Re: Numerical Integrate

*To*: mathgroup at smc.vnet.net*Subject*: [mg73005] Re: Numerical Integrate*From*: "Robert Dodier" <robert.dodier at gmail.com>*Date*: Sun, 28 Jan 2007 01:27:57 -0500 (EST)*References*: <ep7bsu$b7s$1@smc.vnet.net>

On Jan 24, 3:20 am, Bob Hanlon <hanl... at cox.net> wrote: > f1[x_]=PDF[NormalDistribution[0, 1],x]; > > f2[x_]=PDF[LogNormalDistribution[0, 1],x]; > > convolve[f_,g_, t_?NumericQ]:= > NIntegrate[f[u]*g[t-u],{u,0,t}]; This isn't correct -- the integration needs to be over (-inf, +inf). The above definition of convolution is OK in some contexts but not here -- essentially we need to marginalize over the the first variable, so the integration is over the support of f1 (namely -inf to +inf). > NIntegrate[convolve[f1,f2,t],{t,-Infinity,0}]//Re > > 0.5 This is also incorrect -- the convolution yields the sum of normal and lognormal variables, which has a skewed density since the lognormal is skewed (and the normal is symmetric). So the mass to the left of 0 is something less than 1/2. A simpler way to solve this particular problem is to see that Pr(X + Y < 0) is equal to Pr(Y > X) where X is the lognormal variable and Y is the normal. (Note Y and -Y have the same density.) Furthermore Pr(Y > X) = integral(Pr(Y > a) Pr(X = a), a). For Pr(Y > a) there is an exact result in terms of the Gaussian error function, so we only have to compute one numerical integration, not two -- this is a big win. I get Pr(Y > X) = 0.1759 (approximately). Hope this helps, Robert Dodier