# Maximum likelihood estimator of $1/v$

I am confused about the following question.

Let $$X_1, \dots, X_n$$ be i.i.d distributed with probability function $$$$p_v(x) = \lambda \exp(-\lambda x)$$$$ for $$\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!

• 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. Commented Jan 27, 2022 at 23:28
• Are you familiar with the invariance principle for MLEs? Commented Jan 27, 2022 at 23:30
• 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 $\eta=\frac{1}{\theta}$ Commented Jan 27, 2022 at 23:59
• @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. Commented Jan 28, 2022 at 0:19
• Thanks to your comments, it is now clear to me :) Commented Jan 28, 2022 at 8:32

$$\widehat{g(\theta)}=g(\hat \theta)$$
where hats indicate ML estimators. Your setup is the case where $$g:t\rightarrow1/t$$.