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 $b$?


1 Answer 1


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*}

  • $\begingroup$ Thank you. I want to add that the point b is unique because $\gamma$ is convex. $\endgroup$ Apr 27, 2019 at 14:10
  • $\begingroup$ Thanks @DavidNguyen. That is correct, and $b$ is therefore the optimal point. $\endgroup$
    – Gordon
    Apr 27, 2019 at 14:14

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