Upper bound on difference between Gaussian CDFs

Question

May I know if there are any non-trivial upper bounds $f$ on the following:

$$\Phi(a + \Delta) - \Phi(a) \leq f(a, \Delta)$$

for $\Phi$ the CDF of a standard normal and all $a, \Delta > 0$.

Thanks!

Answer 1

Let $X$ be a standard normal random variable. Then you want an upper bound of

$$ P(X>a) - P(X>a+\Delta). $$

Now this post gives $P(X>a) \leq \frac{1}{a\sqrt{2\pi}}\exp(-a^2/2)$ and this other post gives $$P(X>a+\Delta) \geq \frac{e^{-1/2}(1- e^{-(a+\Delta)-(a+\Delta)^2/2})}{\sqrt{2\pi}(a+\Delta)}.$$ Therefore a choice of $f(a,\Delta)$ would be

$$ f(a,\Delta) := \frac{1}{a\sqrt{2\pi}}\exp(-a^2/2) - \frac{e^{-1/2}(1- e^{-(a+\Delta)-(a+\Delta)^2/2})}{\sqrt{2\pi}(a+\Delta)}. $$

Answer 2

According to the Mean value theorem: if $f(x): [a, b] \rightarrow \mathbb{R}$ is a continuous function on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, where $a < b$, there exists some $c \in (a, b)$ such that

$$ f(b)-f(a) = f'(c)(b-a). $$

Applying this theorem to our example ($f(x) = \Phi(x)$, $a = a$, $b = a + \Delta$),

$$ \Phi(a + \Delta) - \Phi(a) = \Phi'(c)\left((a + \Delta) - a\right), \text{ where } c \in (a, a+\Delta). $$

$$ \Phi(a + \Delta) - \Phi(a) = \varphi(c)\Delta, \text{ where } \varphi(c) = \frac{1}{\sqrt{2\pi}}e^{-\frac{-c^2}{2}} \text{ is a p.d.f. of the standard normal distribution} \Rightarrow. $$

$$ \Phi(a + \Delta) - \Phi(a) = \varphi(c)\Delta < \left|\text{ since } 0 < a < c \right| < \varphi(a)\Delta. $$

The choice of $f(a, \Delta)$ is

$$ f(a, \Delta) := \frac{\Delta}{\sqrt{2\pi}}e^{-\frac{a^2}{2}} $$