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I am confused about the following question.

Let $X_1, \dots, X_n$ be i.i.d distributed with probability function \begin{equation} p_v(x) = \lambda \exp(-\lambda x) \end{equation} for $\lambda > 0$. Find the Maximum Likelihood Estimator for $1/ \lambda$.

I don't quite understand which of the following two approaches I should follow:

  1. Compute $v^{*,1} = \arg\max_v L_X(v) = \prod_{i} v \exp(-v x_i)$ and then use $1/v^{*,1}$ as estimator.

  2. Compute $v^{*,2} = \arg\max_v L_X(\frac{1}{v}) = \prod_{i} \frac{1}{v} \exp(-\frac{1}{v} x_i)$ and then use $v^{*,2}$ as estimator.

In this case, they seem to evaluate to the same result. However, I would really like to know which of these two approaches is the correct idea.

Thanks for your help!

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    $\begingroup$ In my opinion, I think both definitely (not only seem to) yield the same result just because this is just change the variable from $v^{*,1}$ to $v^{*,2}$... both of them are correct, actually. $\endgroup$
    – JetfiRex
    Commented Jan 27, 2022 at 23:28
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    $\begingroup$ Are you familiar with the invariance principle for MLEs? $\endgroup$ Commented Jan 27, 2022 at 23:30
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    $\begingroup$ The map $\eta:v\mapsto \frac{1}{v}$ is one to one from $(0,\infty)$ to $(0,\infty)$. This, the exponential family $\{f_v(t)=v e^{-vt}\mathbb{1}_{(0,\infty)}(t): v>0\}$ can also be parametrized by the $\{\tilde{f}_{\eta}(t)=\frac{1}{\eta}e^{-t/\eta}\mathbb{1}_{(0,\infty)}(t): \eta>0\}$. Thus, if $\hat{\theta}$ is a ML estimator of $\theta$, $\eta(\hat{\theta})=\frac{1}{\hat{\theta}}$ is the ML estimator of $\eta=\frac{1}{\theta}$ $\endgroup$
    – Mittens
    Commented Jan 27, 2022 at 23:59
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    $\begingroup$ @Godsbane: For general functions $\tau:v\rightarrow\tau(v)$ if the paremeter $v$ that describes the population, the maximum likelihood estimator for $\tau(v)$ can be defined as in this posting. $\endgroup$
    – Mittens
    Commented Jan 28, 2022 at 0:19
  • $\begingroup$ Thanks to your comments, it is now clear to me :) $\endgroup$
    – Godsbane
    Commented Jan 28, 2022 at 8:32

1 Answer 1

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They are equivalent due to the property of functional equivalence, i.e.

$$\widehat{g(\theta)}=g(\hat \theta)$$

where hats indicate ML estimators. Your setup is the case where $g:t\rightarrow1/t$.

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