Re: Variation on binomial distribution

Ah, sorry. Yes, they are effectively independent. There is actually a
tiny amount of random (roughly normally distributed, zero mean, unsure
of the variance offhand) covariance among any pair of qualities, but it
is safe to ignore it and should be much easier to treat them as
independent. Certainly for my purposes an assumption of independence is
fine.

Ray Koopman wrote:
Are the qualities independent?

digfarenough@xxxxxxxxx wrote:
Hey all. I have a question I've been pondering over and can't quite
solve. There's a clear brute force approach, but I was hoping for a
better solution.

I have a set S of n objects that each have m binary qualities (say
black/white, square/triangle, rough/smooth, etc., represented as 0 or
1, say). Each object s from S has probability ps(m) that its quality m
is 1 (a success, binomially speaking). That is, the probability of a
given quality being found in an object is the same for all objects, but
each quality has a different probability of occurring.

On each of t trials, I select a quality of interest at random with
probability p(m) and note for each object whether its value for that
quality is 1 or 0 (success or failure).

After those t trials, I am interested in the distribution of the number
of noted 1s for each object, ignoring the quality from which the 1
came. That is, I want to know the number of objects for which a 1 came
up only once, the number for which a 1 came up only twice, etc. (number
of successes for each object).

If ps(m) didn't depend on m (that is, if the success probability for
each quality was the same), then I would simply have a binomial
distribution and the distribution of interest would be
t*Binomial[n,ps].

Simulating this shows that, as expected, the distribution is roughly
normal, and I'm quite fine using that as an approximation, but I'm
unclear on how to derive the mean and standard deviation of that
distribution, which is what I'd need if lacking an exact solution.

The brute force approach, which may be the only way, is to calculate
each possibility separately: There are t_choose_x ways in which an
object could have ended up with exactly x 1s, each of which would be
one term in a sum of products of probabilities. To demonstrate the sum,
we can consider two trials with two qualities m1 and m2. I figure the
probability of some s producing a failure in the comparisons in both
trials is (1 - p(m1)*ps(m1))*(1 - p(m2)*ps(m2)). The probability of
some s producing a success on exactly one trial is p(m1)*ps(m1)*(1 -
p(m2)*ps(m2)) + (1 - p(m1)*ps(m1))*p(m2)*ps(m2), and so forth.

As my values of t, n, and m are fairly large, however, that's not a
pretty prospects.
Anyone have an suggestions?
Thanks!

.

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