Probability of ordered sequence of Gaussian Distributions Suppose there are 5 cars racing on a match. Suppose we know that the distributions of their finishing time are roughly in some Gaussian distributions $N(\mu_i,\sigma_i^2), i=1,2,3,4,5$. How can I determine the probability of any ordered sequence of their finishing time, say {2, 1, 3, 5, 4}?
 A: Using independence of $X_1,X_2,X_3,X_4,X_5$ we get that
$$
P(X_1>X_2>X_3>X_4>X_5) = \int_\Omega 1_{\{X_1>X_2>X_3>X_4>X_5)\}}dP
$$
$$
=\int_{\mathbb{R}^5} 1_{\{(x_1>x_2>x_3>x_4>x_5)\}}d(X_1,X_2,X_3,X_4,X_5)(P)(x_1,x_2,x_3,x_4,x_5)
$$
$$
 = \int_{-\infty}^\infty \int_{-\infty}^{x_1} \int_{-\infty}^{x_2} \int_{-\infty}^{x_3} \int_{-\infty}^{x_4} 1dX_5(P)(x_5)dX_4(P)(x_4)dX_3(P)(x_3)dX_2(P)(x_2)dX_1(P)(x_1)
$$
$$
=\int_{-\infty}^\infty \int_{-\infty}^{x_1} \int_{-\infty}^{x_2} \int_{-\infty}^{x_3} \int_{-\infty}^{x_4} \prod_{i=1}^5\frac{1}{\sqrt{2\pi\sigma_i^2}}\exp\left(-\frac{(x_i-\mu_i)^2}{2\sigma_i^2}\right)dx_5dx_4dx_3dx_2dx_1
$$
I don't think you can get any closer to an answer than this.
A: For your specific problem,given that the normal distributions are independent, it holds that  $(X_1,...,X_5)\sim N(\mu,D)$ where $\mu=(\mu_1,...,\mu_5)$ and $D$ is a diagonal matrix with $D_{ii}=\sigma^2_i$. The event {1,2,3,4,5} corresponds to 
$$P(X_5>X_4>X_3>X_2>X_1)\\
=P(X_5-X_4>0,X_4-X_3>0,X_3-X_2>0,X_2-X_1>0).$$
Indeed $Y:=(Y_1,Y_2,Y_3,Y_4)=(X_5-X_4,X_4-X_3,X_3-X_2,X_2-X_1)=f(X)$ where f is a linear map represented by a matrix M. Thus Y is normal distributed as well with parameters
$$Y\sim N(M\mu,MDM^t).$$
Since $Y$ is normal distributed it has a well known density $p(y_1,y_2,y_3,y_4)$, which we use to find the desired probability
$$P(Y_1>0,...,Y_4>0)=\int_{\mathbb{R}_+^4}p(y_1,y_2,y_3,y_4) d(y_1,y_2,y_3,y_4).$$
(Which probably needs a computer)
