# How do I find the MLE of $\theta$ when x is dependent on $\theta$?

Let $X_{1},X_{2},...,X_{n}$ represent a random sample from a distribution with pdf:

$f(x; \theta)=e^{-(x-\theta)}, \theta \le x<\infty, -\infty<\theta<\infty$ | zero elsewhere

I need to find the MLE $\hat {\theta}$ of $\theta$.

Since the support space of the pdf is dependent on $\theta$, do I need to express the pdf in terms of an indicator function?

i.e. $f(x; \theta)=e^{-(x-\theta)}I_{(\theta,\infty)}(x)$

If so, do I find the MLE in the standard manner?

i.e. $L(x;\theta)=\displaystyle \prod^{n}_{i=1} f(X_{i};\theta)=e^{-(\sum^{n}_{i=1}X_{i}-n\theta)}I_{(\theta,\infty)}(X_{(1)})$

$\ln L(x;\theta)=-\displaystyle \sum^{n}_{i=1} X_{i} +n\theta +\ln I_{(\theta,\infty)}(X_{(1)})$

The next step would be to take the partial derivative of the log-likelihood function with respect to $\theta$, but how would I find the partial derivative of the indicator function?

Am I approaching this question in the correct manner? Any help would be greatly appreciated!

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You can't write the log-likelihood like this. Take a look at your likelihood function instead. Which value of $\theta$ would maximize this function? – Patrick Li Jan 29 at 5:51
Partial derivatives are not the only way to maximize a function. Try to rewrite $L$ as $L(x;\theta)=c\mathrm e^{n\theta}\mathbf 1_{\theta\lt y}$ and to think about the shape of the function $\theta\mapsto c\mathrm e^{n\theta}\mathbf 1_{\theta\lt y}$. – Did Jan 29 at 8:43

## 1 Answer

Note that the likelihood function is $$\begin{align}L(\theta |x) &=e^{-(\sum^{n}_{i=1}x_{i}-n\theta)},x_i \geq \theta,\forall i \\ &=e^{-\sum^{n}_{i=1}x_{i}}\cdot e^{n\theta},x_{(1)} \geq \theta \end{align}$$ Now note that $L(\theta |x)$ is maximum

iff $e^{-\sum^{n}_{i=1}x_{i}}\cdot e^{n\theta}$ is maximum subject to the restriction $x_{(1)} \geq \theta$

iff $e^{n\theta}$ is maximum subject to the restriction $x_{(1)} \geq \theta$

iff $\theta =x_{(1)}$.

Here Your MLE is $x_{(1)}=min(x_1,\dots ,x_n)$.

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