# Probabilty of inequality for 3 or more independent random variables

If $$X, Y, Z$$ are 3 independent random variables, and
$$X \sim N(\mu_x,\sigma_x^2)$$, $$Y \sim N(\mu_y,\sigma_y^2)$$, $$Z \sim N(\mu_z,\sigma_z^2)$$

I know how to calculate the inequality probability of any pair of $$X, Y, Z$$ (ex: $$P(X\ge Y$$), $$P(X\ge Z$$), $$P(Y\ge Z$$) ) from the answer of this question

However, from the answer of this question , I can't get $$P(X\ge Y \ge Z)$$ by
$$P(X\ge Y \ge Z) = P(X \ge Y) * P(Y \ge Z)$$

How do I calculate the probability that $$P(X\ge Y \ge Z)$$?

Since $$X,Y,Z$$ are independent, and you know their densities u can form $$W = (X,Y,Z)$$ and the density of $$W$$ is equal to the product of densities of $$X,Y,Z$$.
$$\mathbb P(X\geq Y\geq Z) = \mathbb P(\{\omega\in\Omega : X(\omega) \geq Y(\omega) \geq Z(\omega) \}) = \mathbb P(\{ \omega \in \Omega : W(\omega)\in A\}) = \mu_{_W}(A)$$
Where $$A = \{ (x,y,z) \in \mathbb R^3 : x\ge y\geq z\}$$, and $$\mu_{_W}(A) = \int_{A}g_{_W}(s)d\lambda_3(s)$$,
From what was said before, $$g_{_W}(s) = g_{_X}(s) \cdot g_{_Y}(s) \cdot g_{_Z}(s)$$, where by $$g$$ with indice $$X,Y,Z$$ respectively I mean their densities.
Now you only have to integrate that function over the set $$A$$.