00:01
Now, this problem would have been given the following joint distribution function.
00:04
I would like to begin by finding the marginal distributions for both x and y.
00:09
And the marginal of x just means to integrate out the y.
00:14
So we integrate from 0 to 1, 3x minus y over 11, the y.
00:22
And so this becomes 1 over 11 times 3xy minus 1ā2 y square.
00:33
As y goes from 0 to 1.
00:35
We put you in 1 for y, this is 1 over 11 times 3x minus 1 half.
00:44
And so this is our marginal distribution for x.
00:53
And then for y, our marginal for y, just means integrate out the y, or out the x.
00:59
And so x goes from 1 to 3, so we integrate from 1 to 3 of 3x minus y over 11 d x.
01:11
And so this is 1 over 11 times 3 halves x squared minus yx, evaluated from x is...