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?
 A: 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:

*

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

*$k < \mu$, 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)

A: 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 $t\ge0$.
A: 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*}
