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Suppose that three random variables \({{\bf{X}}_{\bf{1}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}\;{\bf{and}}\;{{\bf{X}}_{\bf{3}}}\)have a continuous joint distribution for which the joint p.d.f. is as follows:

\(f\left( {{x_1},{x_2},{x_3}} \right) = \left\{ \begin{array}{l}8{x_1}{x_2}{x_3}\;for\;0 < {x_i} < 1\left( {i = 1,2,3} \right),\\0,otherwise\end{array} \right.\)

Suppose also that \({{\bf{Y}}_{\bf{1}}}{\bf{ = }}{{\bf{X}}_{\bf{1}}}{\bf{,}}{{\bf{Y}}_{\bf{2}}}{\bf{ = }}{{\bf{X}}_{\bf{1}}}{{\bf{X}}_{\bf{2}}}\;{\bf{and}}\;{{\bf{Y}}_{\bf{3}}}{\bf{ = }}{{\bf{X}}_{\bf{1}}}{{\bf{X}}_{\bf{2}}}{{\bf{X}}_{\bf{3}}}{\bf{.}}\)Find the joint p.d.f. of\({{\bf{Y}}_{\bf{1}}}{\bf{,}}{{\bf{Y}}_{\bf{2}}}\;{\bf{and}}\;{{\bf{Y}}_{\bf{3}}}\).

Short Answer

Expert verified

The joint pdf is:

\(g\left( {{y_1},{y_2},{y_3}} \right) = \left\{ \begin{array}{l}8{y_3}{\left( {{y_1}{y_2}} \right)^{ - 1}}\;\;\;for\;0 < {y_3} < {y_2} < {y_1} < 1\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

Step by step solution

01

Given information

We are given the following p.d.f.

\(f\left( {{x_1},{x_2},{x_3}} \right) = \left\{ \begin{array}{l}8{x_1}{x_2}{x_3}\;for\;0 < {x_i} < 1\left( {i = 1,2,3} \right),\\0,otherwise\end{array} \right.\)

The random variables are:

\(\begin{array}{l}{Y_1} = {X_1}\\{Y_2} = {X_1}{X_2}\;\\{Y_3} = {X_1}{X_2}{X_3}.\end{array}\)

02

 Find the inverse of the random variables.

The inverse transformations of the variables are:

\(\begin{aligned}{X_1} &= {Y_1}\\{X_2} &= \frac{{{Y_2}}}{{{X_1}}}\\ &= \frac{{{Y_2}}}{{{Y_1}}}\end{aligned}\)

\(\begin{aligned}{X_3} &= \frac{{{Y_3}}}{{{X_1}{X_2}}}\\ &= \frac{{{Y_3}}}{{{Y_2}}}.\end{aligned}\)

Hence, the transformed variables are:

\({X_1} = {Y_1},\)

\({X_2} = \frac{{{Y_2}}}{{{Y_1}}},\)

\({X_3} = \frac{{{Y_3}}}{{{Y_2}}}\)

03

State the formula for joint pdf with new variables

The joint pdf of \({{\bf{Y}}_{\bf{1}}}{\bf{,}}{{\bf{Y}}_{\bf{2}}}{\bf{,}}{{\bf{Y}}_{\bf{3}}}\) is given as follows,\({\bf{g}}\left( {{{\bf{y}}_{\bf{1}}}{\bf{,}}{{\bf{y}}_{\bf{2}}}{\bf{,}}{{\bf{y}}_{\bf{3}}}} \right){\bf{ = }}\left\{ \begin{array}{l}f\left( {{{\bf{y}}_{\bf{1}}}{\bf{,}}{{\bf{y}}_{\bf{2}}}{\bf{,}}{{\bf{y}}_{\bf{3}}}} \right)\left| {\bf{J}} \right|\;\;\;\;\;{\bf{for}}\;{\bf{0 < }}{{\bf{y}}_{\bf{3}}}{\bf{ < }}{{\bf{y}}_{\bf{2}}}{\bf{ < }}{{\bf{y}}_{\bf{1}}}{\bf{ < 1}}\\{\bf{0}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\bf{otherwise}}\end{array} \right.\)

The jacobian is determined as follows,

\(\begin{aligned}J &= \left| {\begin{array}{*{20}{c}}{\frac{{\partial {x_1}}}{{\partial {y_1}}}}&{\frac{{\partial {x_2}}}{{\partial {y_1}}}}&{\frac{{\partial {x_3}}}{{\partial {y_1}}}}\\{\frac{{\partial {x_1}}}{{\partial {y_2}}}}&{\frac{{\partial {x_2}}}{{\partial {y_2}}}}&{\frac{{\partial {x_3}}}{{\partial {y_2}}}}\\{\frac{{\partial {x_1}}}{{\partial {y_3}}}}&{\frac{{\partial {x_2}}}{{\partial {y_3}}}}&{\frac{{\partial {x_3}}}{{\partial {y_3}}}}\end{array}} \right|\\ &= \left| {\begin{array}{*{20}{c}}1&{\frac{{ - {y_2}}}{{{y_1}^2}}}&0\\0&{\frac{1}{{{y_1}}}}&{\frac{{ - {y_3}}}{{{y_2}^2}}}\\0&0&{\frac{1}{{{y_2}}}}\end{array}} \right|\\ &= \frac{1}{{{y_1}{y_2}}}\;\end{aligned}\)

Now, the Jacobian of the transformed variable:\(\frac{{\bf{1}}}{{{{\bf{y}}_{\bf{1}}}{{\bf{y}}_{\bf{2}}}}}\)

04

Substitute the values to obtain the pdf

The joint pdf is obtained using the formula as stated below,

\(g\left( {{y_1},{y_2},{y_3}} \right) = f\left( {{x_1},{x_2},{x_3}} \right)\left| J \right|\)

\(\begin{aligned}g\left( {{y_1},{y_2},{y_3}} \right) &= 8\left( {{y_1}} \right)\left( {\frac{{{y_2}}}{{{y_1}}}} \right)\left( {\frac{{{y_3}}}{{{y_2}}}} \right)\left( {\frac{1}{{{y_1}{y_2}}}} \right) \Rightarrow g\left( {{y_1},{y_2},{y_3}} \right)\\ &= \left\{ \begin{array}{l}8{y_3}{\left( {{y_1}{y_2}} \right)^{ - 1}}\;for\;0 < {y_3} < {y_2} < {y_1} < 1\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\end{aligned}\)

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Most popular questions from this chapter

Question:For the joint pdf in example 3.4.7,determine whether or not X and Y are independent.

Suppose that two balanced dice are rolled, and letXdenote the absolute value of the difference between thetwo numbers that appear. Determine and sketch the p.f.ofX.

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

\(\left( {\bf{c}} \right)\)\({\bf{Pr}}\left( {{{\bf{X}}_{\bf{3}}}{\bf{ < }}\frac{{\bf{1}}}{{\bf{2}}}\left| {{{\bf{X}}_{\bf{1}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}{\bf{,}}{{\bf{X}}_{\bf{2}}}{\bf{ = }}\frac{{\bf{3}}}{{\bf{4}}}} \right.} \right){\bf{.}}\)

In a certain city, three newspapersA,B, andC,are published. Suppose that 60 percent of the families in the city subscribe to newspaperA, 40 percent of the families subscribe to newspaperB, and 30 percent subscribe to newspaperC. Suppose also that 20 percent of the families subscribe to bothAandB, 10 percent subscribe to bothAandC, 20 percent subscribe to bothBandC, and 5 percent subscribe to all three newspapersA,B, andC. Consider the conditions of Exercise 2 of Sec. 1.10 again. If a family selected at random from the city subscribes to exactly one of the three newspapers,A,B, andC, what is the probability that it isA?

Suppose that the p.d.f. of X is as given in Exercise 3. Determine the p.d.f. of\(Y = 4 - {X^3}\)

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