# Find the MLE estimator for $\theta$

Let $$Y_1 ,Y_2 ,\ldots,Y_n$$ be a random sample from a distribution with pdf

$$f(y) = e^{-(y -\theta) }$$ for $$y \geq 0$$ and $$0$$ else

a) Find the Method of Moments estimator for $$\theta$$

b) Find the MLE estimator for $$\theta$$

I'm pretty sure I found out how to do a) but b) I'm having trouble with. Everytime I take the logarithm and then take the derivative, $$\theta$$ disappears, any help?

• What you wrote is not a pdf. Do you mean $y\ge\theta$ in the definition instead?
– Ian
Feb 16, 2015 at 2:25
• Note that $f(y)=e^{-(y-\theta)}$ for $y\geq \theta$. So may be the smallest $Y_i$? Feb 16, 2015 at 2:25
• math.stackexchange.com/questions/2019525/… Sep 17, 2019 at 15:11

The MLE estimator is by definition $\hat\theta$ which maximizes $$\prod_{k=1}^n{\mathrm e}^{-(Y_k-\theta)},\quad\theta\in\left(-\infty,\min(Y_1,\dots,Y_n)\right],$$ or equivalently (by taking the logarithm), $$\sum_{k=1}^n(\theta-Y_k),\quad\theta\in\left(-\infty,\min(Y_1,\dots,Y_n)\right].$$ This is an increasing function of $\theta$, so...
• Ha, it is the smallest $Y_n$. Feb 16, 2015 at 2:33
• so the MLE is the $max(Y_1 ,...,Y_n)$ ? Feb 16, 2015 at 8:56
• or is it the $min(Y_1, ... , Y_1)$ ? Being the max would make more sense to me since its an increasing but everyone here seems to say its the min? Feb 16, 2015 at 20:49
• @AlexChavez, $\theta$ is in the interval $(-\infty,\min(Y_1,\dots,Y_n)]$ so it cannot be the maximum which is outside the interval.