Inverse of cumulative normal distribution

APrendergast · March 8th, 2006, 10:31 am

Hello all,I'm trying to generate a normally-distributed random variable by applying the inverse of the cumulative normal distribution to a uniform (0,1) rv. The approximation I'm using seems to be the standard one, of this form:where t = sqrt(ln(1/p^2)) and the c's and d's are ugly constants.My question is this: it seems as if the resulting r.v. will have standard deviation 1. Is there a change of variables we can make to give a normal r.v. whose standard deviation is some chosen \rho?Thanks in advance,a.p.

Alan · March 8th, 2006, 5:03 pm

Here is an alternative method that works well: http://mathworld.wolfram.com/Box-Muller ... on.htmlBut, to answer your original question, justmultiply the r.v. you generate by \rho.regards,

eredhuin · March 20th, 2006, 12:05 am

I second the recommendation by Alan. Go to www.nr.com and search for the numerical recipes code on box-mueller. Alternately, google Algorithm AS241 from Applied Statistics aka PPND16 algorithm 241. It has 10^-16 acccuracy for transforming uniform -> normal.

Halliron · March 29th, 2006, 11:45 am

Box-Mueller isn't suitable for certain monte carlo methods e.g. using stratified samples, or SOBOL.It's definately better for general simulation though,