00:01
In this problem, the joint density function of x and y, that is f of x, comma y, is 12 xy, 1 minus x where 0 less than x less and 1 .0 minus 1 .1 .1.
00:26
And in first part, marginal pdf of x1 is f of x, x is equal to 0 to 1, f of x minus y, dy, that is 12, 0 to 1 ,000, and it is x 12 x y square over 2 minus x square y square over 2 0 to 1 which we get 6 x minus x square now marginal pdf of y is that is f of y is 0 to 1 is 0 to 1 f of 1 f of x plus y d x which is equal to 12 0 to 1 x y minus x square y d x then it becomes 12 x square y over 2 minus x q y over 3 which becomes equal to 2 y since f of x y is f of x x x f of y is f of y f of y so, x and y are independent.
02:16
Independent.
02:18
Now, in second part, in b part, give joint probability mass function of x1 and x2.
02:39
That is, x is 0 to 1x, fdx, d .h.
02:45
Which is equal to 6 0 to 1 x square minus x cube d x and it will be 6 x cube over 3 minus x 4 over 4 0 to 1 and we get half now in c part e of y is integration 0 to 1 y divide d y uh f y d y it is 2, 0 to 1, y squared, then we get y, y, y, cube over 3, 0 to 1.
03:28
Then it is 2 by 3...