Re: variance of product of 2 independent variables

*To*: mathgroup at smc.vnet.net*Subject*: [mg87095] Re: variance of product of 2 independent variables*From*: dh <dh at metrohm.ch>*Date*: Mon, 31 Mar 2008 04:51:41 -0500 (EST)*References*: <fsl27i$gab$1@smc.vnet.net>

Hi Rudy, obvioulsy the equation <a^2><b^2> - <a>^2<b>^2 = v(a)<b> +v(b)<a> + v(a)v(b) is wrong. You can see this easily by considering the units. This also gives a hint how to correct: <a^2><b^2> - <a>^2<b>^2 = v(a)<b>^2 +v(b)<a>^2 + v(a)v(b) hope this helps, Daniel Dankwort, Rudolf C wrote: > Hello Ben - > > > > I have a question about the subject matter. To review, you sent Frank > Brand the following: > > > > <<If a and b were completly uncorrelated (not even non-linear > correlations among them), > then you can compute the variance of their product quite easily > > v(ab) := < a^2b^2 > - < ab >^2 = <a^2><b^2> - <a>^2<b>^2 = v(a)<b> + > v(b)<a> + v(a)v(b); > v(a)=<a^2>-<a>^2, v(b)=<b^2>-<b>^2 > > here v(.) denotes variance, <.> denotes mean. > Note that we do not have to assume normal distributions for a and b, > essential is that their are uncorrelated, hence the means of products > factor into products of means.>> > > If <a> = 1000 and <b> = 0.001, and v(a) = 100 and v(b) = 1e-10 (in other > words, both a and b have 1% standard deviations), then I compute v(ab) = > 100*0.001 + 1e-10*1000 + 100*1e-10 ~ 0.1 > which is obviously wrong (<ab> = 1.000 and std deviation would be > sqrt(0.1) = 0.31. > > Help! Da stimmt was nicht! > > Rudy dankwort > Phoenix AZ USA >