Weighted sums of Bernoulli random variables
in [Math]
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From: atrovantes on 10 May 2010 05:18 Hi, does anyone know any results about weighted sums of Bernoullis? A simple example follows. Let X_n be i.i.d. Bernoulli random variables. What's the distribution of \sum_{k=1}^N X_k k ? Or if not the distribution maybe some other results... Non-trivial of course (e.g. I can find the mean myself). Thanx in advance.
From: danheyman on 10 May 2010 10:55 On May 10, 9:18 am, atrovantes <eba...(a)gmail.com> wrote: > Hi, > does anyone know any results about weighted sums of Bernoullis? > > A simple example follows. Let X_n be i.i.d. Bernoulli random variables. What's the distribution of > \sum_{k=1}^N X_k k ? > > Or if not the distribution maybe some other results... Non-trivial of course (e.g. I can find the mean myself). > > Thanx in advance. The sum of iid Bernoullis is binomial.
From: Ray Vickson on 10 May 2010 11:16 On May 10, 6:18 am, atrovantes <eba...(a)gmail.com> wrote: > Hi, > does anyone know any results about weighted sums of Bernoullis? > > A simple example follows. Let X_n be i.i.d. Bernoulli random variables. What's the distribution of > \sum_{k=1}^N X_k k ? > > Or if not the distribution maybe some other results... Non-trivial of course (e.g. I can find the mean myself). > > Thanx in advance. For the specific case of S = X_1 + 2*X_2 + 3*X_3 + ... + N*X_N the moment-generating function is G(z) = (q + p*z)*(q + p*z^2)*(q +p*z^3)*...*(q+p*z^N), where q = 1-p. For small N one can expand out the product and collect coefficients of z^j to find the probabilities P{S = j}, but for large N it is probably a hopeless task. However, finding some of the moments is relatively easy. For any sum of the form T = sum{i=1}^N a_i*X_i with constants a_i and iid Bernoulli X_j, the mean = sum_{i} a_i * p = p*sum a_i, while for any _independent_ terms, Var(sum) = sum (var), hence Var(T) = sum_{i} (a_i)^2 *p*q = p*q*sum (a_i)^2. For the special case of T = S (with a_i = i) you can use G(z) to find some of the higher moments, while for the general case of S you can use the Laplace transform. R.G. Vickson
From: atrovantes on 10 May 2010 10:40 > On May 10, 6:18 am, atrovantes <eba...(a)gmail.com> > wrote: > > Hi, > > does anyone know any results about weighted sums of > Bernoullis? > > > > A simple example follows. Let X_n be i.i.d. > Bernoulli random variables. What's the distribution > of > > \sum_{k=1}^N X_k k ? > > > > Or if not the distribution maybe some other > results... Non-trivial of course (e.g. I can find the > mean myself). > > > > Thanx in advance. > > For the specific case of S = X_1 + 2*X_2 + 3*X_3 + > ... + N*X_N the > moment-generating function is G(z) = (q + p*z)*(q + > p*z^2)*(q > +p*z^3)*...*(q+p*z^N), where q = 1-p. For small N one > can expand out > the product and collect coefficients of z^j to find > the probabilities > P{S = j}, but for large N it is probably a hopeless > task. That's what I was thinking, but you put it in a beautiful and clearer form using the moment-generating function. Thanx. So now the question is (still for the specific case of S^N = X_1 + 2*X_2 + ... + N*X_N) how does G^N(z) = (q + p*z)*(q + p*z^2)*...*(q + p*z^N) behaves. Well, there is a pattern. We can write G^N(z) as follows: G^N(z) = p^N * (b + z)*(b + z^2)*...*(b + z^N), where b=q/p. And we have: b + z (b + z)*(b + z^2) = b^2 + b*z + b*z^2 + z^3 (b^2 + b*z + b*z^2 + z^3)*(b + z^3) = b^3 + b^2*z + b^2*z^2 + (b+b^2)z^3 + b*z^4 + b*z^5 + z^6 ... [it looks better on paper.] I have a headache now, but I think something will come out of this. > > However, finding some of the moments is relatively > easy. For any sum > of the form T = sum{i=1}^N a_i*X_i with constants a_i > and iid > Bernoulli X_j, the mean = sum_{i} a_i * p = p*sum > a_i, while for any > _independent_ terms, Var(sum) = sum (var), hence > Var(T) = sum_{i} > (a_i)^2 *p*q = p*q*sum (a_i)^2. For the special case > of T = S (with > a_i = i) you can use G(z) to find some of the higher > moments, while > for the general case of S you can use the Laplace > transform. > > R.G. Vickson
From: Robert Israel on 10 May 2010 15:31
atrovantes <ebaerr(a)gmail.com> writes: > Hi, > does anyone know any results about weighted sums of Bernoullis? > > A simple example follows. Let X_n be i.i.d. Bernoulli random variables. > What's the distribution of > \sum_{k=1}^N X_k k ? > > Or if not the distribution maybe some other results... Non-trivial of > course (e.g. I can find the mean myself). > > Thanx in advance. Let S_N = sum_{k=1}^N X_k k. P(S_N = m) = f_N(m)/2^N where f_N(m) is the number of partitions of m into parts that are all <= N. This sort of thing has been well studied by number theorists, I think. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada |