Question

Question:Suppose thatXandYhave a continuous joint distribution

for which the joint p.d.f. is defined as follows:

\({\bf{f}}\left( {{\bf{x,y}}} \right){\bf{ = }}\left\{ \begin{array}{l}\frac{{\bf{3}}}{{\bf{2}}}{{\bf{y}}^{\bf{2}}}\;{\bf{for}}\;{\bf{0}} \le {\bf{x}} \le {\bf{2}}\;{\bf{and}}\;{\bf{0}} \le {\bf{y}} \le {\bf{1}}\\{\bf{0}}\;\,{\bf{otherwise}}\end{array} \right.\)

a. Determine the marginal p.d.f.’s ofXandY.

b. AreXandYindependent?

c. Are the event{X<1}and the event\(\left\{ {{\bf{Y}} \ge \frac{{\bf{1}}}{{\bf{2}}}} \right\}\)independent?

Short answer

1. The marginal p.d.f of X is \({f_X}\left( x \right) = \left\{ \begin{array}{l}\frac{1}{2}\;for\;0 \le x \le 2\\0\;othewise\end{array} \right.\)

The marginal p.d.f of Y is \({f_Y}\left( y \right) = \left\{ \begin{array} {l}3{y^2}\;for\;0 \le y \le 1\\0\;otherwise\end{array} \right.\)

2. X and Y are independent.

c. The event {X<1} and the event \(\left\{ {Y \ge \frac{1}{2}} \right\}\) are independent

Step-by-step solution

Given information

Two random variables, X and Y, have a continuous joint distribution.

The joint probability density function is given by,

\(f\left( {x,y} \right) = \left\{ \begin{array}{l}\frac{3}{2}{y^2}\;for\;0 \le x \le 2\;and\;0 \le y \le 1\\0\;\,otherwise\end{array} \right.\)

(a) Determine the marginal probability density function

The marginal distribution of X is,

\(\begin{array}{c}{f_X}\left( x \right) = \int\limits_0^1 {\frac{3}{2}} {y^2}dy\\ = \frac{3}{2}\int\limits_0^1 {{y^2}} dy\\ = \left( {\frac{3}{2}} \right)\left. {\frac{{{y^3}}}{3}} \right|_0^1\end{array}\)

Thus, the marginal p.d.f of X is,\({f_X}\left( x \right) = \left\{ \begin{array}{l}\frac{1}{2}\;for\;0 \le x \le 2\\0\;othewise\end{array} \right.\)

Determine the marginal probability density function

The marginal probability density function of Y is,

\(\begin{array}{c}{f_Y}\left( y \right) = \int\limits_0^2 {\frac{3}{2}{y^2}dx} \\ = \frac{3}{2}{y^2}\int\limits_0^2 {dx} \\ = \frac{3}{2}{y^2}\left( {2 - 0} \right)\\ = 3{y^2}\end{array}\)

\(\begin{array}{c} = \frac{3}{2}\left( {\frac{1}{3} - 0} \right)\\ = \frac{1}{2}\end{array}\)

Thus, the marginal p.d.f of Y is,\({f_Y}\left( y \right) = \left\{ \begin{array}{l}3{y^2}\;for\;0 \le y \le 1\\0\;otherwise\end{array} \right.\)

(b) Check the independency

We can say that X and Y are independent if\({{\bf{f}}_{{\bf{X,Y}}}}\left( {{\bf{x,y}}} \right){\bf{ = }}{{\bf{f}}_{\bf{X}}}\left( {\bf{x}} \right){{\bf{f}}_{\bf{Y}}}\left( {\bf{y}} \right)\).

Now,

\(\begin{array}{c}{f_X}\left( x \right){f_Y}\left( y \right) = \frac{1}{2} \times 3{y^2}\\ = \frac{3}{2}{y^2}\\ = {f_{X,Y}}\left( {x,y} \right)\end{array}\)

Thus, we can conclude that X and Y are independent.

(c) Check the independency

Consider two events\(\left( {X < 1} \right)\;and\;\left( {Y \ge \frac{1}{2}} \right)\).

So,

\(\begin{array}{c}\Pr \left[ {\left( {X < 1} \right)\;and\;\left( {Y \ge \frac{1}{2}} \right)} \right] = \int\limits_0^1 {\int\limits_{\frac{1}{2}}^1 {f\left( {x,y} \right)dxdy} } \\ = \int\limits_0^1 {\int\limits_{\frac{1}{2}}^1 {{f_X}\left( x \right){f_Y}\left( y \right)dxdy} } \\ = \int\limits_0^1 {{f_X}\left( x \right)dx} \int\limits_{\frac{1}{2}}^1 {{f_Y}\left( y \right)dy} \\ = \Pr \left( {X < 1} \right)\Pr \left( {Y \ge \frac{1}{2}} \right)\end{array}\)

As we can see that the probability of the event\(\left( {X < 1} \right)\)depends only on the values of X,and for the event\(\left( {y \ge \frac{1}{2}} \right)\), it depends only on the values of Y, so we can conclude that the considered events are independent.