2
$\begingroup$

I have a situation where a distribution is dependent on 2 variables, one of which follows the poisson distribution, and the other the normal distribution, and I want to establish the method of calculating the meand and spread for the dependent variable.

Specifically

I have an algortihm whcih matches names against a bad guy list. The bad guy list is static. The matching algorith can identify 0, 1, 2, or more matches. ie a Poisson distribution. So if I only apply one name I can calcualte the lamda for the distribution.

But, I don't just apply one name I apply many names, once per day, so one day one I might have 100, day 2, 150, etc. The number of names applied each day follows a normal distribution. For this I can calculate the mean and s.d.

What I want to be able to find is the number of matches I can expect each day, and the potential spread. ie combining the 2 distributions

The reason to do this is so that I can determine how many people I need to review the matches, given that it take a set amount of time to review each one. Getting the calculation wrong can be costly, or increase the risks that we may not identify correctly a bad guy becasue we dont have enough staff.

$\endgroup$
  • $\begingroup$ Please do not use the tag (poisson-geometry) for questions related to Poisson distributions. The two are unrelated. $\endgroup$ – Did Mar 7 '17 at 7:42
4
$\begingroup$

Some starting points: The ten largest airports in the world average 70,0124,224 passengers per year (2013; ATL, PEK, LHR, HND, ORD, LAX, DXB, CDG, DFW, CGK)$^1$ or 192,121 passengers per day. Assume that 1 per 1,000 passengers are bad guys (adjust as appropriate). It would be reasonable to assume that $A$ = Number of passenger $A$rrivals at the airport's checkpoint is distributed Poisson with mean $\lambda$. Then the the number of $B$ad guys matched $given$ the number of $A$rrivals at the checkpoint will be Binomial and the probability of a bad guy will be:

$P_{Binomial}(B = k | A = j) = \dfrac{j!}{k!(j-k)!} p^k(1−p)^{j−k}$

with probability of success $p$ = 0.001. The unconditional probability distribution of $B$ad guys matched at the checkpoint will be:

$P(B=k) = \sum_{i=k}^{\infty} P_{Binomial}(B = k | A = i)P_{Poisson}(A = i)$

or

$P(B=k) = \sum_{i=k}^{\infty} \dfrac{i!}{k!(i-k)!} p^k(1−p)^{i−k} \dfrac{\lambda^i e^{-\lambda}}{i!}$

Let $m=i−k$. Then

$P(B=k) = \sum_{m=0}^{\infty} p^k(1−p)^m \dfrac{\lambda^{k+m} e^{-\lambda}}{k!m!} = \dfrac{(\lambda p)^ke^{−\lambda}}{k!} \sum_{m=0}^{\infty} \dfrac{(\lambda(1−p))^m}{m!} = \dfrac{(\lambda p)^ke^{−\lambda}}{k!} e^{\lambda (1-p)}$

so that

$P_{Poisson}(B=k) = \dfrac{(\lambda p)^ke^{−\lambda p}}{k!} $

Hence, $B$ is distributed Poisson with mean $\lambda p$ and standard deviation $\sqrt{\lambda p}$. A reasonable range then for the number of $B$ad guys matched at the checkpoint would be $\lambda p \pm 3\sqrt{\lambda p}$ for sufficiently large $\lambda$. For example, if 192,121 passengers are expected to arrive $(\lambda)$ and the probability of any arrival being a bad guy is 0.001 ($p)$, then 192 bad guys are expected plus or minus 42 bad guys.

$^1$wiki http://en.wikipedia.org/wiki/World%27s_busiest_airports_by_passenger_traffic#2013_statistics

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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