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I want to ensure my reasoning is correct.

The sample is distributed with the pdf: $$f_a(x) = e^{-(x-a)}\cdot \chi_{[a, +\infty)}(x)$$ where $a>0$. Find the maximum likelihood estimator for $a$.

The likelihood function would be: $$L(a, x_1, ..., x_n) = e^{-x_1+a}\cdot \chi_{[a, +\infty)}(x_1) \cdot \dots \cdot e^{-x_n+a}\cdot \chi_{[a, +\infty)}(x_n)$$

$$=e^{na-\sum\limits_{i=1}^n x_i} \prod\limits_{i=1}^n \chi_{[a, +\infty)}(x_i)$$

$$= \begin{cases} e^{na-\sum\limits_{i=1}^n x_i} & x_1, \dots, x_n \in [a, +\infty) \\ 0, & \text{otherwise} \end{cases}$$

$$= \begin{cases} e^{na-\sum\limits_{i=1}^n x_i} & \forall i \in \{1, \dots, n\}: x_i \geq a \\ 0, & \exists j \in \{1, \dots, n\}: x_j < a \end{cases}$$

which, if plotted, would probably look like this:

Plot for L

Since all $x_i$ have to be greater than $a$ and the maximum value is reached at $a$, my guess is that the MLE would be:

$$\hat{a}_n = \min{(x_1, \dots, x_n)}$$

Is this correct?

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    $\begingroup$ Yes, this seems correct! And nice graph that you have there! $\endgroup$
    – Jimmy R.
    Jun 26, 2017 at 16:13
  • $\begingroup$ Which is the variable (horizontal axis) in your graph? $\endgroup$
    – leonbloy
    Jun 26, 2017 at 17:49
  • $\begingroup$ @JimmyR. As pointed out in my answer (and in the accepted one) this is not correct, the graph is definitely wrong because it uses $x_i$ as variable instead of $a$. $\endgroup$
    – leonbloy
    Jun 27, 2017 at 12:46
  • $\begingroup$ math.stackexchange.com/q/2019525/321264 $\endgroup$ Feb 9, 2020 at 20:01

2 Answers 2

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What you graphed is the density function, and that for only a single observation rather than a joint density for $n$ observations, not the likelihood function. You have $$ L(a) = \left.\begin{cases} e^{n(a-\bar x)} & \text{if for } i = 1, \ldots,n,\ a\le x_i \\ 0 & \text{otherwise} \end{cases}\right\} = \begin{cases} e^{n(a-\bar x)} & \text{if } a \le \min\{x_1,\ldots,x_n\}, \\ 0 & \text{otherwise.} \end{cases} $$ Thus $L(a)$ increases as $a$ increases, until $a$ gets as big as $\min\{x_1,\ldots,x_n\}.$

Note that $\min\{x_1,\ldots,x_n\}$ is a point on the $a$-axis.

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Your graph and your reasoning is wrong. The likelihood must be drawn/analyzed as a function of the parameter (in this case, $a$).

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  • $\begingroup$ How am I supposed to estimate a parameter if you want me to use this parameter in the estimation? $\endgroup$
    – Angie
    Jun 26, 2017 at 19:28
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    $\begingroup$ You use the paramenter as a variable (not as a fixed value). And you maximize with respect of that variable. $\endgroup$
    – leonbloy
    Jun 27, 2017 at 3:19

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