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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
>
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