# How to maximize this likelihood function to find the MLE? [duplicate]

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?

• Search the site please. Dec 15, 2019 at 12:31

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