How can I compute an expression for $P(X_1>X_2>X_3>X_4)$ if $X_1,X_2,X_3,X_4$ are normal and mututally independent?

Question

How can I compute an expression for $P(X_1>X_2>X_3>X_4)$ if $X_1,X_2,X_3,X_4$ are normal and mututally independent?

I managed to compute $P(X_1>X_2>X_3)$ as follows but the same technique does not seems to be working for 4 variables since I cannot change variables in an efficient way. For the case with 3 random variables I can show that

$P(X_1>X_2>X_3)=\int \int \int_{x_1>x_2>x_3} f_1(x_1)f_2(x_2)f_3(x_3)dx_1dx_2dx_3 =\int_{-\infty}^{\infty}dx_3 \int_{x_3}^{\infty}dx_2 \int_{x_2}^{\infty}f_1(x_1)f_2(x_2)f_3(x_3)=\int_{-\infty}^{\infty}dx_3 \int_{x_3}^{\infty}f_2(x_2)f_3(x_3)[1-F_1(x_2)]dx_2=\int_{-\infty}^{\infty}dx_2\int_{-\infty}^{x_2}f_2(x_2)f_3(x_3)[1-F_1(x_2)]dx_3=\int_{-\infty}^{\infty} F_3(x_2)[1-F_1(x_2)]f_2(x_2)dx_2=\int_{-\infty}^{\infty}F_3(y)[1-F_1(y)]f_2(y)dy$

In the case of 4 variables I am unable to change the variables in a nice way to get a simplifies expression as above. Can somebody help me here?

Answer 1

\begin{align} &\mathbb{P}\left(X_1>X_2>X_3>X_4\right)\\ &=\int_{x_1>x_2>x_3>x_4}f(x_1)f(x_2)f(x_3)f(x_4){\rm d}x_1{\rm d}x_2{\rm d}x_3{\rm d}x_4\\ &=\int_{-\infty}^{\infty}f(x_1){\rm d}x_1\int_{-\infty}^{x_1}f(x_2){\rm d}x_2\int_{-\infty}^{x_2}f(x_3){\rm d}x_3\int_{-\infty}^{x_3}f(x_4){\rm d}x_4\\ &=\int_{-\infty}^{\infty}f(x_1){\rm d}x_1\int_{-\infty}^{x_1}f(x_2){\rm d}x_2\int_{-\infty}^{x_2}F(x_3)f(x_3){\rm d}x_3\\ &=\int_{-\infty}^{\infty}f(x_1){\rm d}x_1\int_{-\infty}^{x_1}f(x_2){\rm d}x_2\int_{-\infty}^{x_2}F(x_3){\rm d}F(x_3)\\ &=\int_{-\infty}^{\infty}f(x_1){\rm d}x_1\int_{-\infty}^{x_1}\frac{1}{2}F^2(x_2)f(x_2){\rm d}x_2\\ &=\int_{-\infty}^{\infty}f(x_1){\rm d}x_1\int_{-\infty}^{x_1}\frac{1}{2}F^2(x_2){\rm d}F(x_2)\\ &=\int_{-\infty}^{\infty}\frac{1}{6}F^3(x_1)f(x_1){\rm d}x_1\\ &=\int_{-\infty}^{\infty}\frac{1}{6}F^3(x_1){\rm d}F(x_1)\\ &=\frac{1}{24}F^4(x)|_{-\infty}^{\infty}\\ &=\frac{1}{24}. \end{align}

Answer 2

It is an iterated integral such as this

$$P(X_1\ge X_2\ge X_3\ge X_4) = \int_{x_4=-\infty}^{\infty}\int_{x_4}^{\infty}\int_{x_3}^{\infty}\int_{x_2}^{\infty} \dfrac{1}{\sqrt{16\pi^4\sigma_1\sigma_2\sigma_3\sigma_4}}e^{-{\sum_{i=1}^{4} \dfrac{(x_i-\mu_i)^2}{2\sigma_i^2}}}dx_1dx_2dx_3dx_4$$

$\int_{c=-\infty}^{\infty}\int_{c}^{\infty}\int_{b}^{\infty} \dfrac{1}{\sqrt{8\pi^3}}e^{-{\frac{(a^{2}+b^{2}+c^{2})}{2}}}dadbdc =\frac{1}{6}$

http://www.wolframalpha.com/input/?i=%5Cint_%7Bc%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cint_%7Bc%7D%5E%7B%5Cinfty%7D%5Cint_%7Bb%7D%5E%7B%5Cinfty%7D+(1%2Fsqrt(8%5Cpi%5E3))e%5E%7B-%7B(a%5E%7B2%7D%2Bb%5E%7B2%7D%2Bc%5E%7B2%7D)%2F2%7D%7Ddadbdc

when all $a, b, c$ are $N(0,1)$ just like somebody mentioned above for four variables.

Answer 3

Interestingly, if $X_1,X_2,\ldots,X_n$ are normal and mutually independent, you can show that $$\mathbf P(X_1 > X_2>X_3 > \ldots > X_n) = \frac{1}{n!}$$ using essentially the same approach as in @hypernova's answer.