Is the Chi-squared divergence $\sum_{i} \frac{(x(i)-y(i))^2}{x(i)}$ a Bregman divergence? I.e., can it be written as $\phi(x) - \phi(y) - \langle\phi'(y),x-y\rangle$?

If so, what is the potential function $\phi(x)$ from which it is generated? If not, does there exists a Bregman divergence which bounds Chi-square divergence from above?

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    $\begingroup$ This seems more appropriate for Math, but still, why the downvote? $\endgroup$ – Sasho Nikolov Oct 12 '19 at 15:37
  • $\begingroup$ Not a proof, but since this paper introduces a generalization of Bregman divergences and then show this generalization includes the chi-squared divergence, it's fair to assume the "usual" Bregman divergences do not. As for the last question: KL divergence? $\endgroup$ – Clement C. Oct 12 '19 at 18:41
  • $\begingroup$ Thanks! I it an interesting work.. however, I didn't spot there proof of non-existence (they indeed show their generalization includes the chi-squared, however, they don't show -- as far I understood -- it cannot be realized as a Bregman divergence). + I think that the Chi square upper bounds the KL and not vice-versa: cstheory.stackexchange.com/questions/30693/… $\endgroup$ – Yonatan Oct 13 '19 at 2:04
  • $\begingroup$ @Yonatan Of course, my bad :( $\endgroup$ – Clement C. Oct 13 '19 at 17:17

$\chi^2$-divergence is not a Bregman divergence.

I'll show it for sample size $n=1$. We would have $$ (x-y)^2/x=f(x)-f(y)-f'(y)(x-y)$$ If $y=0$ and $x>0$ this says $$x=f(x)-f(0)-xf'(0),$$ $$1=\frac{f(x)-f(0)}x-f'(0).$$ Taking $x\to 0^+$ this gives the contradiction $1=0$.

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