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

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

Short Answer

Expert verified

.X and Y have the same marginal pdf

Step by step solution

01

Given Information

The joint pdf of X and Y is

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

02

Computing the probability

As each point inside the circle can\({x^2} + {y^2} \le 1,\,\,f\left( {x,y} \right)\)can be factored.

This factorisation cannot be satisfied at every point outside this circle.

This shows X and Y are not independent.

As the joint pdf of X and Y is

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

The Marginal pdf of\(X,f\left( x \right)\)

\(\begin{array}{c}X,f\left( x \right) = \int\limits_{ - 1}^1 {K{x^2}{y^2}dy} \\ = 2k{x^2}\frac{{{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}}}{3}\end{array}\)

Therefore,

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

Similarly the marginal density of\(Y,f\left( y \right)\)

\(\begin{array}{c}Y,f\left( y \right) = \int\limits_{ - 1}^1 {k{x^2}{y^2}dx} \\ = 2k{y^2}\frac{{{{\left( {1 - {y^2}} \right)}^{\frac{3}{2}}}}}{3}\end{array}\)

Therefore,

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

Hence we can say that X and Y are not independent and have the same marginal pdf’s

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Most popular questions from this chapter

Question: Suppose that the joint p.d.f. of two random variablesXandYis as follows:

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

Determine (a) the value of the constantc;

\(\begin{array}{l}\left( {\bf{b}} \right)\,{\bf{Pr}}\left( {{\bf{0}} \le {\bf{X}} \le {\bf{1/2}}} \right){\bf{;}}\,\left( {\bf{c}} \right)\,{\bf{Pr}}\left( {{\bf{Y}} \le {\bf{X + 1}}} \right)\\\left( {\bf{d}} \right)\,{\bf{Pr}}\left( {{\bf{Y = }}{{\bf{X}}^{\bf{2}}}} \right)\end{array}\)

The unique stationary distribution in Exercise 9 is \({\bf{v = }}\left( {{\bf{0,1,0,0}}} \right)\). This is an instance of the following general result: Suppose that a Markov chain has exactly one absorbing state. Suppose further that, for each non-absorbing state \({\bf{k}}\), there is \({\bf{n}}\) such that the probability is positive of moving from state \({\bf{k}}\) to the absorbing state in \({\bf{n}}\) steps. Then the unique stationary distribution has probability 1 in the absorbing state. Prove this result.

A civil engineer is studying a left-turn lane that is long enough to hold seven cars. LetXbe the number of cars in the lane at the end of a randomly chosen red light. The engineer believes that the probability thatX=xis proportional to(x+1)(8−x)forx=0, . . . ,7 (the possible values ofX).

a. Find the p.f. ofX.

b. Find the probability thatXwill be at least 5.

Suppose that a random variable X has a uniform distribution on the interval [0, 1]. Determine the p.d.f. of (a)\({{\bf{X}}^{\bf{2}}}\), (b) \({\bf{ - }}{{\bf{X}}^{\bf{3}}}\), and (c) \({{\bf{X}}^{\frac{{\bf{1}}}{{\bf{2}}}}}\).

Suppose that a fair coin is tossed 10 times independently.

Determine the p.f. of the number of heads that will be obtained.

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