3{Y}_{2})$

Text Transcription:

f(y_1,y_2)={_0, elsewhere ^6(1-y_2), 0</=y_1</=y_2</=1,

E(Y_1)

E(Y_2)

V(Y_1)

V(Y_2)

E(Y_1-3Y_2)

ANSWER:

Solution:

Step 1 of 3:

The given joint probability density function is

a). To find $E({Y}_{1})\ and\ E({Y}_{2}).$

Hence the marginal density functions for ${Y}_{1}$ is,

${f}_{1}({y}_{1})\ =\int\limits_{{y}_{1}}^{1}6(1-{y}_{2})d{y}_{2}\ =3(1\ -{{y}_{1}{)}^{2},\ }_{\ \ \ \ \ }0\ {\leq{}\ y}_{1}\leq{}\ 1$

$\$ ${f}_{1}({y}_{1})\ =\ 3({{y}_{1}\ -1{)}^{2},\ }_{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }0\ {\leq{}\ y}_{1}\leq{}\ 1$

The marginal density functions for ${Y}_{2}$ is,

${f}_{2}({y}_{2})\ =\int\limits_{0}^{{y}_{2}}6(1-{y}_{2})d{y}_{1}$

$\ =\int\limits_{0}^{{y}_{2}}6(1-{y}_{2})d{y}_{1}\ =6{y}_{2}(1-{y}_{1}),$

${f}_{2}({y}_{2})\ =\ 6{y}_{2}(1-{y}_{2}),\ \ \ \ 0\ {\leq{}\ y}_{2}\leq{}\ 1$

${f}_{1}({y}_{1})\ {f}_{2}({y}_{2})\ne{}f({y}_{1},{y}_{2}).$

Hence they are not independent.

$E({Y}_{1})=\int\limits_{-\infty{}}^{\infty{}}x\ f(x)\ dx$

$=\int\limits_{0}^{1}x\ 3({x\ -1{)}^{2}dx\ }_{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$

$=\int\limits_{0}^{1}3{x}^{3}+3x-6{x}^{2}dx$

$=3(\frac{{x}^{4}}{4}){\ }_{0}^{1}+3(\frac{{x}^{2}}{2}){\ }_{0}^{1}-6(\frac{{x}^{3}}{3}){\ }_{0}^{1}$

$=\frac{1}{4}$

$E({Y}_{2})=\int\limits_{-\infty{}}^{\infty{}}x\ f(x)\ dx$

$=\int\limits_{0}^{1}x\ \ 6x\ (1-x)\ dx$

$=\int\limits_{0}^{1}6{x}^{2}-6{x}^{3}dx$

$=6(\frac{{x}^{3}}{3}){\ }_{0}^{1}-6(\frac{{x}^{4}}{4}){\ }_{0}^{1}$

$=\frac{1}{2}$

Therefore, $E({Y}_{1})=\frac{1}{4}$ , $\ E({Y}_{2})=\frac{1}{2}$ .

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