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.

  • $\begingroup$ Take the derivative with respect to $a$ $\endgroup$ Dec 21, 2022 at 10:17
  • 1
    $\begingroup$ @AnneBauval Thanks! $\endgroup$
    – OnoL
    Dec 21, 2022 at 15:13
  • $\begingroup$ You are welcome. I didn't find it worth typing it as a more detailed answer ;-) $\endgroup$ Dec 21, 2022 at 15:16
  • 1
    $\begingroup$ @AnneBauval Fair enough~ $\endgroup$
    – OnoL
    Dec 21, 2022 at 15:17

1 Answer 1


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 $\theta=0$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .