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Question:Suppose that two random variables X and Y have the joint p.d.f.\(f\left( {x,y} \right) = \left\{ \begin{array}{l}k{x^2}{y^2}\,\,\,\,\,\,\,\,\,\,\,\,for\,{x^2} + {y^2} \le 1\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise\end{array} \right.\). Compute the conditional p.d.f. of X given

Y = y for each y.

Short Answer

Expert verified

Conditional pdf of x given y=y for each y is

\(\left\{ \begin{array}{l}1.5{x^2}{\left( {1 - {y^2}} \right)^{^{ - \frac{3}{2}}\,}}\,\,\,\,{\rm{for}}\, - {\left( {1 - {y^2}} \right)^{\frac{1}{2}}} < x < {\left( {1 - {y^2}} \right)^{\frac{1}{2}}}\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{otherwise}}\end{array} \right.\,\,\)

Step by step solution

01

Given Information.

For the random variables x and y

\(f\left( {x,y} \right) = \left\{ \begin{array}{l}k{x^2}{y^2}\,\,\,\,\,\,\,\,\,\,\,\,for\,{x^2} + {y^2} \le 1\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise\end{array} \right.\)

02

Computing the Marginal probability

The marginal pdf\({f_1}\)and\({f_2}\)of x and y is

\({f_1}\left( x \right) = \left\{ \begin{array}{l}{e^{ - x}}\,\,\,\,for\,x \ge 0\\0\,\,\,\,\,\,\,\,\,otherwise\end{array} \right.\)and\({f_2}\left( y \right) = \left\{ \begin{array}{l}2{e^{ - 2y}}\,\,\,\,for\,y \ge 0\\0\,\,\,\,\,\,\,\,\,otherwise\end{array} \right.\)

When we multiply \({f_1}\left( x \right)\)times \({f_2}\left( y \right)\)and compare the product to \(f\left( {x,y} \right)\)we get \(k = 2\)

03

Computing the conditional probability

Now to compute the conditional pdf of\({f_{x|y}}\left( {x|y} \right)\)

\(\begin{array}{l}{f_{X|Y}}\left( {x|y} \right) = \frac{{{f_{X|Y}}\left( {x|y} \right)}}{{{f_Y}\left( y \right)}}\\ = \left\{ \begin{array}{l}1.5{x^2}{\left( {1 - {y^2}} \right)^{^{ - \frac{3}{2}}\,}}\,\,\,\,for\, - {\left( {1 - {y^2}} \right)^{\frac{1}{2}}} < x < {\left( {1 - {y^2}} \right)^{\frac{1}{2}}}\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise\end{array} \right.\,\,\end{array}\)

Hence,

\({f_{X|Y}}\left( {x|y} \right) = = \left\{ \begin{array}{l}1.5{x^2}{\left( {1 - {y^2}} \right)^{^{ - \frac{3}{2}}\,}}\,\,\,\,for\, - {\left( {1 - {y^2}} \right)^{\frac{1}{2}}} < x < {\left( {1 - {y^2}} \right)^{\frac{1}{2}}}\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise\end{array} \right.\) is the Conditional pdf of x given y=y for each y.

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Most popular questions from this chapter

Suppose that a random variableXhas a discrete distribution

with the following p.f.:

\(f\left( x \right) = \left\{ \begin{array}{l}cx\;\;for\;x = 1,...,5,\\0\;\;\;\;otherwise\end{array} \right.\)

Determine the value of the constantc.

There are two boxes A and B, each containing red and green balls. Suppose that box A contains one red ball and two green balls and box B contains eight red balls and two green balls. Consider the following process: One ball is selected at random from box A, and one ball is selected at random from box B. The ball selected from box A is then placed in box B and the ball selected from box B is placed in box A. These operations are then repeated indefinitely. Show that the numbers of red balls in box A form a Markov chain with stationary transition probabilities, and construct the transition matrix of the Markov chain.

Question:In example 3.5.10 verify that X and Y have the same Marginal pdf and that

\({f_1}\left( x \right) = \left\{ \begin{array}{l}2k{x^2}\frac{{{{\left( {1 - {x^2}} \right)}^{\frac{2}{3}}}}}{3} for - 1 \le x \le 1\\0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,otherwise\end{array} \right.\) .

Question:Prove Theorem 3.5.6.

Let X and Y have a continuous joint distribution. Suppose that

\(\;\left\{ {\left( {x,y} \right):f\left( {x,y} \right) > 0} \right\}\)is a rectangular region R (possibly unbounded) with sides (if any) parallel to the coordinate axes. Then X and Y are independent if and only if Eq. (3.5.7) holds for all\(\left( {x,y} \right) \in R\)

Verify the rows of the transition matrix in Example 3.10.6 that correspond to current states\(\left\{ {AA,Aa} \right\}\)and\(\left\{ {Aa,aa} \right\}\)

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