Consider two random variables, $x$ and $y$, which are independent draws from the same normal probability distribution: $x,y \sim N(\mu, \sigma^2)$.
I would like to calculate the probability that $x$ and $y$ fall on the same side of some threshold $c$. (That is, $x>c$ and $y>c$ OR $x<c$ and $y<c$.)
I think I am right that this equates to the sum of two probabilities:
1) $P(x>c,y>c) = (1 - \Phi(c))\times(1 - \Phi(c)) = (1 - \Phi(c))^2$
2) $P(x<c,y<c) = \Phi(c)\times\Phi(c) = \Phi(c)^2$
where $\Phi$ is the cumulative distribution function of the normal distribution. The answer I am looking for is therefore $(1 - \Phi(c))^2 + \Phi(c)^2$.
Questions:
1) Does the above make sense? I think that my presentation of this is probably somewhat laborious.
2) I would like to be able to relate this probability to the parameters $\mu$ and $\sigma^2$ as well as the threshold $c$, but I'm stuck with this! I can see from simulations that as $\mu$ approaches $c$, the probability decreases, as it does when $\sigma^2$ increases, but I would like to be able to express this analytically if possible. Any suggestions?
EDIT:
3) I would also be interested in knowing whether there is a way of generalising this idea to other probability distributions. I have a feeling that this is probably something quite general, and would appreciate being pointed towards other things I could read about problems like this.
If $X, Y \sim \operatorname{Normal}(\mu, \sigma^2)$, then $$\Pr[(X < c) \cap (Y < c)] = \Pr\left[\left(\frac{X - \mu}{\sigma} < \frac{c-\mu}{\sigma}\right) \cap \left(\frac{Y - \mu}{\sigma} < \frac{c-\mu}{\sigma}\right)\right],$$ and since these are independent, this becomes $$\Phi((c-\mu)/\sigma)^2.$$ Similarly, $$\Pr[(X > c) \cap (Y > c)] = (1 - \Phi((c - \mu)/\sigma))^2.$$ Therefore, the total probability is $$1 - 2 \Phi((c - \mu)/\sigma) + 2 \Phi((c - \mu)/\sigma)^2.$$