Compute the maximum likelihood estimator for the unknown (one or two dimensional) parameter, based on a sample of n i.i.d. random variables with that distribution. In each case, is the Fisher information well defined ? If yes, compute it.

We have a shihifted exponential distribution with parameters $\alpha \in \mathbb{R},\:\lambda >0:$

$\:f_{\alpha ,\lambda }\left(x\right)=\lambda e^{-\lambda \left(x-\alpha \right)}1_{x\ge \alpha },\:\forall x\in \mathbb{R}$

I want to find fisher information for this pdf. How can I do that?

I tried to find the second derivative of a log-likelihood function of $a$ but it is zero, so fisher information of $a$ is zero?

  • $\begingroup$ According to Wikipedia, the Fisher information is defined in terms of the first partial: $\mathcal{I}(\alpha) = \mathbf{E}\left[\left(\dfrac{\partial}{\partial \alpha} \log f(x; \alpha) \right)^2\big| \alpha \right]$ $\endgroup$ – Joe Jun 23 '20 at 10:35
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    $\begingroup$ @Joe : when the model is a regular model, which is not the case.... $\endgroup$ – tommik Jun 23 '20 at 10:44
  • $\begingroup$ @tommik, I’m not familiar with Fisher information, so maybe the Wikipedia article is incorrect, but it says the formula in my previous comment is the definition, which under certain regularity conditions can also be expressed in terms of the second partial derivative: en.m.wikipedia.org/wiki/Fisher_information $\endgroup$ – Joe Jun 23 '20 at 10:48
  • $\begingroup$ @Joe : yes, under "certain regularity conditions" that are not satisfied in the model of the example: they are very general condition about derivative under intergral sign...but when the domain depends on the parameter (that is our exercise), these conditions are not satisfied $\endgroup$ – tommik Jun 23 '20 at 10:52
  • $\begingroup$ @user9102437: can we have the complete text of the exercise? $\endgroup$ – tommik Jun 23 '20 at 10:54

Hint for the solution

  • First define the Likelihood fuction, that is

$$L(\alpha;\lambda)=\lambda^ne^{-\lambda \sum_i x_i}e^{n \alpha \lambda}\mathbb{1}_{(-\infty; x_{(1)}]}(\alpha)$$

  • Find the MLE estimator for $\alpha$;

Observing that

$$L(\alpha)\propto e^{n \alpha \lambda}\mathbb{1}_{(-\infty; x_{(1)}]}(\alpha)$$

this likelihood is strictly increasing in $\alpha$ so the MLE is


  • Fix $\alpha$ with $\hat{\alpha}$ and find with the usual procedure the MLE for $\lambda$

  • the fisher information is well defined only for $\lambda$....calculate it with the definition. That is because the general regularity conditions are not satisfied in this model, with respect to $\alpha$


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