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Statistics / Mathematical Statistics with Applications 7 / Chapter 5 / Problem 26E

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

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In Exercise 5.8, we derived the fact that is a valid joint

Chapter 5, Problem 26E

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QUESTION:

In Exercise 5.8, we derived the fact that

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

4 y_{1} y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1, \\

0, & \text { elsewhere }

\end{array}\right.

is a valid joint probability density function. Find

a the marginal density functions for $Y_{1}$ and $Y_{2}$.

b $P\left(Y_{1} \leq 1 / 2 \mid Y_{2} \geq 3 / 4\right)$.

c the conditional density function of $Y_{1}$ given $Y_{2}=y_{2}$.

d the conditional density function of $Y_{2}$ given $Y_{1}=y_{1}$.

e $P\left(Y_{1} \leq 3 / 4 \mid Y_{2}=1 / 2\right)$.

Equation Transcription:

$f({y}_{1},{y}_{2})=$

$4{y}_{1}{y}_{2},\ 0\leq{}{y}_{1}\leq{}1,0\leq{}{y}_{2}\leq{}1,$

$0,\ elsewhere$

${Y}_{1}$

${Y}_{2}$

$P({Y}_{1}\leq{}1/2{|Y}_{2}\geq{}3/4)$

${Y}_{1}$

${Y}_{2}={Y}_{2}$

${Y}_{2}$

${Y}_{1}={y}_{1}$

$P({Y}_{1}\leq{}3/4|{Y}_{2}=1/2)$

Text Transcription:

f(y1,y2)=

4y1y2, 0y11,0y21,

0, elsewhere

P(Y11/2|Y23/4)

Y2=Y2

Y1=y1

P(Y13/4|Y2=1/2)

Questions & Answers

QUESTION:

In Exercise 5.8, we derived the fact that

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

4 y_{1} y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1, \\

0, & \text { elsewhere }

\end{array}\right.

is a valid joint probability density function. Find

a the marginal density functions for $Y_{1}$ and $Y_{2}$.

b $P\left(Y_{1} \leq 1 / 2 \mid Y_{2} \geq 3 / 4\right)$.

c the conditional density function of $Y_{1}$ given $Y_{2}=y_{2}$.

d the conditional density function of $Y_{2}$ given $Y_{1}=y_{1}$.

e $P\left(Y_{1} \leq 3 / 4 \mid Y_{2}=1 / 2\right)$.

Equation Transcription:

$f({y}_{1},{y}_{2})=$

$4{y}_{1}{y}_{2},\ 0\leq{}{y}_{1}\leq{}1,0\leq{}{y}_{2}\leq{}1,$

$0,\ elsewhere$

${Y}_{1}$

${Y}_{2}$

$P({Y}_{1}\leq{}1/2{|Y}_{2}\geq{}3/4)$

${Y}_{1}$

${Y}_{2}={Y}_{2}$

${Y}_{2}$

${Y}_{1}={y}_{1}$

$P({Y}_{1}\leq{}3/4|{Y}_{2}=1/2)$

Text Transcription:

f(y1,y2)=

4y1y2, 0y11,0y21,

0, elsewhere

P(Y11/2|Y23/4)

Y2=Y2

Y1=y1

P(Y13/4|Y2=1/2)

ANSWER:

Answer:

Step 1 of 5:

(a)

We have given the joint probability density function.

$f({y}_{1},\ {y}_{2})\ =\ \ 4{y}_{1}\ {y}_{2},\ \ \ \ \ \ \ \ \ \ \ \ \ \ 0\ {\leq{}\ y}_{1}\leq{}\ 1,\ 0{\ \leq{}y}_{2}\leq{}1\ \ and\ 0\ else\ where$

We need to find the marginal density functions for ${Y}_{1}$ and ${Y}_{2}$ .

Let ${Y}_{1}$ and ${Y}_{2}$ jointly continuous random variables with the joint (or bivariate) probability function $f({y}_{1},\ {y}_{2}).$ Then the marginal density functions of ${Y}_{1}$ and ${Y}_{2},$ respectively, are given by

${f}_{1}({y}_{1})\ =\int\limits_{-\infty{}}^{\infty{}}f({y}_{1},\ {y}_{2})d{y}_{2}\ \ and\$ ${f}_{2}({y}_{2})\ =\int\limits_{-\infty{}}^{\infty{}}f({y}_{1},\ {y}_{2})d{y}_{1}$