Re: weighted sum of bernoulli random variables

From: Herman Rubin (hrubin_at_odds.stat.purdue.edu)
Date: 01/16/05 

Date: 16 Jan 2005 15:40:38 -0500

In article <zxtrdtp03rn3@legacy>, Fred <farengeti@yahoo.com> wrote:
>On Fri, 14 Jan 2005 09:54:22 -0000, David Jones wrote:
>>fred wrote:
>>> I'm trying to find the distribution of the following random
>>variable,
>>> Y

>>> n-1
>>> Y = Sum( X(i)*m^i )
>>> i=0

>>> Here m is a constant parameter in (0,1) and the X(i) are
>distributed
>>> Bernoulli( p ) and are independent.

>>> Alternatively, do you know of any distributions for X(i) for which
>>the
>>> distribution of Y is well defined? A location-scale family
>perhaps?

>>(i) the result for X(i) distributed as Normal is easy
>>(ii) similarly, results for stable distributions are easy
>>(iii) By "well-defined", do you mean having an explicit form for
>>distribution function, density, or probabilities? There is always a
>>"well-defined" answer if you work in terms of characteristic
>functions
>>or moment generating functions.
>>(iv) A guess is that you might be interested in n tending to infinity
>>for abs(m)<1. For the Bernoulli case, the limit distribution can be
>>one of thise cases where the distribution function is continuous but
>>not differentiable (e.g. for the case m=1/3, p=1/2). If you were
>>interested in the case m=1/2, p=1/2, the limiting distribution is
>>uniform on (0,2).There has been discussion of this here in the past
>>(one or two years ago).

>>David Jones

>There are many things I need to do with Y, so having an explicit form
>for it's distribution function would be ideal. Motivating this
>specific post, however, was an attempt to derive the expected value of

>Zn = log( Yn ).

P(Yn = 0) = (1-p)^n > 0. So the logarithm is minus
infinity with positive probability. Even the logarithm
of a binomial random variable is difficult.

>If I understand you correctly, in (iv) you are saying that for
>Bernoulli X(i) and general m,p in (0,1), the limiting distribution of
>Yn is not differentiable, which then means that the CLT doesn't hold
>(??) and I can't use the delta method to estimate moments of Zn.

The CLT never holds if |m| is not 1. If |m| < 1, the series
converges, and Y_n)(m) = m^{n-1}Y_n(1/m) if |m| > 1.

>So I need to estimate moments of Zn, and was hoping to do so by using
>the distribution of Yn. So I guess when I said "well-defined" I meant
>converging to a p.d.f. that I can compute expectations of using either
>straight integration or delta method and slutsky's or what not.

There is no problem with moments, and as above, the series
converges for |m| < 1, to a limiting distribution, which is
entirely singular or entirely absolutely continuous.

>Sorry if this is unclear, it's been a while since I've done
>statistics. Thanks again for any help.

There is no statistics involved; there is probability and
analysis. 

-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558
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