00:01
Hello students, according to the given question, we have to find the given bits.
00:06
So they have given the pmf, p of x will be equal to theta into 1 minus theta whole power x minus 1.
00:17
Here x is equal to 1, 2, so on and 0 less than theta less than 1.
00:23
In the first bit, we have to find the likelihood function of theta.
00:27
So l of theta will be equal to pi, i is equal to 1 to n, p of xi.
00:35
So which is equal to pi, i is equal to 1 to n, so by substituting theta into 1 minus theta whole power xi minus 1.
00:44
Which is equal to theta power n into 1 minus theta power summation x i is equal to 1 to n minus n minus n.
00:53
So in the second bit we have to find the maximum likelihood estimator of theta.
01:00
So l will be equal to log of l of theta.
01:06
So from the above by applying log we get n log theta plus summation xi is equal to 1 to n minus n into log of 1 minus theta.
01:21
So because mla requires minimization of l with respect to theta.
01:30
Now, do l by do theta will be equal to 0.
01:36
That implies n by theta minus summation xi minus n by 1 minus theta is equal to 0.
01:45
So n by theta will be equal to summation xi minus n by 1 minus theta.
01:51
So 1 minus theta by theta will be equal to 1 by n summation xi minus 1.
02:01
Now 1 by theta will be equal to x bar minus 1 plus 1.
02:08
So 1 by theta will be equal to x bar.
02:11
Therefore, mla of theta is theta cap will be equal to 1x bar.
02:26
In the third bit, we have to find the maximum likelihood estimator of zai will be equal to 1 by theta.
02:35
So from invariance property of mle, that is maximum likelihood estimator, so we know that theta be the mle of theta and phi of theta be any continuous function.
02:54
So then mle, that is maximum likelihood estimator of zhi will be the mlea, which is maximum likelihood estimator be equal to 1 by theta.
03:02
So that implies zy cap will be equal to 1 by theta cap so which is equal to x bar.
03:10
So in the fourth bit from the given pmf expectation of x will be equal to 1 by theta from the exponential...