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Let $X_1,X_2,...,X_n$ be a random sample of size $n$ from a population with density

$f(x) = \left\{ \begin{array}{lr} e^{\theta-x} & , x \geq\theta\\ 0 & , \text{otherwise} \end{array} \right.$

Find the maximum likelihood estimator for $\theta$.

Here is my attempt:

$\begin{align*} L(\theta)&=\prod_{k=1}^{n}e^{\theta-x_k}\\ &=\prod_{k=1}^{n}e^\theta e^{-x_k}\\ &=e^{n\theta}\prod_{k=1}^{n}e^{-x_k}\\ &=e^{n\theta} e^{-\sum_{k=1}^{n}} \end{align*}$

Then $\ln L(\theta)=n\theta-\sum_{k=1}^{n}x_k$

Take the derivative with respect to $\theta$ to get

$n$

But there is no $\theta$ in this expression, and so how can I find the MLE of $\theta$?

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  • $\begingroup$ Here $E[X]=\theta+1$ by direct computation, so $\theta=E[x] -1. Doesn't this mean that the MLE is $\bar X -1$? Admittedly, I'm guessing here. $\endgroup$ – MPW Sep 26 '14 at 1:46
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    $\begingroup$ Your likelihood function is incompletely specified. The unknown parameter $\theta$ must be smaller than each and every one of the $x_k$, no? $\endgroup$ – Dilip Sarwate Sep 26 '14 at 2:14
  • $\begingroup$ MPW, I believe you are thinking about the method of moments estimator. $\endgroup$ – nonremovable Sep 26 '14 at 12:54
  • $\begingroup$ Dilip Sarwate, you are correct. Although I can't quite see how I should account for that fact in my work. $\endgroup$ – nonremovable Sep 26 '14 at 12:57
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    $\begingroup$ Without using the "differentiate w.r.t $\theta$ and set the derivative equal to $0$" method of finding the locations of extrema, where would you say the function $$L(\theta) = \left(\prod_{k=1}^{n}e^{-x_k}\right)e^{n\theta}, ~0 < \theta \leq \min_i x_i$$ has maximum value? Further hint: Since the $x_i$ are known quantities, that thingy inside the large parentheses is a constant that affects only the value of the maximum, not the location of the maximum. If you do work it out, can you think why I might have suggested that you not use the "differentiate...." approach? $\endgroup$ – Dilip Sarwate Sep 26 '14 at 22:39
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Your expression for likelihood is not completely correct. It should be

$$L(\theta) = \exp(n\theta) \exp(-\sum_1^n x_k) 1_{\theta \leq X_{(1)}}$$ where $X_{(1)}$ is the smallest of all $X_i$'s. We need the extra term because the likelihood will be zero if any $X_i$ is smaller than $\theta$. Now, we know that exponential is increasing function and we want $\theta \leq X_{(1)} $ for $L$ to be nonzero. Hence, the maximum should occur at $\hat{\theta} = X_{(1)}$

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