4
$\begingroup$

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 $b$?

$\endgroup$
3
$\begingroup$

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*}

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

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.