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 $p$.


1 Answer 1


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.


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