Analytical solution to the multivariate CDF given multivariate pdf Is there any way of approximating or analytically solving the below CDF (let's say even for $n\to\infty$)?
I am trying to find the below probability:
\begin{align}
     &P\left[X_{2}-X_{1} \leq 0,X_{3}-X_{1} \leq 0, X_{4}-X_{1} \leq 0 \right]\\ &= F_{\tilde{X_{2}},\tilde{X_{3}},\tilde{X_{4}}}(0,0,0)
\end{align}
Where we know all the distributions (normal with known mean and variance) and we also know that the random variables are correlated.This method of solving seems very tedious and does not generalize well if we have even more random variables. 
Edit: removed the integral term, since that implies that $X_{2}-X_{1}$ etc are independent. Sorry for the confusion
 A: Sorry, I think it is not true that $(X_2-X_1, X_3-X_1,...)$ are independent. Nevertheless, you can get an expression for the density of these variables: If $\Phi$ is a $C^1$-diffeomorphism (i.e. bijection and $\Phi$ and $\Phi^{-1}$ are both differentiable) then for any random variable $X$ with density $f_X$,
$$f_{\Phi(X)}(\tilde{x}) = f(\Phi^{-1}(\tilde{x})) |D\Phi^{-1}(\tilde{x})|$$
The map $(x_1,x_2,x_3) \mapsto (x_1, x_2 - x_1, x_3-x_1)$ is such a map. Hence,
$$f_{X_1,X_2-X_1,X_3-X_1}(x_1,\tilde{x}_2, \tilde{x}_3) = f_{X_1,X_2,X_3}(x_1,\tilde{x}_2+x_1, \tilde{x}_3+x_1) = f_{X_1}(x_1) f_{X_2}(\tilde{x}_2+x_1) f_{X_3}(\tilde{x}_3+x_1)$$
so
$$f_{X_2-X_1,X_3-X_1}(\tilde{x}_2, \tilde{x}_3) = \int_{\mathbb{R}} f_{X_1}(x_1) f_{X_2}(\tilde{x}_2+x_1) f_{X_3}(\tilde{x}_3+x_1) dx_1$$
from here you can also figure out the densities of $X_2-X_1$ and $X_3-X_1$ and it does not seem as if $f_{X_2-X_1,X_3-X_1} = f_{X_2-X_1} * f_{X_3-X_1}$...
However, now that we know $f_{X_2-X_1,X_3-X_1}$ we can write down an explicit expression for
$$P[X_2-X_1 < a, X_3-X_1 < b] = \int_{-\infty}^{a}\int_{-\infty}^{b} f_{X_2-X_1,X_3-X_1}(\tilde{x}_2, \tilde{x}_3) d\tilde{x}_2 d\tilde{x}_3$$
namely,
$$P[X_2-X_1 < a, X_3-X_1 < b] = \int_{-\infty}^{a}\int_{-\infty}^{b} \int_{\mathbb{R}} f_{X_1}(x_1) f_{X_2}(\tilde{x}_2+x_1) f_{X_3}(\tilde{x}_3+x_1) dx_1 d\tilde{x}_2 d\tilde{x}_3$$
Now you can reorganize integrals to
$$\int_{\mathbb{R}} f_{X_1}(x_1) \left(\int_{-\infty}^{a} f_{X_2}(\tilde{x}_2+x_1)d\tilde{x}_2 \right) \left(\int_{-\infty}^{b} f_{X_3}(\tilde{x}_3+x_1) d\tilde{x}_3 \right) dx_1 $$
and by substitution (for example for $i=2$) we get
$$\int_{-\infty}^{a} f_{X_2}(\tilde{x}_2+x_1)d\tilde{x}_2 = \int_{-\infty}^{a+x_1} f_{X_2}(x_2)dx_2 = \Phi(a+x_1)$$
so what you get is
$$\int_{\mathbb{R}} f_{X_1}(x_1) \Phi(a+x_1) \Phi(b+x_1) dx_1 $$
Doesn't look to good though... :-( Does that help?
