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?

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.
