Re: delayed function assignment
- To: mathgroup at smc.vnet.net
- Subject: [mg88187] Re: delayed function assignment
- From: Szabolcs Horvát <szhorvat at gmail.com>
- Date: Sun, 27 Apr 2008 07:00:55 -0400 (EDT)
- Organization: University of Bergen
- References: <fv1fec$eru$1@smc.vnet.net>
leigh wrote: > Dear Mathgroup, > > I am trying to integrate the pdf of a chi-square mixture distribution. > If I define the pdf as > > pdf2[0] := .25 > \!\(pdf2[x] := .5*\(\(1/2\^\(1/ > 2\)\)\/Gamma[1/2]\) \(x\^\(1/2 - 1\)\) \[ExponentialE]\^\(\(-x > \)/ > 2\) + .25*\(1\/\(2*Gamma[1]\)\) \[ExponentialE]\^\(\(-x > \)/2\)\) > > I can then obtaing the cumulative distribution function symbolically > as > > In[14]:=\!\($B"i(Bpdf2[x] \[DifferentialD]x\) > > Out[14]=\!\(\(-0.25`\)\ 2.718281828459045`\^\(\(-0.5`\)\ x\) + 0.5`\ \ > Erf[0.7071067811865476`\ \@x]\) > > I can get numerical values using for example > > In[31]:=\!\(1 - .25 - $B"i(B\_0\%12 pdf2[x] \[DifferentialD]x\) > > Out[31]=0.000885691 > > However if I try to define this as a function of the statistic > obtained, say > > \!\(cum2[x] := 1 - .25 - $B"i(B\_0\%x pdf2[t] \[DifferentialD]t\) > > Then > > In[33]:=cum2[9] > > Out[33]=cum2[9] > > Doesn't evaluate. What should I do to obtain the value as output. Please read the tutorial on defining functions and the introduction to patterns: http://reference.wolfram.com/mathematica/tutorial/DefiningFunctions.html http://reference.wolfram.com/mathematica/tutorial/Introduction-Patterns.html The proper function definition in your application has the form f[x_] := x and not f[x] := x (The latter has a different meaning.)