# Is $P(|X_1|>k)\le P(|X_2|> k)$ when $X_i\sim N(\mu_i,\sigma^2)$ and $|\mu_2| \ge |\mu_1|$?

Suppose $$X_1\sim N(\mu_1,\sigma^2)$$ and $$X_2\sim N(\mu_2,\sigma^2)$$ where $$\mu_2\ge \mu_1$$.

Since $$\mu_2\ge \mu_1$$, based on a characterization of stochastic ordering, we can say that

$$P(X_1>c)\le P(X_2> c) \quad \text{ for any constant }c\,.$$

Now suppose $$|\mu_2| \ge |\mu_1|$$ instead of $$\mu_2 \ge \mu_1$$.

Can we also say that $$P(|X_1|>k)\le P(|X_2|> k)\quad \text{ for any constant }k\,?$$

This is certainly the case if $$E|X_2|\ge E|X_1|$$, but does this hold?

$$|X_i|$$ has a folded normal distribution and its mean is given by

$$E|X_i|=\sigma\sqrt{\frac 2{\pi}}\exp\left\{-\frac{\mu_i^2}{2\sigma^2}\right\}+\mu_i\left(1-2\Phi\left(-\frac{\mu_i}{\sigma}\right)\right)$$

But it is not immediately clear to me if $$E|X_2|\ge E|X_1|$$. Is there any simpler way to approach this?

We can show this without having to deal with integrals, or the normal density function. It arises from a symmetry property, and the normal density of $$x$$ decreasing as $$|x|$$ increases.

First, note that this is trivial for $$k \leq 0$$, since then $$\mathbb{P}(|X| > k) = 1$$ for all $$\mu$$. We therefore only need to examine the case where $$k > 0$$.

Let $$f(\mu; k) := \mathbb{P}(|X| > k \,;\, \mu) = 1 - \Phi\left(\frac{k-\mu}{\sigma}\right) + \Phi\left(\frac{-k-\mu}{\sigma}\right).$$

This is symmetric in $$\mu$$, since $$\Phi(x) = 1 - \Phi(-x)$$:

$$f(-\mu; k) = 1 - \Phi\left(\frac{k+\mu}{\sigma}\right) + \Phi\left(\frac{-k+\mu}{\sigma}\right) = 1 + \Phi\left(\frac{-k-\mu}{\sigma}\right) - \Phi\left(\frac{k-\mu}{\sigma}\right) = f(\mu;k).$$

We can therefore restrict ourselves to considering positive values of $$\mu$$.

We now differentiate by $$\mu$$:

$$f'(\mu;k) = \frac{\textrm{d}}{\textrm{d}\mu} \mathbb{P}(|X| > k \,;\, \mu) = \phi\left(\frac{k-\mu}{\sigma}\right) - \phi\left(\frac{-k-\mu}{\sigma}\right),$$

where $$\phi$$ is the normal density function. For the proposed property to hold, we require $$f'$$ to be positive for all $$\mu \geq 0$$, which is equivalent to

$$\phi\left(\frac{k+\mu}{\sigma}\right) \leq \phi\left(\frac{k-\mu}{\sigma}\right) \quad \textrm{for all } \mu \geq 0.$$

This is true if and only if $$|k+\mu| \geq |k-\mu|$$, since $$\phi(x)$$ decreases as $$|x|$$ increases. $$k + \mu > 0$$, since $$k > 0$$ and $$\mu \geq 0$$, so we have two cases:

1. $$k \geq \mu$$, and we require $$k+\mu \geq k-\mu$$, i.e. $$\mu \geq 0$$, which is true.
2. $$k < \mu$$, and we require $$k+\mu \geq \mu-k$$, i.e. $$k \geq 0$$, which is true.

Therefore, the property is true for all $$k$$.

Some examples using R:

f <- function(mu, k) 1 - pmax(0, pnorm(k, mu, 1) - pnorm(-k, mu, 1))
curve(f(x, 0), from = -5, to = 5)
curve(f(x, 1), from = -5, to = 5)
curve(f(x, 10), from = -5, to = 5)


This also holds for any other density function that's "symmetric-decreasing". Here are some examples for the Cauchy distribution:

f <- function(mu, k) 1 - pmax(0, pcauchy(k, mu, 1) - pcauchy(-k, mu, 1))
curve(f(x, 0), from = -5, to = 5)
curve(f(x, 1), from = -5, to = 5)
curve(f(x, 10), from = -5, to = 5)

• Do you actually prove that $P(|X|>k)$ is an increasing function of $|\mu|$ when $X$ is normal with mean $\mu$? Jul 4, 2021 at 13:28
• Yes. In fact, this proves it's an increasing function for any continuous distribution that's "symmetric-decreasing" around its centre point. I'd say "around its mean", but this applies to the Cauchy distribution, that doesn't have a mean. See the code examples at the end. Jul 4, 2021 at 13:29

A possible approach: wlog we may assume $$\mu_1 \ge 0$$, $$k \ge 0$$.

$$P(|X_2|>k) - P(|X_1|>k) = \int_0^{\infty}(f_2(k+t) - f_1(k+t)) - (f_1(-k-t) - f_2(-k-t)) \,dt$$

and it would suffice to show that the integrand is nonnegative for all $$t\ge0$$.

You can reduce your question to the case you understand. Let $$s_i$$ be the sign of $$\mu_i$$, that is, $$s_i \in \{-1,1\}$$ and $$s_i \mu_i = |\mu_i|$$. Since the question is about the marginals of $$(X_1,X_2)$$ we can use any coupling between them.

Take $$X_1 = \mu_1 + Z$$ and $$X_2 = \mu_2 + s_1 Z$$ where $$Z \sim N(0,\sigma^2)$$. These will have the correct marginals. Let $$X'_1 = |\mu_1| + s_1 Z$$ and $$X'_2 = |\mu_2| + s_1 s_2 Z$$.

\begin{align*} \mathbb P(|X_1| > k ) = \mathbb P(|s_1 X_1 | > k) &= \mathbb P(|s_1 \mu_1 +s_1 Z| > k) \\ &= \mathbb P(|X'_1| > k) \\ &\le \mathbb P(|X'_2| > k) \\ &= \mathbb P(|s_2 \mu_2 + s_2 s_1 Z| > k) \\ &= \mathbb P(|\mu_2 + s_1 Z| > k) \\ &= \mathbb P(|X_2| > k). \end{align*}