# Is the Chi-square divergence a Bregman divergence?

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?

• This seems more appropriate for Math, but still, why the downvote? – Sasho Nikolov Oct 12 '19 at 15:37
• 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? – Clement C. Oct 12 '19 at 18:41
• 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/… – Yonatan Oct 13 '19 at 2:04
• @Yonatan Of course, my bad :( – 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 $$1=0$$.