1
$\begingroup$

Is there any way of approximating or analytically solving the below CDF (let's say even for $n\to\infty$)?

I am trying to find the below probability: \begin{align} &P\left[X_{2}-X_{1} \leq 0,X_{3}-X_{1} \leq 0, X_{4}-X_{1} \leq 0 \right]\\ &= F_{\tilde{X_{2}},\tilde{X_{3}},\tilde{X_{4}}}(0,0,0) \end{align}

Where we know all the distributions (normal with known mean and variance) and we also know that the random variables are correlated.This method of solving seems very tedious and does not generalize well if we have even more random variables.

Edit: removed the integral term, since that implies that $X_{2}-X_{1}$ etc are independent. Sorry for the confusion

$\endgroup$
  • $\begingroup$ If the whole vector $(X_1,...,X_4)$ is a multivariare Gaussian then you could diagonalise the covariace matrix, i.e. assume that they are independent one dimensional Gaussians. Then I think (not sure here) that $X_i - X_1$ are also independent (Gaussians). Then $P[X_2-X_1<a,X_3-X_1<b,...]=P[X_2-X_1<a]*...$ is a product of $\Phi$ functions... $\endgroup$ – Fabian Werner Jul 10 at 9:54
  • $\begingroup$ The point that makes this tricky is that they are not independent and that is a main feature of the problem. Assuming that they are independent solves a different problem $\endgroup$ – Highness Jul 10 at 10:45
0
$\begingroup$

Sorry, I think it is not true that $(X_2-X_1, X_3-X_1,...)$ are independent. Nevertheless, you can get an expression for the density of these variables: If $\Phi$ is a $C^1$-diffeomorphism (i.e. bijection and $\Phi$ and $\Phi^{-1}$ are both differentiable) then for any random variable $X$ with density $f_X$, $$f_{\Phi(X)}(\tilde{x}) = f(\Phi^{-1}(\tilde{x})) |D\Phi^{-1}(\tilde{x})|$$

The map $(x_1,x_2,x_3) \mapsto (x_1, x_2 - x_1, x_3-x_1)$ is such a map. Hence,

$$f_{X_1,X_2-X_1,X_3-X_1}(x_1,\tilde{x}_2, \tilde{x}_3) = f_{X_1,X_2,X_3}(x_1,\tilde{x}_2+x_1, \tilde{x}_3+x_1) = f_{X_1}(x_1) f_{X_2}(\tilde{x}_2+x_1) f_{X_3}(\tilde{x}_3+x_1)$$

so

$$f_{X_2-X_1,X_3-X_1}(\tilde{x}_2, \tilde{x}_3) = \int_{\mathbb{R}} f_{X_1}(x_1) f_{X_2}(\tilde{x}_2+x_1) f_{X_3}(\tilde{x}_3+x_1) dx_1$$

from here you can also figure out the densities of $X_2-X_1$ and $X_3-X_1$ and it does not seem as if $f_{X_2-X_1,X_3-X_1} = f_{X_2-X_1} * f_{X_3-X_1}$...

However, now that we know $f_{X_2-X_1,X_3-X_1}$ we can write down an explicit expression for

$$P[X_2-X_1 < a, X_3-X_1 < b] = \int_{-\infty}^{a}\int_{-\infty}^{b} f_{X_2-X_1,X_3-X_1}(\tilde{x}_2, \tilde{x}_3) d\tilde{x}_2 d\tilde{x}_3$$ namely, $$P[X_2-X_1 < a, X_3-X_1 < b] = \int_{-\infty}^{a}\int_{-\infty}^{b} \int_{\mathbb{R}} f_{X_1}(x_1) f_{X_2}(\tilde{x}_2+x_1) f_{X_3}(\tilde{x}_3+x_1) dx_1 d\tilde{x}_2 d\tilde{x}_3$$

Now you can reorganize integrals to

$$\int_{\mathbb{R}} f_{X_1}(x_1) \left(\int_{-\infty}^{a} f_{X_2}(\tilde{x}_2+x_1)d\tilde{x}_2 \right) \left(\int_{-\infty}^{b} f_{X_3}(\tilde{x}_3+x_1) d\tilde{x}_3 \right) dx_1 $$

and by substitution (for example for $i=2$) we get

$$\int_{-\infty}^{a} f_{X_2}(\tilde{x}_2+x_1)d\tilde{x}_2 = \int_{-\infty}^{a+x_1} f_{X_2}(x_2)dx_2 = \Phi(a+x_1)$$

so what you get is

$$\int_{\mathbb{R}} f_{X_1}(x_1) \Phi(a+x_1) \Phi(b+x_1) dx_1 $$

Doesn't look to good though... :-( Does that help?

$\endgroup$
  • $\begingroup$ I am horribly sorry for the mistake that I made in my problem statement. I have removed the integral now to not imply independence between the variables. So the main problem becomes how to derive an expression for $F_{\tilde{X_{2}},\tilde{X_{3}},\tilde{X_{4}}}$ $\endgroup$ – Highness Jul 10 at 10:50
  • $\begingroup$ The integral still stays valid: The general definition of 'having a density' for a touple $X,Y$ of random variables is that $P[(X,Y) \in W] = \int_W f_{X,Y}(x,y) d(x,y)$ and for $W = (-\infty,a] \times (-\infty,b]$ you get the integral that you wrote earlier... so that seems to be fine, right? $\endgroup$ – Fabian Werner Jul 10 at 10:55
  • $\begingroup$ Yes.You are right. $\endgroup$ – Highness Jul 10 at 11:11
  • $\begingroup$ That is basically just an application of the substitution theorem: en.m.wikipedia.org/wiki/Integration_by_substitution. If you don‘t see it let me know then I will extend the answer... $\endgroup$ – Fabian Werner Jul 10 at 12:05
  • $\begingroup$ On second thought, I am not sure about you going from $f_{x_{1},..x_{3}}(x_{1},x_{2}+x_{1},..)$ to $f_{x_{1}}(x_{1})f_{x_2}(x_{2}+x_{1})..$. Doesn't that assume independence between the terms which is not the case? $\endgroup$ – Highness Jul 10 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.