28 Let X and Y have joint density Determine the distribution

28. Let X and Y have joint density Determine the distribution of X^2/Y.

Solution

The joint pdf of X and Y is

f(x,y) = cx   if 0 <x2 < y < sqrt(x) < 1

          = 0    otherwise

Let us transform U = X2

X=sqrt(U)

We have 0 <x2 < y < sqrt(x) < 1

Then 0 < u < y   and y < u1/4 < 1

=> 0 < u < y and y4 < u < 1

=> 0<y4 < u < y < 1

And |J| = 1/(2*sqrt(u))

Then the joint pdf of U and Y is

f(u,y) = f(sqrt(u),y)*|J|

          = c*sqrt(u)/(2*sqrt(u))           if 0<y4 < u < y < 1

              0 Otherwise

          = c/2     if 0<y4 < u < y < 1

              0   Otherwise

Now the pdf of Y is given by

f(y) = Integration(from=y4,to=y, c/2 du)

          =c/2 * [u,from=y4,to=y]

          =c(y-y4)/2                      if 0<y<1

             0    Otherwise

Now the conditional distribution of X2|Y i.e. U|Y is

f(u|y)= f(u,y)/f(y)

          = (c/2)/( c(y-y4)/2)        if 0<y4 < u < y < 1

            0     otherwise

          = 1/(y-y4)   if 0<y4 < u < y < 1

             0    Otherwise

Thus X2|Y ~ Uniform(y4,y)    where 0<y<1.

 28. Let X and Y have joint density Determine the distribution of X^2/Y. SolutionThe joint pdf of X and Y is f(x,y) = cx if 0 <x2 < y < sqrt(x) < 1

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