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 $\theta$?
$\endgroup$ 71 Answer
$\begingroup$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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