Calculating probability of an event involving two normally distributed random variables I have two normally distributed, independent random variables $S^{(1)}$ and $S^{(2)}$:
$$S^{(1)} \sim \mathcal{N}(\mu_1, \sigma_1^2),\ S^{(2)} \sim \mathcal{N}(\mu_2, \sigma_2^2)$$
Given two positive real numbers, $a_i$ and $a_j$, I need to calculate the probability of the event
$$ S^{(1)} \ge a_i\ \land\ (S^{(1)} - a_i > S^{(2)} -a_j) $$
Any clue about how I could achieve this? My main difficulty is that the two sub-events I am composing are not independent.
 A: Writing $X$, $Y$, $a_1$, and $a_2$ for $S^{(1)}$, $S^{(2)}$, $a_i$, and $a_j$, and writing $F_X$ and $F_Y$ for the CDFs of $X$ and $Y$:
$$\begin{align*}
P(X ≥ a_1 \text{ and } X - a_1 ≥ Y - a_2)
&= P(X ≥ a_1) P(X - a_1 ≥ Y - a_2 | X ≥ a_1) \\
&= (1 - F_X(a_1) P(X - Y ≥ a_1 - a_2 | X ≥ a_1) \\
&= (1 - F_X(a_1) [P(X - Y ≥ a_1 - a_2 | X ≥ a_1, Y < a_2) P(Y < a_2) +  P(X - Y ≥ a_1 - a_2 | X ≥ a_1, Y ≥ a_2) P(Y ≥ a_2)] \\
&= (1 - F_X(a_1) [F_Y(a_2) + (1 - F_Y(a_2)) P(X - Y ≥ a_1 - a_2 | X ≥ a_1, Y ≥ a_2)] .
\end{align*}$$
The last equality works because when $X ≥ a_1$ and $Y < a_2$, $X - Y ≥ a_1 - a_2$ follows algebraically.
This is a closed-form solution with the exception of the expression $P(X - Y ≥ a_1 - a_2 | X ≥ a_1, Y ≥ a_2)$, which can be thought of as a call to the CDF of the difference of two one-sided truncated normal distributions. Expressing this CDF in terms of $F_X$ and $F_Y$ or the standard normal CDF would likely be extremely messy (if it is even possible to evaluate all the integrals analytically), given two answers (1, 2) by wolfies on other problems involving truncated normals.
