# Bregman Mean of a Distribution

In a paper (link), author writes, given that $$\gamma:R\rightarrow \bar{R}$$ is a convex function, $$dom_{\gamma}:=\{x\in R:\gamma(x)<+\infty\}$$ is a non-empty open set and $$\gamma$$ a closed proper differentiable function in the interior of $$dom_{\gamma}$$, $$d$$ is Bregman divergence $$d_{\gamma}(x,x')=\gamma(x)-\gamma(x')-\gamma'(x')(x-x')$$ Define the Bregman mean as the unique point $$b$$ in the support of $$\mu$$ satisfying $$\int d_{\gamma}(b,x)\mu(dx)=\min_{m\in dom_{\gamma}}\int d_{\gamma}(m,x)\mu(dx)$$.

He says that it is very easy to obtain $$b$$ by differentiating: $$b=\gamma'^{-1}[\int\gamma'(x)\mu(dx)]$$.

Can anyone explain to me the definition and how he gets the formula for $$b$$?

Note that \begin{align*} f(m) &= \int d_{\gamma}(m,x)\mu(dx)\\ &=\int \big[\gamma(m)-\gamma(x)-\gamma'(x)(m-x)\big]\mu(dx)\\ &=\gamma(m) - \int \big[\gamma(x)+\gamma'(x)(m-x)\big]\mu(dx). \end{align*} Then, \begin{align*} \frac{df}{dm} = \gamma'(m) - \int \gamma'(x)\mu(dx), \end{align*} and the critical point is given by \begin{align*} b = \big(\gamma'\big)^{-1}\Big( \int \gamma'(x)\mu(dx)\Big). \end{align*}

• Thank you. I want to add that the point b is unique because $\gamma$ is convex. – David Nguyen Apr 27 at 14:10
• Thanks @DavidNguyen. That is correct, and $b$ is therefore the optimal point. – Gordon Apr 27 at 14:14