Consider the probability according to the distribution of X of landing outside an interval around
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
$f$ is what increasing with $|\mu_X-\mu_Y|$,
$f$ is called decreasing with $\sigma_Y$, and
$f=0$ 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 $\sigma_Y$.
