Question

Suppose that the joint p.d.f. of two random variables X and Y is as follows:

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

Determine

(a) the conditional p.d.f. of X for every given value of Y, and

(b) \({\rm P}\left( {X > \frac{1}{2}|Y = \frac{3}{2}} \right)\).

Short answer

1. The conditional pdf of x for every value of y is \({g_1}\left( {x|y} \right) = \left\{ {\frac{{x + {y^2}}}{{\frac{1}{2} + {y^2}}}\,for\,\;} \right.0 \le x \le 1\,and\,0 \le y \le 1\)
2. \(\frac{1}{3}\)

Step-by-step solution

Given information

The joint pdf of two random variables X and Y:

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

Calculating Conditional pdf

a)

Marginal pdf of Y For \(0 \le y \le 1\):

\(\begin{aligned}{f_2}\left( y \right) = \int\limits_0^1 {f\left( {x,y} \right)dx} \\ = c\left( {\frac{1}{2} + {y^2}} \right).\end{aligned}\)

Therefore, for\(0 \le x \le 1\)and\(0 \le y \le 1\)the conditional pdf of X given that Y=y:

\(\begin{aligned}{g_1}\left( {x|y} \right) = \frac{{f\left( {x,y} \right)}}{{{f_2}\left( y \right)}}\\ = \frac{{x + {y^2}}}{{\frac{1}{2} + {y^2}}}\end{aligned}\) .

Hence,

\({g_1}\left( {x|y} \right) = \left\{ {\frac{{x + {y^2}}}{{\frac{1}{2} + {y^2}}}\,for\,\;} \right.0 \le x \le 1\,and\,0 \le y \le 1\)

When: \({\rm P}\left( {X > \frac{1}{2}|Y = \frac{3}{2}} \right)\)

b)

When\(Y = {\raise0.7ex\hbox{$1$} \!\mathord{\left/

{\vphantom {1 2}}\right.\kern-\nulldelimiterspace}

\!\lower0.7ex\hbox{$2$}}\)it follows part a, which means

\({g_1}\left( {x|y = \frac{1}{2}} \right) = \left\{ \begin{aligned}{l}\frac{4}{3}\left( {x + \frac{1}{4}} \right)\,\,\,\,\,for\,0 \le x \le 1\\0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,othrwise\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{aligned} \right.\,\,\,\,\)

Therefore,

\(\begin{aligned}{\rm P}\left( {X < \frac{1}{2}|Y = \frac{1}{2}} \right) = \int\limits_0^{\frac{1}{2}} {{g_1}\left( {x|y = \frac{1}{2}} \right)} dx\\ = \frac{1}{3}\end{aligned}\)