From: hrubin@odds.stat.purdue.edu (Herman Rubin) Subject: Re: How to find distribution for... Date: 4 Sep 1999 15:33:13 -0500 Newsgroups: sci.math Keywords: combinations of independent random variables In article <37D16CB6.B39DBBE7@eunet.yu>, Filip Miletic wrote: >Hello all. >I'm having trouble with the following problem: >Given three independent random variables X, Y, Z with standard normal >distributions > N(0,1), >find the distribution of: >U = ( X + YZ) / ( SQRT( 1 + Z^2) >Is there an (easy) way to do this? Seems that applying the inversion >theorem gets >you nowhere since you are stuck with common distribution density which >seems impossible >to integrate. I suspect there might be an elegant ad-hoc solution. There is a very simple way to do this, and it should also drop out with integration. If A_j are independent N(0,1) and c_j are constants, then \sum A_j*c_j / \sqrt(\sum c_j^2) is N(0,1). If now the c_j, or some of them, are replaced by random variables independent of the A_j, the result remains from the fact that probability is the expectation of conditional probability. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399 hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 ============================================================================== From: "Noel Vaillant" Subject: Re: How to find distribution for... Date: Sat, 4 Sep 1999 22:50:02 +0100 Newsgroups: sci.math > There is a very simple way to do this, and it should also drop > out with integration. > > If A_j are independent N(0,1) and c_j are constants, then > \sum A_j*c_j / \sqrt(\sum c_j^2) is N(0,1). If now the c_j, or > some of them, are replaced by random variables independent of > the A_j, the result remains from the fact that probability is > the expectation of conditional probability. This is excellent ! (I was sweating heavily) Specifically, the independence between Z and (X,Y) is useful when writing that for any bounded borel function f, the conditional expectation E[f(X,Y,Z)|Z]=g(Z) where, g is defined as g(z)=E[f(X,Y,z)]. Given a borel set A in R, take f(x,y,z)=1_A((x+yz)/sqrt(1+z^2)) where 1_A is the characteristic of A. Then: P(U in A) = E[f(X,Y,Z)]=E[ E[f(X,Y,Z)|Z] ] = E[g(Z)] = \int g(z) dm(z) (m being the distribution of Z, (measure on R)) But: g(z) = P(U_z in A) , where U_z = (X+zY)/(1+z^2) As pointed out by Mr Rubin , U_z is N(0,1) (Using the independence between X and Y). So g(z) = N(0,1)(A) (this is a constant) Finally P(U in A) = N(0,1)(A). Hence the distribution of U is N(0,1). Note that I am just stating more explictely what was already said in the previous post. Nothing original. I hope this may be useful to some. Regards. Noel. ------------------------------------------- Dr Noel Vaillant http://www.probability.net vaillant@probability.net