Suppose that:
$\Delta \sim G$, where $G$ is a distribution that is symmetric about the origin
I get a normally-distributed signal:
$\hat{\delta} \: |\Delta, \tau^2 \sim N(\Delta, \tau^2)$
How can I prove that the posterior mean $E(\Delta | \hat{\delta})$ is positive if and only if the signal $\hat{\delta}$ is positive.
(Intuition: The prior mean is zero, so the posterior mean should assume the sign of the signal $\hat{\delta}$)
This may be helpful. Note
$\hat{\delta}=\Delta+\epsilon$,
where $\epsilon$ is normally distributed with mean zero and variance $\tau^2$. Let this normal density be given by $\phi$. Note
$$P(\hat{\delta} \leq c)=\int_{-\infty}^{\infty}\int_{-\infty}^{y-c} \phi(y) g(x) dxdy $$
Here $g$ is the density function of $G$ (if this does not exist you can still use more general integrals).Density is obtained via Leibniz rule
$$f(c)=-\int_{-\infty}^{\infty} \phi(y) g(y-c) dy $$
The conditional density is
$$f(\Delta| \hat{\delta})=\frac{\phi(\hat{\delta}-\Delta) g(\Delta)}{f(\delta)}$$
And then
$$\mathbb{E}(\Delta| \hat{\delta})=\int_{-\infty}^{\infty}\Delta f(\Delta| \hat{\delta})d\Delta$$
$\Delta \phi(\hat{\delta}-\Delta) g(\Delta)$ is symmetric around $\hat{\delta}=0$. You need to show that
$$\int_{-\infty}^{\infty}\Delta\phi(\hat{\delta}-\Delta) g(\Delta)d\Delta>0$$
iff $\hat{\delta}>0$