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

1. ## MLE of expected value of pdf e^-(x-a)

Hi guys

I'm trying to solve this

I have this pdf

$f(x)=e^{-(x-a)} , x > a$

$f(x)=0 , x \leq a$

I must find the maximum likelihood estimator of the expected value

I did this

$f(x_1|a)...f(x_n|a)=e^{-(x_1-a)}...e^{-(x_n-a)}=e^{\sum_{i=1}^{n}-x_i}e^{na}$

The log of the last expression is

${\sum_{i=1}^{n}-x_i}+na$

In its derivative the $a$ disappears

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

How can I proceed?

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

$1e^{-1x}e^{a}$

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!

2. 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 $\forall i, x_i>a$

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 $a=\min \{x_1,\dots,x_n\}$ ?

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 !

3. Thank you Moo

So x is defined from a to inf

$E(x)=\int_{a}^{+\infty }xf(x)dx=\int_{a}^{+\infty }xe^{a-x}dx=a+1$

The estimator of a is

$\hat{a}=min(x_1,...,x_n)$

So the estimator of the expected value should be

$\hat{a}+1=min(x_1,...,x_n)+1$

Is it correct?

Thanks

4. Yes

And the estimator by the method of moments will be $\hat{a}=\frac 1n \cdot \sum_{i=1}^n x_i-1$

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