Intuition for Prediction Under Bregman Loss

Elements of the Bregman family of loss functions, denoted $B(y, \hat{y})$, take the form:
$$B(y, \hat{y}) = \phi(y) - \phi(\hat{y}) - \phi'(\hat{y}) (y-\hat{y})$$ where $\phi: \mathcal{Y} \rightarrow R$ is any strictly convex function, and $\mathcal{Y}$ is the support of $Y$.

Several readers have asked for intuition for equivalence between the predictive optimality of $E[y|\mathcal{F}]$ and Bregman loss function $B(y, \hat{y})$.  The simplest answers come from the proof itself, which is straightforward.

First consider $B(y, \hat{y}) \Rightarrow E[y|\mathcal{F}]$.  The derivative of expected Bregman loss with respect to $\hat{y}$ is
$$\frac{\partial}{\partial \hat{y}} E[B(y, \hat{y})] = \frac{\partial}{\partial \hat{y}} \int B(y,\hat{y}) \;f(y|\mathcal{F}) \; dy$$
$$=  \int \frac{\partial}{\partial \hat{y}} \left ( \phi(y) - \phi(\hat{y}) - \phi'(\hat{y}) (y-\hat{y}) \right ) \; f(y|\mathcal{F}) \; dy$$
$$=  \int (-\phi'(\hat{y}) - \phi''(\hat{y}) (y-\hat{y}) + \phi'(\hat{y})) \; f(y|\mathcal{F}) \; dy$$
$$= -\phi''(\hat{y}) \left( E[y|\mathcal{F}] - \hat{y} \right).$$
Hence the first order condition is
$$-\phi''(\hat{y}) \left(E[y|\mathcal{F}] - \hat{y} \right) = 0,$$
so the optimal forecast is the conditional mean, $E[y|\mathcal{F}]$.

Now consider $E[y|\mathcal{F}] \Rightarrow B(y, \hat{y})$. It's a simple task of reverse-engineering. We need the f.o.c. to be of the form
$$const \times \left(E[y|\mathcal{F}] - \hat{y} \right) = 0,$$
so that the optimal forecast is the conditional mean, $E[y|\mathcal{F}]$. Inspection reveals that $B(y, \hat{y})$ (and only $B(y, \hat{y})$) does the trick.

One might still want more intuition for the optimality of the conditional mean under Bregman loss, despite its asymmetry.  The answer, I conjecture, is that the Bregman family is not asymmetric! At least not for an appropriate definition of asymmetry in the general $L(y, \hat{y})$ case, which is more complicated and subtle than the $L(e)$ case.  Asymmetric loss plots like those in Patton (2014), on which I reported last week, are for fixed $y$ (in Patton's case, $y=2$ ), whereas for a complete treatment we need to look across all $y$. More on that soon.

[I would like to thank -- without implicating -- Minchul Shin for helpful discussions.]