Question

Question:Prove Theorem 3.5.6.

Let X and Y have a continuous joint distribution. Suppose that

\(\;\left\{ {\left( {x,y} \right):f\left( {x,y} \right) > 0} \right\}\)is a rectangular region R (possibly unbounded) with sides (if any) parallel to the coordinate axes. Then X and Y are independent if and only if Eq. (3.5.7) holds for all\(\left( {x,y} \right) \in R\)

Short answer

\(f\left( {x,y} \right) = {h_1}\left( x \right){h_2}\left( y \right)\)for all \(\left( {x,y} \right) \in R\)

Step-by-step solution

Given information

R be the rectangular region:

\(R = \left\{ {\left( {x,y} \right):{x_0} < x < {x_1},{y_0} < y < {y_1}} \right\}\) with \({x_0}\)

Computing probability

We assume that \(f\left( {x,y} \right) = \frac{{{h_1}\left( x \right)}}{{{h_2}\left( y \right)}}\) for all (x,y) that satisfy f(x,y)>0.

Then

Hence,

Again,\(f\left( {x,y} \right) = {h_1}\left( x \right){h_2}\left( y \right)\)for all (x,y) and

\(f\left( {x,y} \right) = {h_1}\left( x \right){h_2}\left( y \right)\)for all \(\left( {x,y} \right) \in R\)

Hence, proved.