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?

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    $\begingroup$ All 24 orderings of the list $X_1,X_2,X_3,X_4$ are equally likely, so $P(X_1>X_2>X_3>X_4)=1/24$. $\endgroup$ – Mike Earnest May 15 '18 at 13:21
  • $\begingroup$ Do you have that the means and standard deviations of each random variable are equal? Without that, this problem becomes much more complicated. $\endgroup$ – InterstellarProbe May 15 '18 at 13:25
  • $\begingroup$ yes they are all different $\mathcal{N}(\mu_i,{\sigma^2}_i), i=1,2,3,4$. I want an expression just like above $\endgroup$ – user3503589 May 15 '18 at 13:29
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    $\begingroup$ I see. I cannot think of a way to write this as a single integral, but you can write it as a double integral:$$P(X_1>X_2>X_3>X_4)=\int_{-\infty}^\infty\int_{\infty}^{x_2} (1-F_1(x_2))f_2(x_2)f_3(x_3)F_4(x_3)\,dx_3 \,dx_2$$ $\endgroup$ – Mike Earnest May 15 '18 at 16:59
  • $\begingroup$ Thank you @MikeEarnest , but I already can write it as a double integral . I was trying to get an expression using just a single integral . $\endgroup$ – user3503589 May 15 '18 at 17:01

\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}

  • $\begingroup$ But it’s not a standard Normal? $\endgroup$ – user3503589 May 15 '18 at 14:10
  • $\begingroup$ @user3503589: This applies to all i.i.d. random variables. $\endgroup$ – hypernova May 15 '18 at 14:11
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    $\begingroup$ In the case they are standard normal , I can immediately conclude that it’s 1/n! for the case of n ordered mutually independent standard normals using the symmetricity $\endgroup$ – user3503589 May 15 '18 at 14:13
  • $\begingroup$ The distribution functions are all different if the random variables have different means . $\endgroup$ – user3503589 May 15 '18 at 14:14
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    $\begingroup$ @user3503589: You are right. In that case, the formula would be involved. As you need to figure out $F_4(x_3){\rm d}F_3(x_3)$ artificially, which does not yield an elegant expression... $\endgroup$ – hypernova May 15 '18 at 14:19

It is an iterated integral such as this

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

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


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


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