Thread: MLE of expected value of pdf e^-(x-a)

Hi guys

I'm trying to solve this

I have this pdf





I must find the maximum likelihood estimator of the expected value

I did this



The log of the last expression is



In its derivative the disappears

Without taking the log I end maximizing the same expression (the exponent of )

How can I proceed?

I see that I can express the pdf in this way:



This is the exponential distribution multiplied by a constant (but i know that it is defined for different values of x)

Is this a wrong path?

Thanks!

Hello,

You get n as the derivative.
Which is always positive.
Thus the likelihood function is always increasing.

In order to maximise the likelihood function, make a the greatest possible.

But you know that

Since the xi's are observations, they're already realised. So find a with respect to xi. Such that it's the greatest possible.

Would you agree that ?



A similar reasoning is made for the MLE of a uniform distribution
I've seen that on a wikipedia (can't remember in which language), so you should find an example over the internet !

Thank you Moo

So x is defined from a to inf



The estimator of a is



So the estimator of the expected value should be



Is it correct?

Thanks

Yes


And the estimator by the method of moments will be

By the way, we always write E(X), not E(x).