Let $\mu_Y$ 0<\alpha<1$(typically$\alpha=0.05$) and choose$\varepsilon>0$such thathas probability 95% according to the distribution of Y. In a sense this $$\alpha = \mathbb P(\left |X-\mu_X\right | > \varepsilon)$$ Now let $$f = \mathbb P(\left |Y-\mu_X\right | > \varepsilon) - \alpha$$ Then 1.$f$is what increasing with$|\mu_X-\mu_Y|$, 2.$f$is called decreasing with$\sigma_Y$, and 3.$f=0$when$\mu_X=\mu_Y$and$\sigma_X=\sigma_Y$. This approach comes not so much from information theory as from hypothesis testing, in particular the notion of statistical power. Then subtract 5% Some generalizing from this probability. Only problem is this can be 0 in some undesired cases. So it only satisfies 3 the case of your 4 conditions.the normal distribution seems possible, but note that in (3) we need the distribution to be determined by$\mu_Y$and$\sigma_Y$. 1 Consider the probability according to the distribution of X of landing outside an interval around$\mu_Y\$ that has probability 95% according to the distribution of Y. In a sense this is what is called statistical power. Then subtract 5% from this probability. Only problem is this can be 0 in some undesired cases. So it only satisfies 3 of your 4 conditions.