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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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