00:01
Here we are given f of x1 equal to 1 plus theta x1 power theta and f of x2 equal to 1 plus theta, x2 power theta, etc, f of x n equal to 1 plus theta xan power theta.
00:17
Therefore the likelihood function is l of theta which is equal to f of x1, f of x2 etc, f of x n which is equal to 1 plus theta power n x 1 .x1.
00:30
Theta, x2 power theta, etc.
00:33
X n power theta.
00:35
Now take log on both sides.
00:37
So log of l of theta, which is equal to log of 1 plus theta, the whole power n into x1 theta, x2 power theta, etc, x n power theta, which is equal to n into log of 1 plus theta, plus theta times x1 plus x2 plus etc plus x n.
01:00
Now differentiate this equation on both sides...