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 $A$.


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