Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What distribution do you obtain when you multiply a poisson distribution and a binomial distribution, and why? I'm assuming you obtain a poisson distribution.

share|improve this question
1  
What do you call to multiply two distributions? –  Did Oct 13 '12 at 0:41
    
The problem is underspecified. Are the two distributions independent? (I also don't know what it means to multiply two distributions. Are you taking the product of the corresponding random variables?) –  Qiaochu Yuan Oct 13 '12 at 1:04
    
Obviously the OP does not intend to answer the two comments above. Exquisite courtesy... –  Did Oct 13 '12 at 8:23
add comment

1 Answer 1

up vote 2 down vote accepted

Here is a physical example:

If we interpret $\mu$ and the probability that a photon produces an electron, then for a given number of photons entering the photo detector (say $n$) the probability distribution of electrons coming out is a binomial distribution with n trials and a probability of success of $\mu$.

$$ P(m)=\frac{n!}{m!(n-m)!}\mu^m (1-\mu)^{n-m}_{} =\binom{n}{m}\mu^m (1-\mu)^{n-m}_{} $$

The number of photons that go into the binomial distribution is the output of a Poisson distribution. We cannot get more electrons out than photons that went into the photo detector. We can sum up all the possible binomial distributions with a Poisson distribution weighting factor.

$$ P(m,\lambda,\mu)=\sum_{j=m}^{\infty} \frac{j!}{(j-m)!m!}\mu^m(1-\mu)^{j-m} \frac{\lambda^j e^{-\lambda}}{j!} $$ Simplify and bring terms that do not depend on j outside of the sum.

$$ P(m,\lambda,\mu)= \frac{\mu^m e^{-\lambda}}{m!} \sum_{j=m}^{\infty} \frac{\lambda^j(1-\mu)^{j-m}}{(j-m)!} $$ Let $n=j-m$

$$ P(m,\lambda,\mu)= \frac{\mu^m e^{-\lambda}}{m!} \sum_{n=0}^{\infty} \frac{\lambda^{n+m}(1-\mu)^{n}}{n!} $$

$$ P(m,\lambda,\mu)= \frac{(\lambda \mu)^m e^{-\lambda}}{m!} \sum_{n=0}^{\infty} \frac{(\lambda (1-\mu))^{n}}{n!} $$

$$ P(m,\lambda,\mu)= \frac{(\lambda \mu)^m e^{-\lambda \mu}}{m!} $$

This is a Poisson distribution with a mean of $\lambda \mu$!

Note: the "!" in the last line of text should be understood as excitement and a not a factorial.

share|improve this answer
    
Thank you for the help. –  shmiggens Oct 13 '12 at 1:54
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.