Ben, Ravi, Chuck, and Haneef, Note that the standard normal density can be written (in pseudo-TeX) as 1/sqrt(2*pi) * exp[- ( x/sqrt(2) )^2].
The exponential on the right is a special case of the stretched exponential function exp[- (x/sqrt(tau))^beta]. The stretched exponential function has a nontrivial density and distribution. However, the nth moment of this density is tau^(n)/beta * Gamma[(n)/beta]. The substitution of n=3/2 (not 1/2), tau=sqrt(2), and beta=2, and multiplying by 1/sqrt(2*pi) yields the Mathematica result below. See http://en.wikipedia.org/wiki/Stretched_exponential_function. The use of n=3/2 rather than 1/2 is required because the stretched exponential function is already a function for the first moment. Joe Ben Bolker <bol...@ufl.edu> Sent by: r-help-boun...@r-project.org 04/11/2010 01:09 AM To Ravi Varadhan <rvarad...@jhmi.edu> cc "r-h...@stat.math.ethz.ch" <r-h...@stat.math.ethz.ch> Subject Re: [R] Expectation of E(x^1/2) And Mathematica says In[2]:= 1/Sqrt[2 Pi] Integrate[Exp[-x^2/2] Sqrt[x],{x,0,Infinity}] 3 Gamma[-] 4 Out[2]= ------------- 3/4 2 Sqrt[Pi] (I suppose there's probably a change-of-variables trick to do this ...) in R: > gamma(3/4)/(2^(3/4)*sqrt(pi)) [1] 0.4110895 Ravi Varadhan wrote: > Chuck showed how to do this: > > fn <- function(x) sqrt(x) * dnorm(x) > >> integrate(fn, 0, Inf) > 0.4110895 with absolute error < 4.7e-05 > > So the (almost) exact answer is 0.4110895 + 1i * 0.4110895 > ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
