Convergence rate of the ratio of two normal distributions? I am interested in the behaviour of  $\log \frac{\Phi(A-x)}{\Phi(B-x)}$, when ${x\rightarrow -\infty}$ and $\Phi$ is the normal cumulative distribution function, $A<B$, and $A,B\in {\mathbb R}$. It seems like this logratio converges to $0$, but I was wondering if it is possible to check the convergence rate?
 A: The normal CDF $\Phi(x)$ approaches 1 for $x\longrightarrow \infty$ and approaches 0 for $x\longrightarrow -\infty$.
Since the CDF is strictly monotonic $\Phi(x) < \Phi(x+\epsilon)$ for $\epsilon > 0$, the ratio $\Phi(A-x)/\Phi(B-x)$ is always smaller than 1. To determine the value of the ratio for $x\longrightarrow -\infty$, just differentiate (L'Hospital's rule) to obtain an analytical expression, from which you'll get that $$\frac{\Phi(A-x)}{\Phi(B-x)}\longrightarrow 1\quad \mathrm{for} \quad x\longrightarrow -\infty$$
So, the logarithmic value approaches 0.
Edit:
The convergence rate can be calculated from the integral description of the CDF. Take  $$\Phi(x) = \int_{-\infty}^x c*e^{-(x-\mu)^2/2\sigma^2} dx$$
where c is just some constant ($1/\sqrt{2\pi\sigma^2}$) Without limiting generality, take the shifted standard normal distribution for $\Phi(A-x)$ (or $B$). The ratio is then just 
$$\frac{\Phi(A-x)}{\Phi(B-x)} = \frac{\int_{-\infty}^x c*e^{-(x-A)^2/2\sigma^2} dx}{\int_{-\infty}^x c*e^{-(x-B)^2/2\sigma^2} dx}$$
Since both have the same the same variance, most things vanish after expanding the term. Omitting the constant terms, since they're not interesting for convergence it reduces to:
$$\frac{\Phi(A-x)}{\Phi(B-x)} = \beta e^{-\alpha x(B-A)}$$
where $\alpha$ and $\beta$ are some constants.
When you take the logarithm of the ratio, you'll get a linear term. Ergo the convergence is linear.
