00:01
Hello students here a f of xy will be valid pdf if first one the value of f of xy is non -negative for its domain.
00:35
The domain is 0 less than x less than 1 and 0 less than y less than 1.
00:43
F of x comma y is always greater than or equal to zero therefore first condition satisfied now the second condition is the double integration x y into f of x comma y d x x into d y is equal to one so we have to show this one which is equal to integration x integration y into f of x comma y x x into dy where integration x from 0 to 1 y from 0 to 1 4 x y d x therefore 4 into integration 0 to 1 x d x d x into 0 to 1 integration y d y d y that is equal to 4 into x square minus by 2 that interval from 0 to 1 into y square from divided by two interval from 0 to 1 which is equal to 4 into 1 by 2 into 1 by 2 that is 1.
01:58
So both the conditions are satisfied.
02:03
So f of x y is a valid pdf.
02:06
Now we have to find the marginal pdf of x that is f of x that is f of x equal to integration.
02:19
Y into f of x comma y into d y that is equal to integration 0 to 1 4 into x into y dy which is equal to 4x into integration 0 to 1 y dy which is equal to 4x into integration 0 to 1 dy that is 4x into 1 by 2 which is equal to 2 x now the marginal pdf of y is given as f of y is equal to integration x, f of x comma y, dx, that is equal to integration 0 to 1, 4x, y into dx, that is 4x integration 021 1 into x, dx, that is 4x integration 021 into x, dx, that is 4y into 1 by 2, which is equal to 2y.
03:27
Therefore the marginal pdf of x that is f of x is 2x where 0 x less than 1 and marginal fdf of y is 2y that is 0 less than y less than 1 now we have to find if x and y are independent so if x and y are independent, then f of x given y is equal to f of x...