How to show that the CDFs of two Normal distributions cross? 
If one normal distribution has a larger mean and a larger standard deviation, prove that the CDFs of the two distributions have a crossing point.

If the assertion is false, can you provide a counter-example?
 A: Let $X_1 \sim N(\mu_1,\sigma_1)$ and $X_2 \sim N(\mu_2,\sigma_2)$, where $\mu_1 < \mu_2$ and $\sigma_1 < \sigma_2$. Let $F_1(x)$ and $F_2(x)$ be the CDF of $X_1$ and $X_2$, respectively.

*

*Note that the mean and median of normal distributions are equal. Then $F_1(\mu_1) = 0.5$ and $F_2(\mu_2) = 0.5$, which implies that $F_1(\mu_2) >  F_2(\mu_2)$, since $F_1$ is an increasing functions.


*Define $z_1 = \frac{x_0-\mu_1}{\sigma_1}$ and $z_2 = \frac{x_0-\mu_2}{\sigma_2}$ and take $x_0$ such $z_1 < z_2$. Solving $z_1 < z_2$ you will obtain $x_0 < \frac{\sigma_2\mu_1 - \sigma_1\mu_2}{\sigma_2+\sigma_1}$.
For any $x_0$ that satisfies the inequity above, $F_1(x_0) <  F_2(x_0)$, by construction. Note also that $x_0 <  \mu_2$.
Now, as $F_1$ and $F_2$ are continuous functions and
a)  $F_1(x_0) <  F_2(x_0)$
b) $F_1(\mu_2) >  F_2(\mu_2)$.
where $x_0 <  \mu_2$, then exist a $x \in (x_0,\mu_2)$ such $F_1(x) = F_2(x)$. Therefore $F_1$ and $F_2$ have a crossing point.
A: It can be fun and useful to identify the key idea behind a result like this.  To do so, let's generalize the situation as much as possible so that incidental characteristics, peculiar to Normal distributions, are stripped away.
Let $F$ and $G$ be two distribution functions.  To say that they have a "crossing point" means there are numbers $a \lt b$ for which $F(a) \le G(a)$ and $F(b) \ge G(b).$  What minimal assumptions are needed to assure this condition?
When $F$ and $G$ are absolutely continuous, they have density functions $f = F^\prime$ and $g = G^\prime.$  If there is a number $a$ for which $x\lt a$ implies $f(x)\le g(x),$ then clearly
$$F(a) = \int_{-\infty}^a f(x)\,\mathrm{d}x \le \int_{-\infty}^b g(x)\,\mathrm{d}x = G(a).$$
In the same way, if there is a number $b$ for which $x \gt b$ implies $f(x) \le g(x),$ we deduce $F(b)\ge G(b),$ whence there will be a crossing point.
To summarize:

When $F$ and $G$ have densities, and one of the densities eventually dominates (exceeds) the other for sufficiently large values of $|x|,$ then $F$ and $G$ have a crossing point.

Let's apply this simple observation to the question.  In the case of two Normal distributions, $f$ and $g$ are never zero, allowing us to compare them by taking their logarithms.  Letting the parameters of these distributions be $(\mu_F,\sigma_F)$ and $(\mu_G,\sigma_G)$ we are thereby motivated to compare $-\log(f(x))$ = $\log(\sigma_F) + (x-\mu_F)^2/(2\sigma_F^2)$ to its counterpart for $G.$
It's simplest to do the comparison by subtracting one log from the other (which, by the properties of logarithms, is equivalent to evaluating the ratio of the densities $g(x)/f(x)$):
$$\begin{aligned}
\log(g(x)) - \log(f(x)) &=  \log(\sigma_F) - \log(\sigma_G) + \frac{(x-\mu_F)^2}{2\sigma_F^2} - \frac{(x-\mu_G)^2}{2\sigma_G^2}\\
&= \left(\frac{1}{2\sigma_F^2} - \frac{1}{2\sigma_G^2}\right)x^2 + Bx + C
\end{aligned}$$
where $B$ and $C$ are numbers determined by the parameters -- the details don't matter.
As we all well know, any polynomial in $x$ like this one eventually behaves like its leading term--all smaller terms are asymptotically negligible in comparison.  That leading term is the coefficient of $x^2$ provided $\sigma_F \ne \sigma_G.$  In such cases, the difference of logs becomes either extremely negative as $x^2$ grows large or it becomes extremely positive, depending on whether $\sigma_F$ exceeds $\sigma_G$ or not, respectively.  In either case this shows the log of one of the densities eventually dominates the log of the other for sufficiently large values of $|x|.$  The desired conclusion follows.
