I am a undergraduate in the introductory mathematical statistics sequence and I'm having problems finding the MLE for this probability density: $f(y|\theta)=e^{-(y-\theta)}, y>\theta$

When I try to solve it, I find the log-likelihood: $log(\mathcal{L}([y_1,...,y_n]|\theta))=-\Sigma y_i + n\theta$

and I take the derivative w.r.t. $\theta$ and equate it to zero to arrive at: $n=0$.

Which is incorrect let alone useless for finding the estimator.

I'm very confused. Am I missing something?

  • $\begingroup$ See some of the advice here, particularly in relation to drawing the likelihood or log likelihood. A more general version of the problem is discussed here $\endgroup$
    – Glen_b
    Commented Jul 31, 2017 at 12:28

1 Answer 1


Since this is a self study question, I'm only going to take it part of the way there.

The log likelihood is actually more complicated that the expression you have here. You're implicitly constraining the space of $y$'s that are applicable, but you need to be a bit more explicit to see what is going on.

$$ \mathcal{L}(y_1, \ldots, y_n; \theta) = \prod_i e^{-(y - \theta)} \ \ \ \text{if}\ y_1, \ldots, y_n \geq \theta $$ $$ \mathcal{L}(y_1, \ldots, y_n; \theta) = 0 \ \ \ \text{otherwise} $$

Your computation is

$$\frac{d}{d\theta} \log \left( \mathcal{L}(y_1, \ldots, y_n; \theta) \right) = n \ \ \ \text{if} \ y_1, \ldots, y_n > \theta $$

This is simply telling you that there is no local optima in the region $y_1, y_2, \ldots, y_n > \theta$. So if it's not there, where can it be?

  • $\begingroup$ Thank you, I figured it out. I chose the largest $\theta$ such that the likelihood is non-zero, $Y_{(0)}$. $\endgroup$ Commented Jul 30, 2017 at 22:19
  • $\begingroup$ Yes, but it's not necessarily $y_0$. $\endgroup$ Commented Jul 31, 2017 at 5:02

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