# Divergence based robust inference

The term 'divergence' means a function $D$ which takes two probability distributions $g,f$ as input and puts out a non-negative real number $D(g,f)$. I have learnt that the inference based on minimizing the following divergence is robust against those data points which are not compatible with the assumed model (called outliers), for some specific range of $\alpha \in \mathbb{R}^n$. $$D_{\alpha}(g,f) = \frac{1}{\alpha-1}\log \int g^{\alpha}f^{1-\alpha}~dx,$$ where $g$ stands for the data driven density and $f$ stands for the model density (see, for example here). There are several names to this $D_\alpha$ in the literature viz., Renyi divergence, power divergence, $\alpha$-divergence etc. Can someone give a justification of how this method is robust?

Update: (as suggested by @Bruce Trumbo)

Some Preliminaries:

Suppose that $x_1,\dots,x_n$ i.i.d. samples drawn according to a particular member of a parametric family of probability distributions $\mathcal{F} = \{f_\theta\}$ (model). (Let us assume, for simplicity, that the $f_{\theta}$'s have a common support set $\mathbb{X}$ which is finite and also that $\mathbb{X}$ is the undelying sample space.)

Our objective is to choose a special member $f_{\theta^*}$ of $\mathcal{F}$ which "best" explains the observed samples.

The Maximum Liklihood Estimate (MLE) is a widely used inference method which asks us to choose the $f_\theta$ for which $\Pi_{i=1}^n f_\theta(x_i)$ is maximum.

Let $\hat{f}$ be the empirical measure of the $x_i$'s. Then observe that \begin{eqnarray} \frac{\prod_{i=1}^n f_\theta(x_i)}{\prod_{i=1}^n \hat{f}(x_i)} & = & \prod_{x\in\mathbb{X}} \left(\frac{f_\theta(x)}{\hat{f}(x)}\right)^{n\hat{f}(x)}\\ & = &\exp\{-nD(\hat{f}\|f_\theta)\}, \end{eqnarray} where $D(\hat{f}\|f_\theta)=\sum_x \hat{f}(x)\log(\hat{f}(x)/f_\theta(x))$ is the well-known Kullback-Leibler divergence.

Hence, MLE is a minimizer of $D(\hat{f}\|f_\theta)$ over $\theta$. For minimization, $(\partial/\partial_\theta) D(\hat{f}\|f_\theta) = 0$. This implies $$\sum_{x\in \mathbb{X}} \hat{f}(x) \frac{\partial}{\partial\theta}\log f_\theta(x) = 0.$$ That is, $$\sum_{i=1}^n \frac{\partial}{\partial\theta}\log f_\theta(x_i) = 0,$$ called the score equation. Thus, one needs to solve the score equation for finding the MLE.

However, MLE is not robust when few of the $x_i$ are outliers.

The inference based on minimizing the power divergence is known to be robust against the outliers. Note that the power divergence $D_\alpha(\cdot\|\cdot)\to D(\cdot\|\cdot)$ as $\alpha\to 1$.

I roughly remember that the "improved" (or generalized) score equation corresponding to the power divergence to be something like $$\sum_{i=1}^n f_{\theta}(x_i)^c\frac{\partial}{\partial\theta}\log f_\theta(x_i) = 0,$$ for some $c>0$, potentially a function of $\alpha$.

The thing I would like to know is whether there is any way by which this generalized score equation can be shown to be derived from the power divergence as in the case of MLE as the score equation is derived from the Kullback-Leibler divergence.

PS: I asked a question along the same lines in stats.stackexchange which did not get much attention. The link is here.

• Might help if you defined terms and gave some context. – BruceET Sep 1 '15 at 20:44
• @BruceTrumbo: Thank you for your suggestion. I have now edited the question and elaborated. – Ashok Sep 5 '15 at 7:50
• Looks like you got your answer on Stats.SE? – user237392 Sep 9 '15 at 10:32
• @Bey: I am not fully convinced by that response. However, I accepted that answer as it atleast gave some justification and started some discussion. I am keeping it here to see if someone else gives a better explanation. If it is violation of etiquette, I can remove this post. – Ashok Sep 10 '15 at 6:36