InverseFunction of a CDF

*To*: mathgroup at smc.vnet.net*Subject*: [mg102767] InverseFunction of a CDF*From*: "Christian Winzer" <cw420 at cam.ac.uk>*Date*: Wed, 26 Aug 2009 07:55:06 -0400 (EDT)

Hello, I am calculating the cumulated density function of a variable x based on the variables m and z using the following code: (* Distribution of the stochastic variable m*) M = NormalDistribution[0, 1]; mPDF = PDF[M]; mCDF = CDF[M]; (* Weight of m in x*) w = 0.5; (* Distribution of the stochastic variable z*) Z = NormalDistribution[0, 1]; zPDF = PDF[Z]; zCDF = CDF[Z]; (*This section calculates the distribution of x = w * m + z *) (*Conditional cumulated density xCDFm[x,m] of the stochastic variable x conditional on m *) xCDFm = Evaluate[zCDF[(#1 - w * #2)/(Sqrt[1 - w^2])]] &; (*Unconditional cumulated density xCDF[x] of the stochastic variable x *) xCDF = NIntegrate[xCDFm[#1, m]* mPDF[m], {m, -\[Infinity], \[Infinity]}] &; The calculation seems to work, because I can plot xCDF correctly. But for some reason, I can not get the inverse function. I have tried 3 different ways: In[78]:= x = InverseFunction[xCDF][#1] &; Evaluate[x[0.5]] Out[79]= InverseFunction[NIntegrate[xCDFm[#1, m] mPDF[m], {m, -\[Infinity], \[Infinity]}] &][0.5] In[80]:= Evaluate[Quantile[xCDF, 0.5]] Out[81]= Quantile[NIntegrate[xCDFm[#1, m] mPDF[m], {m, -\[Infinity], \[Infinity]}] &, 0.5] In[82]:= Solve[xCDF[x] == 0.5, x] During evaluation of In[73]:= NIntegrate::inumr: The integrand (E^(-(m^2/2)) (1+Erf[0.816497 (-0.5 m+((xCDF^(-1))[Slot[<<1>>]]&))]))/(2 Sqrt[2 \[Pi]]) has evaluated to non-numerical values for all sampling points in the region with boundaries {{-\[Infinity],0.}}. >> During evaluation of In[73]:= NIntegrate::inumr: The integrand (E^(-(m^2/2)) (1+Erf[0.816497 (-0.5 m+((xCDF^(-1))[Slot[<<1>>]]&))]))/(2 Sqrt[2 \[Pi]]) has evaluated to non-numerical values for all sampling points in the region with boundaries {{-\[Infinity],0.}}. >> During evaluation of In[73]:= NIntegrate::inumr: The integrand (E^(-(m^2/2)) (1+Erf[0.816497 (-0.5 m+((xCDF^(-1))[Slot[<<1>>]]&))]))/(2 Sqrt[2 \[Pi]]) has evaluated to non-numerical values for all sampling points in the region with boundaries {{-\[Infinity],0.}}. >> During evaluation of In[73]:= General::stop: Further output of NIntegrate::inumr will be suppressed during this calculation. >> During evaluation of In[73]:= Solve::ifun: Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >> Out[83]= {} In this particular example I know without calculation, that xCDF is the CDF[NormalDistribution[0,Sqrt[1.25]]], however I want to use different distributions for m and z in the future, and then I can't always tell the distribution of x, so I will need to calculate it. Any ideas how I could do that? Thank you very much in advance and best wishes, Christian

**Follow-Ups**:**Re: InverseFunction of a CDF***From:*DrMajorBob <btreat1@austin.rr.com>

**Re: InverseFunction of a CDF***From:*DrMajorBob <btreat1@austin.rr.com>