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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?

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

1 Answer 1

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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...

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    $\begingroup$ Ha, it is the smallest $Y_n$. $\endgroup$
    – Brian Ding
    Feb 16, 2015 at 2:33
  • $\begingroup$ so the MLE is the $max(Y_1 ,...,Y_n)$ ? $\endgroup$ Feb 16, 2015 at 8:56
  • $\begingroup$ 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? $\endgroup$ Feb 16, 2015 at 20:49
  • $\begingroup$ @AlexChavez, $\theta$ is in the interval $(-\infty,\min(Y_1,\dots,Y_n)]$ so it cannot be the maximum which is outside the interval. $\endgroup$
    – Ian
    Feb 17, 2015 at 2:09
  • $\begingroup$ oh okay makes sense thank you! $\endgroup$ Feb 17, 2015 at 3:38

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