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Let $X_1,\,X_2,\,X_3,\ldots ,X_n$ represents a random sample from each of the distributions.

I have a certain function :

$$f(x;\theta)=\begin{cases}e^{\theta-x}& x\geq\theta \\ 0 & x<\theta\end{cases}$$

for $\theta\in(-\infty,\infty)$.

Finding the MLE!

My answer is the following :

$$\begin{align} L(\theta;\,x_i)&=e^{-\sum_{i=1}^n(x_i-\theta)}\\ \ln{L(\theta;\,x_i)}&=-\sum_{i=1}^n(x_i-\theta)\\ \ln{L(\theta;\,x_i)}&=n\theta-\sum_{i=1}^n x_i\\ D_\theta\ln{L(\theta;\,x_i)}&=n>0\\ D_{\theta\theta}\ln{L(\theta;\,x_i)}&=0 \end{align} $$

But i'm having trouble to find $\hat \theta$

My professor said that we must use ordered statistics when we meet this case. I mean, when the second derivative is zero or greater than zero.

But how to use ordered statistics for finding the estimator?

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  • $\begingroup$ But it might not be a pdf.. it might be pmf. My teacher said, that f(x) is pdf or pmf. And it's true cz for example, we can find the estimator from poisson dist. (pmf). Sorry for my bad english $\endgroup$
    – user516076
    Sep 10, 2019 at 1:04
  • $\begingroup$ Just for information, i took this question from Introduction to Mathematical Statistics 6th edition sect. 6 number 3 (if i'm not wrong) $\endgroup$
    – user516076
    Sep 10, 2019 at 1:06
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    $\begingroup$ You did not write the definition of the function correctly. I'm updating your post. $\endgroup$ Sep 10, 2019 at 1:10
  • $\begingroup$ Ok thanks for the correction. $\endgroup$
    – user516076
    Sep 10, 2019 at 1:13
  • $\begingroup$ Differentiation is not the only way of searching for MLE. See for example math.stackexchange.com/questions/651553/…, math.stackexchange.com/questions/2019525/…. This has been asked numerous times here, so do search the site. $\endgroup$ Sep 10, 2019 at 5:56

2 Answers 2

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I think the second derivative is unnecessary.

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  • $\begingroup$ It's needed. Bcz we must to know and ensure that the second derivative is less than zero. From that, we allowed to set the first derivative is equal to zero. But if the second derivative isn't LT zero, we can't. $\endgroup$
    – user516076
    Sep 9, 2019 at 23:00
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    $\begingroup$ @user516076 Forget second derivative, even first derivative is unnecessary here. Notably $\frac{d}{d\theta}\ln L(\theta)$ (your $L(\theta)$ is not completely correct) does not exist at $\theta=\min(x_1,\ldots,x_n)$. $\endgroup$ Sep 10, 2019 at 6:03
  • $\begingroup$ Yes.. on my last comment below i've said that it use ordered statistics. $\endgroup$
    – user516076
    Sep 10, 2019 at 8:47
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Hint:

  • The likelihood is zero if any of the $x_i$s is less than $\theta$

  • The likelihood is a positive and increasing function of $\theta$ (you have shown it has a positive first derivative), provided that $\theta$ is less than or equal to all the $x_i$s

  • So the likelihood is maximised when $\theta$ is ...

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