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As a non-statistician, I have a real world statistical/probability problem that I'm having trouble framing. The software I rely on in inventory management interprets the 'movements' (number of times inventory is used) in a strange way. It offers the number of months, out of 24 months, that the item has been used. For example, a moving code of 20 means that the item was used at least once for 20 out of 24 months..

What I need to be able to do is translate that to find the most probable number of movements over that 24 month period.

If movements randomly fall into 20 out of 24 months with no limitations, obviously the number of movements that really occur is likely to be much greater than 20. How much greater?

Sorry, this question is extra challenging because I have no ideas on how to begin tackling this. Any help is much appreciated.

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  • $\begingroup$ I doubt you can really get an answer to this... could be 20 as well as 5000, there's no way of telling unless the software gave other clues, that is (some logs maybe?). $\endgroup$
    – nico
    Commented Jul 24, 2011 at 11:05
  • $\begingroup$ I think it's quite simple if (and only if) we can assume that each move is independent of the others. Would also help to know the distribution of probabilities of move across different months (presumably some months will be more likely to see a move than others? Or is it safe to assume uniformity?) $\endgroup$
    – crayola
    Commented Jul 24, 2011 at 15:52

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As crayola says, the distribution is unlikely to be independent. Even if it is, as assumed below, the analysis is not simple.

It is possible to work out the probability of $N$ items happening in $M$ unique months out of $24$ with the following recursion:

$$\Pr(M=m|N=n) = \tfrac{(24-m+1)\Pr(M=m-1|N=n-1) + m\Pr(M=m|N=n-1)}{24} $$

starting at $\Pr(M=0|N=0)=1$, and $\Pr(M=0|N=n)=0$ for non-zero $n$. So, for example, $\Pr(M=20|N=42) = 0.2676\ldots$, and for $M=20$ this is the value of $N$ which gives the greatest likelihood.

It is also possible to work out the expected number of unique months $$E[M|N=n] = 24 \left(1-\left(\frac{24-1}{24}\right)^n\right).$$

If we set this equal to 20 and solved for $n$ we would get $$n=\frac{\log(24)-\log(24-20)}{\log(24)-\log(24-1)} \approx 42.09999\ldots$$ though $N$ needs to be an integer. So this too hints at $42$.

This is all looking hopeful, but hides the true horror. For example if $M=20$ then I think the 95% confidence interval for $N$ is $[28,68]$ which is rather wide. And if $M=24$ for a popular item then a similar confidence interval for $N$ could be $[50,\infty )$ while the maximum likelihood and expectation methods which took us from an observation of $20$ to an estimate of $42$ would take us from an observation of $24$ to an estimate of $\infty$.

Even is you used Bayesian methods to deal with the $M=24$ observation case you would still face very wide credible intervals.

The real answer is to get better inventory management reporting.

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  • $\begingroup$ Thank you Henry, I'll be studying your answer! And as you suggested, the error behind this is so wide that to use a firm number is quite a crime. While asking this I was also curious if there was some kind of answer.. if you're correct, well done. $\endgroup$
    – Nick
    Commented Jul 25, 2011 at 12:25

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