Finding the probability X > Y I am curious on how to find the probability a random variable X is greater than Y (given their means and variance, and also that they are independent.)
I stumbled across this post: probability of one random variable being greater than another
but am unsure how to answer this question with not necessarily normal distributions.

$$P(X > Y) = P(X-Y > 0)  = 1-P(X-Y \leq 0)$$
So this problem is equivalent to finding the cumulative distribution function $F(Z \leq z)$ with $Z = X + (-Y)$, for the value $z=0$.

*

*The general expression for the distribution function of a sum of two variables is a convolution $$P(X-Y \leq z) = \int_{-\infty}^{\infty} f_X(x) F_Y(z+x) dx$$


*In special cases the integral has a closed form and often you can look up the solution.
In particular a stable distribution family, which is closed under addition, will have a way to describe the distribution of a linear sum (but the normal distribution is the only case which has finite variance).
The difference between two Poisson distributions is a Skellam distribution
The difference between two Gamma distributions is often more difficult. Difference of Gamma random variables bit special cases like the pdf of the difference of two exponentially distributed random variables is simple.
The difference between two t-distributions is a special case of the Behrens-Fisher distribution.


*If you only know the variance and means then you can not find an exact distribution, but you could find bounds using limits for inequalities which come in several forms (which one to use depends on your situation and goal).
