Question

Question:For the joint pdf in example 3.4.7,determine whether or not X and Y are independent.

Short answer

X and Y are not independent.

Step-by-step solution

Given information.

Given

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

Calculating value of C

Firstly, we will find the value of c

\(\begin{array}{c}\int\limits_{ - \infty }^\infty {\int\limits_{ - \infty }^\infty {f\left( {x,y} \right)dxdy = \int\limits_s {\int {f\left( {x,y} \right)} dxdy} } } \\ = \int\limits_{ - 1}^1 {\int\limits_{{x^2}}^1 {c{x^2}ydydx} } \\ = \frac{4}{{21}}c\\c = \frac{{21}}{4}\end{array}\)

Step 3:  Compute the probability

The marginal density of\(X,f\left( x \right)\)is

\(\begin{array}{c}X,f\left( x \right) = \int\limits_0^1 {\frac{{21}}{4}{x^2}ydy} \\ = \frac{{21{x^2}}}{8}.......(1)\end{array}\)

The marginal density of\(Y,f\left( y \right)\)is

\(\begin{array}{c}Y,f\left( y \right) = \int\limits_{{x^2}}^1 {\frac{{21}}{4}{x^2}ydx} \\ = - \frac{7}{4}\left( {{x^6} - 1} \right)y......\left( 2 \right)\end{array}\)

From (1) and (2) we can say that

\(\begin{array}{c}f\left( x \right) \times f\left( y \right) = \frac{{147y\left( {1 - {x^2}} \right){x^6}}}{{32}}\\ \ne f(x,y)\end{array}\)

Hence we can conclude that X and Y are not independent.