Thread: MLE

**ankuoli** · March 16th 2010, 08:39 AM

Find the MLE of the distribution having the following pdf

$f(x;\theta)=e^-(x-\theta)$ , $\theta \le x<\infty$ , $-\infty<\theta<\infty$

Attempt to a solution:

Likelihood function
$L(\theta;X_i)=e^[-\sum X_i+\theta]$

Log likelihood function
$l(\theta)=-\sum X_i+\theta$

But then taking the derivative with respect to $\theta$ it makes $\theta$ disappear.

I feel like i missing something especially with the bounds of x and $\theta$

**Moo** · March 16th 2010, 12:56 PM

Hello,

It's like for a uniform distribution. The log likelihood function is always increasing, for any theta (since its derivative is n, always positive). So the maximum is attained at the maximum possible value of theta.
But since, for any i, $x_i\geq\theta$ (that's the info in the pdf), we have that $\theta\leq\min(x_i)$
Hence, the MLE is $\min(x_i)$

**matheagle** · March 16th 2010, 04:52 PM

The likelihood function is....

$e^{n\theta}e^{-\sum x_i}I(x_{(1)}\ge \theta)$

you want to maximize this, hence you want $e^{n\theta}$
as large as possible and also that indicator function to be one.

The largest theta can be in that case is $X_{(1)}$

There's no need to take the log here, it only makes it messier.

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