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Let $X_1,..., X_n$ be iid with pdf $f(x|\mu) = e^{-(x-\mu)}I_{[\mu, \infty)}(x)$ where I is $1$ if $x \in [\mu, \infty)$ and $0$ otherwise.

Then the likelihood and log likelihood functions are given by ...

$$f(X|\mu) = \prod_{i=1}^{n} f(x_i|\mu) = e^{-\sum_{i=1}^{n}x_i + n\mu}\prod_{i=1}^{n}I_{[\mu, \infty)}(x_i) = e^{-\sum_{i=1}^{n}x_i + n\mu}I_{[\mu, \infty)}(y)$$

where $y = min\{x_1, ..., x_n\}$

$$log(f(X|\mu)) = -\sum_{i=1}^{n}x_i + n\mu + log(I_{[\mu, \infty)}(y))$$

I don't know how to maximize this function in terms of $\mu$ to find $\hat{\mu}_{MLE}$. Taking the derivative doesn't help. Any tips?

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  • $\begingroup$ Search the site please. $\endgroup$
    – StubbornAtom
    Dec 15, 2019 at 12:31

1 Answer 1

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Hint: The likelihood function is $$L(\mu):=e^{-\sum_{i=1}^{n}x_i + n\mu}I_{[\mu, \infty)}(\min\left\{x_1,\dots,x_n\right\}).$$ (Remember, this is a function of $\mu$, with the data $\{x_1,\ldots,x_n\}$ treated as fixed.)

What is the largest value that $\mu$ can take here before the likelihood becomes $0$ (due to the indicator function)? How does the likelihood function behave as a function of $\mu$ for smaller values of $\mu$ than this largest value?

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    $\begingroup$ Got it thanks. $\mu$ cannot be larger than the $min(x_1, ..., x_n)$ otherwise the likelihood function becomes 0. So the MLE is $min(x_1, ..., x_n)$ $\endgroup$
    – Arnold Doveman
    Dec 15, 2019 at 10:34

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