00:01
Hello students, here in the given problem given the joint pdf that is k and x lies between 0 to 2 and y lies between 0 to 1 and here the 2y is less than equal to x.
00:10
First we find out the value of the k.
00:16
Here we know that the integration y goes from 0 to 1 and integration from x is 2y to 2.
00:30
Therefore, k is taken outside 0 to 1.
00:34
Here take the integration of the x that is x constant integration is x 2y to 2 of dy which is equal to 1.
00:44
Therefore, the integration 0 to 1 here put the limit 2 minus 2y into dy which is equal to 1.
00:52
Therefore, the k here take the integration of the constant of the 1 that is y 0 to 1 minus 2 into integration of y square by 2 0 to 1 which is equal to 1.
01:06
Therefore, k of 2 into 1 minus 0 minus 1 minus 0 here 2 to get cancel is equal to 1.
01:18
Therefore, k into 2 minus 1 which is equal to 1.
01:23
Therefore, k is equal to 1.
01:25
So, the final value of the k which is equal to 1.
01:32
Now we part to find out the marginal pdf of the x and y.
01:40
Therefore, f of x which is equal to integration from y is equal to 0 to 1 into dy which is equal to y of 0 to 1 which is equal to 1.
01:52
And marginal pdf of the y which is equal to integration from 0 to 2 1 into dx which is equal to 1.
02:01
X from here integration is 0 to replace with 2y to 2.
02:07
Therefore, the put the limit 2 minus 2y.
02:11
This is a marginal pdf of the y.
02:16
Here find out the conditional density of the x given y is equal to 0 .8.
02:24
Therefore, the f of x given y which is equal to 0 .8 which is equal to we can write it as f of x 0 .8 divided by f of y 0 .8 which is equal to this value is nothing but marginal pdf of the x that is 1 and marginal pdf of the y here 2 minus y is 0 .8.
02:48
So, put the value of the y...