The question is already posted here : https://math.stackexchange.com/questions/2530202/prove-that-mle-does-not-depend-on-the-dominating-measure

Let $(X,\mathbb{F})$ be a measurable space, $\left\{P_{\theta}:\theta \in \Theta\right\}$ be a family of distributions on $\mathbb{F},$ which is dominated by a $\sigma$-finite measure $\mu_0$ on $\mathbb{F}$ (i.e. all $P_{\theta}'$s are absolutely continuous w.r.t. $\mu_0$). Let $p(x,\theta)=\frac{dP_{\theta}}{d\mu_o}(x)$ be the density (radon nicodym derivative) of $P_{\theta}$ w.r.t. $\mu_0$.

Given a sample $(x_1,\ldots,x_n)$ from $P_{\theta}$ (unknown $\theta$) the maximum likelihood estimator (MLE) of $\theta$ is defined by $$\hat{\theta}_{MLE}=\arg max_{\theta \in \Theta} ~\left\{\prod_{i=1}^n p(x_i,\theta)\right\}$$

To show that $\hat{\theta}_{MLE}$ DOES NOT depend on the choice of the dominating measure $\mu_0,$ although the definition of $\hat{\theta}_{MLE}$ is connected to $\mu_0$ through the density $p$.


This is really simple: Let $\nu$ be another measure on the same space, such that $\mu_0$ and $\nu$ are absolutely continuous with respect to each other. (this means that this measures has the exactly same events with measure zero, and garantees the existence of Radon-Nikodym derivatives). Then the density with respect to $\nu$ is $$ p_\nu(x,\theta) = \frac{d P_\theta}{d \nu}(x) = \frac{d P_\theta}{d \mu_0}(x) \frac{d \mu_0}{d \nu}(x) $$ (and the point is that the last factor do not depend on $\theta$).

Now we can write the likelihood as $$ \prod_{i=1}^n \left( p(x,\theta)\cdot \frac{d \mu_0}{d \nu}(x)\right) = \prod_{i=1}^n p(x_i,\theta) \prod_{i=1}^n \frac{d\mu_0}{d\nu}(x_i) $$ and since the last factor do not depend on $\theta$ it have no influence on the maximum.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.