91Ó°ÊÓ

The pdf of random variables X and Y is given by,

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

a.

The properties of joint pdf of X and Y is given by,

\(\int_y {\int_x {f\left( {x,y} \right)dxdy = 1} } \)

\(\begin{array}{c}\int_x {\int_y {c\left( {{x^2} + y} \right)dxdy} } = \int_{ - 1}^1 {\int_0^{1 - {x^2}} {c\left( {{x^2} + y} \right)dydx} } \\ = c\int_{ - 1}^1 {\int_0^{1 - {x^2}} {\left( {{x^2} + y} \right)dydx} } \\ = c\int_{ - 1}^1 {\left( {y{x^2} + \frac{{{y^2}}}{2}} \right)_0^{1 - {x^2}}dx} \\ = c\int_{ - 1}^1 {\left( {{x^2} - {x^4} + \frac{1}{2} - {x^2} + \frac{{{x^4}}}{2}} \right)dx} \\ = c\int_{ - 1}^1 {\left( { - \frac{{{x^4}}}{2} + \frac{1}{2}} \right)dx} \\ = c\left( { - \frac{{{x^5}}}{{10}} + \frac{x}{2}} \right)_{ - 1}^1\\ = c\left( {\left( { - \frac{1}{{10}} + \frac{1}{2}} \right) - \left( {\frac{1}{{10}} - \frac{1}{2}} \right)} \right)\\ = \frac{4}{5}c\end{array}\)

\(\)\(\begin{array}{l}\int_y {\int_x {f\left( {x,y} \right)dxdy = 1} } \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{4}{5}c = 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c = \frac{5}{4}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c = 1.25\end{array}\)

The value of the constant is 1.25

b.

\(\)\(\begin{array}{c}\Pr \left( {0 \le X \le 1/2} \right) = \int_0^{1/2} {\int_0^{1 - {x^2}} {1.25\left( {{x^2} + y} \right)dydx} } \\ = 1.25\int_0^{1/2} {\int_0^{1 - {x^2}} {\left( {{x^2} + y} \right)dydx} } \\ = 1.25\int_0^{1/2} {\left( {y{x^2} + \frac{{{y^2}}}{2}} \right)_0^{1 - {x^2}}dx} \\ = 1.25\int_0^{1/2} {\left( {{x^2} - {x^4} + \frac{1}{2} - {x^2} + \frac{{{x^4}}}{2}} \right)dx} \\ = 1.25\int_{ - 1}^1 {\left( { - \frac{{{x^4}}}{2} + \frac{1}{2}} \right)dx} \\ = 1.25\left( { - \frac{{{x^5}}}{{10}} + \frac{x}{2}} \right)_0^{1/2}\\ = 1.25\left( {\left( { - \frac{1}{{320}} + \frac{1}{4}} \right) - 0} \right)\\ = 0.308594\end{array}\)

The probability is 0.308594

c.

\(\begin{array}{c}\Pr \left( {Y \le X + 1} \right) = 1 - \Pr \left( {Y > X + 1} \right)\\ = 1 - \int_{ - 1}^0 {\int_{x + 1}^{1 - {x^2}} {1.25\left( {{x^2} + y} \right)dydx} } \\ = 1 - 1.25\int_{ - 1}^0 {\int_{x + 1}^{1 - {x^2}} {\left( {{x^2} + y} \right)dydx} } \\ = 1 - 1.25\int_{ - 1}^0 {\left( {y{x^2} + \frac{{{y^2}}}{2}} \right)_{x + 1}^{1 - {x^2}}dx} \\ = 1 - 1.25\int_{ - 1}^0 {\left( {{x^2} - {x^4} + \frac{1}{2} - {x^2} + \frac{{{x^4}}}{2} - {x^3} - {x^2} - \frac{{{x^2}}}{2} - x - \frac{1}{2}} \right)dx} \\ = 1 - 1.25\int_{ - 1}^0 {\left( { - \frac{{{x^4}}}{2} - {x^3} - \frac{{3{x^2}}}{2} - x} \right)dx} \\ = 1 - 1.25\left( { - \frac{{{x^5}}}{{10}} - \frac{{{x^4}}}{4} - \frac{{{x^3}}}{2} - \frac{{{x^2}}}{2}} \right)_{ - 1}^0\\ = 1 - 1.25\left( {0 - \left( {\frac{1}{{10}} - \frac{1}{4} + \frac{1}{2} - \frac{1}{2}} \right)} \right)\\ = 1 - 0.1875\\ = 0.8125\end{array}\)

The probability is 0.8125

d.

The probability that (X,Y) will lie in the curve \(y = {x^2}\) is 0 for every continuous joint distribution.

\(\Pr \left( {Y = {X^2}} \right) = 0\)

The probability is 0