Re: convolution of lognormals (or other broad distributions)
From: bq (fouad_qaqish_at_yahoo.com)
Date: 08/18/04
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Date: 18 Aug 2004 13:31:07 -0700
nmdc69@hotmail.com (Eric) wrote in message news:<v6rk3kwchkcw@legacy>...
> On Wed, 18 Aug 2004 01:05:08 -0700, Frederick Umminger wrote:
> >
> >"Eric" <nmdc69@hotmail.com> wrote in message
> ><a
> href="news://ef4a7818.0408170614.6487a1e@posting.google.com...">news://ef4a7818.0408170614.6487a1e@posting.google.com...>
> >> Hello, I need a numerical method to compute the convolution
> >> of a probability distribution function p(x) with itself N times,
> >> when p(x) is broad (namely, it extends over several decades of x)
> >> for example p(x) cab be a broad lognormal distribution. Any ideas
> >> on how to do this?
> >>
> >
> >snip
> >
> >> Actually all I need is the first moment of the convoluted
> distribution,
> >> not the whole function, so there may be better methods...
> >> thanks,
> >> Eric
> >
> >If you really only need the first moment, then this is trivial. If
> X1,..XN
> >are independent identically distributed random variables with
> distribution
> >p(x), then X1+..+XN is a random variable with distribution p(x)
> convolved
> >with itself N times. So the first moment is
> <X1+...+XN>=<X1>+...+<XN>=N<X1>.
> >If you want higher moments, then you can compute cumulants in a
> similar
> >manner and use those to compute the moments.
> >
> >-Frederick Umminger
>
>
> Thanks. Sorry, I wrote something incorrectly. What I need is the
> expectation value of the LOGARITHM of the sum X1+...+XN.
> I don't think you can approximate this well with the moments
> if the distribution is large.
Suppose that each X_i has mean \mu and variance \sigma^2,
If N is large, write
log(sum) = log(N) + log(mean),
then use a Taylor expansion of log(mean) to get
E [log(sum)] approximately = log N + log(\mu) - c^2 / (2 N)
where c := \sigma / \mu
This should work well as long as c is small and N is large.
bq
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