[Scilab-users] n dimensional normal distribution

fujimoto2005 fujimoto2005 at gmail.com
Tue Mar 27 00:05:51 CEST 2018


Dear all.
Thank you for your answers.
I am sorry that the reply is delayed.

My original problem is v=∫∫..∫f(X1,X2,..,Xn,Y)n(X1,X2,..,Xn,Y)dX1dX2..dXndY
where n(X1,X2,..,Xn,Y) is a n+1 dimensional normal density function.

This is a problem of a mathematical finance and this integral is usually
calculated by numerical integration using random numbers.
In my problem f() has the following special form.
f(X1,X2,..,Xn,Y)=g(Y)I(x1>=k1)...I(Xn>=Kn) where I() is an indication
function.
In this case, v can be decomposed as v=∫g(Y){∫..∫n(X1,X2,..,Xn
|Y)dX1dX2..dXn}n(Y)dY= ∫g(Y)Prob[(X1>=K1)& ..&(Xn>=Kn)|Y}n(Y)dY where
n(X1,X2,..,Xn |Y) is a normal density conditiona to Y, Prob(..|Y) is a
probability conditional to Y and n(Y) is a one dimensional normal density
function. 
This integration can be calculated firstly when n = 1 since Prob(x1>=K1|Y)
has an analytical approximate function.
When n = 2, it works still fine because Prob[(X1>=K1
&X2>=K2)|Y}=1-Prob[X1<=K1]-Prob[X2<=K2]+Prob[X1<=K1&X2<=K2] and I hear that
Prob[X1<=K1&X2<=K2] have an analytical approximate functions.

I posted the question because I can calculate it first if there is an
analytical approximate function of Prob[(X1>=K1)& ..,(Xn>=Kn)|Y} for large
n.
However, as Samuel demonstrated explicitly and Stéphane implicitly by
referring to R, it seems that there is no choice but to calculate
Prob[(X1>=K1)& ..,(Xn>=Kn)|Y} for large n by using random numbers.
I understand my prospects doesn’t go well because there is no such function
for large n.
When using random numbers, it is the best way to calculate v in the original
form.

Best regards.




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