Duncan Murdoch writes: >On Mon, 13 Oct 2003 10:35:17 +0200, "Caroline Le Gall" ><[EMAIL PROTECTED]> wrote: > >>Hello all, >> >>Is someone can help me ? >>If X is distributed as a gaussian distribution with m as mean and s as >>standard deviation, what is the distribution of 1/X ? >>Particularly, what is the mean and the standard deviation of 1/X ? > >Neither the mean nor the s.d. exist, since the density of X is >positive at 0. I don't think the distribution has a standard name. >You can use standard transformation techniques to work out the density >or CDF from the Gaussian one.
True enough, though you might be able to get a reasonable approximation if most of the distribution (say plus or minus two standard deviations) is bounded well away from zero. You could then use a Taylor series approximation to the function f(x)=1/x to get f(x)=f(m)+(x-m)f'(m)+other terms. We hope that these other terms are insignificant. Then the expectation is Ef(x)=E[f(m)+(x-m)f'(m)]=f(m)+(m-m)*f'(m)=1/m and Var(f(x))=Var(f(m)+(x-m)f'(m))= (f'(m))^2*Var(x) =s^2/m^4. My math may be a bit rusty, but someone will jump in if I made a mistake. I would expect the distribution to be skewed left, with the skewness being most evident when the bulk of the distribution creeps closer to zero. Steve Simon, [EMAIL PROTECTED], Standard Disclaimer. The STATS web page has moved to http://www.childrens-mercy.org/stats. . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
