If x and y are independent and identically distributed uniform random variables on (0,1), compute each of the following joint densities. (a) u=5x, v=4x/y. fu, v(u, v)= (b) u=3x+y, v=5x/(x+y). fu, v(u, v)=

Use the method of transformations for both exercises.

a.

The Jacobian for this transformation is

Then the joint density of and is

Both and are i.i.d with density

(where is either or ) so we have

We can clean up the support a bit:

The first case suggests either or can be negative, but with , that is impossible. This leaves us with

so the joint PDF is

b.

The Jacobian is

and so the joint PDF is

We have

and

which together tell us that

As for , we know that at most we can have , so that can have a maximum value of 5/2, so we can simplify the support to write the PDF as

the answer is 24

step-by-step explanation:

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads