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# Find maximum likelihood estimation (MLE) for theta and prove that it maximizes likelihood function.?

Given X1,X2,...Xn=f(x;theta), where:

f(x;theta)=(theta)(x)^(theta-1) for 0<x<1 and 0<theta<infinity

=0 for otherwise

Find maximum likelihood estimation (MLE) for theta and prove that it maximizes likelihood function.

The final answer is -n/(theta^2)

I would really appreciate the steps explained thanks.

### 1 Answer

- nyc_kidLv 79 years agoFavorite Answer
First, your "final answer" is wrong!

The likelihood function of X1,X2,...Xn is given by ((theta)^n)(X1*X2*....*Xn)^(theta-1).

One common method to find theta that maximizes the likelihood function is to first take the log of the likelihood function::

n(ln(theta)) + (theta-1)ln((X1*X2*....*Xn).

Then, differentiate w.r. to theta: (n/theta) + ln((X1*X2*....*Xn) and equate it to 0.

(n/theta) + ln((X1*X2*....*Xn) = 0 ==> theta = -n/ln((X1*X2*....*Xn) .

So, the MLE of theta is -n/ln((X1*X2*....*Xn).

Note that MLE of theta, can not be a function of theta(after all, theta is the unknown parameter that you are trying to estimate based on the observations X1,X2,...Xn. So it has to be a function of these observations)

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