Q 1E Suppose that three random variab... [FREE SOLUTION]

91影视

Probability and Statistics

Mark J Schervish, Morris H DeGroot

$Math Studyset 91影视 Explanations$ Math

4th Edition 路 850 pages

ISBN: 9780321500465

Chapter 3: Q 1E (page 166)

Suppose that three random variables X1, X2, and X3 have a continuous joint distribution with the following joint p.d.f.:
${\bf{f}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{,}}{{\bf{x}}_{\bf{2}}}{\bf{,}}{{\bf{x}}_{\bf{3}}}} \right){\bf{ = }}\left\{ {\begin{align}{}{{\bf{c}}\left( {{{\bf{x}}_{\bf{1}}}{\bf{ + 2}}{{\bf{x}}_{\bf{2}}}{\bf{ + 3}}{{\bf{x}}_{\bf{3}}}} \right)}&{{\bf{for0}} \le {{\bf{x}}_{\bf{i}}} \le {\bf{1}}\,\,\left( {{\bf{i = 1,2,3}}} \right)}\\{\bf{0}}&{{\bf{otherwise}}{\bf{.}}}\end{align}} \right.$
Determine$\left( {\bf{a}} \right)$the value of the constant c;
$\left( {\bf{b}} \right)$the marginal joint p.d.f. of${{\bf{X}}_{\bf{1}}}$and${{\bf{X}}_{\bf{3}}}$; and
$\left( {\bf{c}} \right)$${\bf{Pr}}\left( {{{\bf{X}}_{\bf{3}}}{\bf{ < }}\frac{{\bf{1}}}{{\bf{2}}}\left| {{{\bf{X}}_{\bf{1}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{ = }}\frac{{\bf{3}}}{{\bf{4}}}} \right.} \right){\bf{.}}$

Short Answer

Expert verified

$\left( a \right)$The value of$c = \frac{1}{3}$

$\left( b \right)$The marginal joint p.d.f. of${X_1}$and${X_3}$is,$f\left( {{x_1},{x_3}} \right)$$ = \frac{1}{3}\left( {{x_1} + 1 + 3{x_3}} \right)$

$\left( c \right)$ The probability is, $\Pr \left( {{X_3} < \frac{1}{2}\left| {{X_1} = \frac{1}{4},{X_2} = \frac{3}{4}} \right.} \right) = 0.384615$

Step by step solution

Given information

There are three random variables ${X_1},{X_2},{X_3}$

The joint p.d.f. of${X_1},{X_2},{X_3}$is given by,

$f\left( {{x_1},{x_2},{x_3}} \right) = \left\{ {\begin{align}{}{c\left( {{x_1} + 2{x_2} + 3{x_3}} \right)}&{for0 \le {x_i} \le 1\,\left( {i = 1,2,3} \right)}\\0&{otherwise.}\end{align}} \right.$

(a) Probability calculations

For finding the value of c,

$\begin{align}\int\limits_0^1 {\int\limits_0^1 {\int\limits_0^1 {f\left( {{x_1},{x_2},{x_3}} \right)d{x_1}d{x_2}d{x_3} = 1} } } \\ \Rightarrow \int\limits_0^1 {\int\limits_0^1 {\int\limits_0^1 {c\left( {{x_1} + 2{x_2} + 3{x_3}} \right)d{x_1}d{x_2}d{x_3} = 1} } } \\ \Rightarrow c\int\limits_0^1 {\int\limits_0^1 {\left( {\frac{{{x_1}^2}}{2} + 2{x_2}{x_1} + 3{x_3}{x_1}} \right)} } _0^1d{x_2}d{x_3} = 1\\ \Rightarrow c\int\limits_0^1 {\int\limits_0^1 {\left( {\frac{1}{2} + 2{x_2} + 3{x_3}} \right)} } d{x_2}d{x_3} = 1\end{align}$

$\begin{align} \Rightarrow c\int\limits_0^1 {\left( {\frac{1}{2}{x_2} + 2\frac{{{x_2}^2}}{2} + 3{x_3}{x_2}} \right)_0^1d{x_3} = 1} \\ \Rightarrow c\int\limits_0^1 {\left( {\frac{1}{2} + 1 + 3{x_3}} \right)d{x_3} = 1} \\ \Rightarrow c\int\limits_0^1 {\left( {\frac{3}{2} + 3{x_3}} \right)d{x_3} = 1} \\ \Rightarrow 3c\int\limits_0^1 {\left( {\frac{1}{2} + {x_3}} \right)d{x_3} = 1} \end{align}$

$\begin{align} \Rightarrow 3c\left( {\frac{{{x_3}}}{2} + \frac{{{x_3}^2}}{2}} \right)_0^1 &= 1\\ \Rightarrow \frac{3}{2}c\left( {1 + 1} \right) &= 1\\ \Rightarrow \frac{3}{2} \times 2 \times c &= 1\\ \Rightarrow c &= \frac{1}{3}\end{align}$

(b) The joint p.d.f. ${{\bf{X}}_{\bf{1}}}$and ${{\bf{X}}_{\bf{2}}}$ is given by,

$\begin{align}f\left( {{x_1},{x_3}} \right) &= \int\limits_0^1 {f\left( {{x_1},{x_2},{x_3}} \right)d{x_2}} \\ &= \int\limits_0^1 {\frac{1}{3}} \left( {{x_1} + 2{x_2} + 3{x_3}} \right)d{x_2}\\ &= \frac{1}{3}\left( {{x_1}{x_2} + \frac{{2{x_2}^2}}{2} + 3{x_3}{x_2}} \right)_0^1\\ &= \frac{1}{3}\left( {{x_1} + 1 + 3{x_3}} \right)\end{align}$

Therefore, the joint p.d.f. ${X_1}$ and ${X_3}$ is $f\left( {{x_1},{x_3}} \right)$$ = \frac{1}{3}\left( {{x_1} + 1 + 3{x_3}} \right)$

(c) First, we have to calculate the joint p.d.f. ${{\bf{X}}_{\bf{1}}}$and${{\bf{X}}_{\bf{2}}}$,

$\begin{align}f\left( {{x_1},{x_2}} \right) &= \int\limits_0^1 {f\left( {{x_1},{x_2},{x_3}} \right)d{x_3}} \\ &= \int\limits_0^1 {\frac{1}{3}} \left( {{x_1} + 2{x_2} + 3{x_3}} \right)d{x_3}\\ &= \frac{1}{3}\left( {{x_1}{x_3} + 2{x_2}{x_3} + \frac{{3{x_3}^2}}{2}} \right)_0^1\\ &= \frac{1}{3}\left( {{x_1} + 2{x_2} + \frac{3}{2}} \right)\end{align}$

$\begin{align}f\left( {{x_1} = \frac{1}{4},{x_2} = \frac{3}{4}} \right) &= \frac{1}{3}\left( {\frac{1}{4} + \frac{{2 \times 3}}{4} + \frac{3}{2}} \right)\\ &= \frac{1}{3}\left( {\frac{1}{4} + \frac{3}{2} + \frac{3}{2}} \right)\\ &= \frac{{13}}{{12}}\end{align}$

\(\begin{align}\Pr \left( {{X_3} < \frac{1}{2}\left| {{X_1} = \frac{1}{4},{X_2} = \frac{3}{4}} \right.} \right) &= \frac{{\int\limits_0^{\frac{1}{2}} {f\left( {{x_1} = \frac{1}{4},{x_2} = \frac{3}{4},{x_3}} \right)d{x_3}} }}{{f\left( {{x_1} = \frac{1}{4},{x_2} = \frac{3}{4}} \right)}}\\ &= \frac{{\int\limits_0^{\frac{1}{2}} {\frac{1}{3}\left( {\frac{1}{4} + \frac{{2 \times 3}}{4} + 3{x_3}} \right)d{x_3}} }}{{\frac{{13}}{{12}}}}\\ &= \frac{{\int\limits_0^{\frac{1}{2}} {\frac{1}{3}\left( {\frac{7}{4} + 3{x_3}} \right)d{x_3}} }}{{\frac{{13}}{{12}}}}\\ &= \frac{{\frac{1}{3}\left( {\left( {\frac{7}{4}{x_3} + \frac{{3{x_3}^2}}{2}} \right)} \right)_0^{\frac{1}{2}}}}{{\frac{{13}}{{12}}}}\\ &= \frac{{\frac{1}{3}\left( {\left( {\frac{{\left( {7 \times 1} \right)}}{{\left( {4 \times 2} \right)}} + \frac{{3{{\left( {\frac{1}{2}} \right)}^2}}}{2}} \right)} \right)}}{{\frac{{13}}{{12}}}}\\ &= \frac{{\frac{1}{3}\left( {\left( {\frac{7}{8} + \frac{3}{8}} \right)} \right)}}{{\frac{{13}}{{12}}}}\\ &= \frac{{\left( {\frac{1}{3} \times \frac{{10}}{8}} \right)}}{{\frac{{13}}{{12}}}}\\ &= \frac{{\frac{5}{{12}}}}{{\frac{{13}}{{12}}}}\\ &= \frac{5}{{13}}\\ &= 0.384615\end{align}\)

Therefore, the probability is, $\Pr \left( {{X_3} < \frac{1}{2}\left| {{X_1} = \frac{1}{4},{X_2} = \frac{3}{4}} \right.} \right) = 0.384615$