# An inequality involving a normal integral

The following inequality "seems" true by simulation, but an analytic justification/disproof doesn't seem straightforward. The integral part does not admit a closed-form expression in general, although one can show that the LHS and RHS are equal when $$\theta=0$$. Viewing the inequality as a function of $$\theta$$ and taking derivative does not seem useful. So I was wondering if there is any technique or counterexample that can help. Any hints/suggestions will be highly appreciated.

For all $$\theta<0$$ and each $$a\geqslant 0$$, we have $$\Phi(\theta)-\Phi(\theta-a)>\int_{-a}^a \Phi(x+\theta)\phi(x-\theta)dx,$$ where $$\Phi(\cdot)$$ and $$\phi(\cdot)$$ denote the standard normal cdf and pdf, respectively.

• Take the derivative with respect to $a$ Dec 21, 2022 at 10:17
• @AnneBauval Thanks!
– OnoL
Dec 21, 2022 at 15:13
• You are welcome. I didn't find it worth typing it as a more detailed answer ;-) Dec 21, 2022 at 15:16
• @AnneBauval Fair enough~
– OnoL
Dec 21, 2022 at 15:17

Let $$I(\theta, a) = \Phi(\theta) - \Phi(\theta-a) - \int_{-a}^a \Phi(x+\theta)\phi(x-\theta)dx$$ You want to show that $$I(\theta, a)>0$$ $$\forall a\ge 0$$.

In the case where $$a=0$$: $$I(\theta, 0) = \Phi(\theta) - \Phi(\theta) - 0 \equiv 0$$.

Note that $$\frac{\partial}{\partial a} \left( \int_{-a}^a f(x)dx \right) = f(-a)+f(a)$$

In the case where $$a > 0$$, differentiate $$I(\theta, a)$$ with respect to $$a$$: $$\frac{\partial}{\partial a}I(\theta, a) = \phi(\theta-a) - \Phi(-a+\theta)\phi(-a-\theta) - \Phi(a+\theta)\phi(a-\theta) \\ = \phi(\theta-a) - \Phi(\theta-a)\phi(-\theta-a) - \Phi(\theta+a)\phi(\theta-a) \\ = (1-\Phi(\theta+a))\phi(\theta-a) - \Phi(\theta - a)\phi(-\theta-a)$$

We must show that $$(1-\Phi(\theta+a))\phi(\theta-a) - \Phi(\theta - a)\phi(-\theta-a) \ge 0 \\ \Leftrightarrow \frac{1-\Phi(\theta+a)}{\phi(-\theta-a)} \ge \frac{\Phi(\theta - a)}{\phi(\theta-a)} \\ \Leftrightarrow \frac{\Phi(-\theta-a)}{\phi(-\theta-a)} \ge \frac{\Phi(\theta - a)}{\phi(\theta-a)}$$

(We have applied the identities $$\phi(-x) \equiv \phi(x)$$ and $$\Phi(-x) \equiv 1-\Phi(x)$$ in several places.)

The LHS and RHS of the above inequality are mirror images of each other about the y-axis. Since $$G(\theta) := {\Phi(\theta-a)} / {\phi(\theta-a)}$$ is a positive, increasing function, the LHS is indeed greater than the RHS when $$\theta\le 0$$, with equality iff $$\theta=0$$.