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Sunday, July 6, 2014

Embarrassing Questions, German Tanks and Estimations


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Q: You are conducting a survey and want to ask an embarrassing yes/no question to subjects. The subjects wouldn't answer that embarrassing question honestly unless they are guaranteed complete anonymity. How would you conduct the survey?

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A: One way to do this is to assign a fair coin to the subject and ask them to toss it in private. If it came out heads then answer the question truthfully else toss the coin a second time and record the result (heads = yes, tails = no). With some simple algebra you can estimate the proportion of users who have answered the question with a yes.

Assume total population surveyed is \(X\). Let \(Y\) subjects have answered with a "yes". Let \(p\) be the sort after proportion. The tree diagram below shows the user flow.


The total expected number of "yes" responses can be estimated as
$$
\frac{pX}{2} + \frac{X}{4} = Y
$$
which on simplification yields
$$
p = \big(\frac{4Y}{X} - 1\big)\frac{1}{2}
$$

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Q: A bag contains unknown number of tiles numbered in serial order \(1,2,3,...,n\). You draw \(k\) tiles from the bag without replacement and find the maximum number etched on them to be \(m\). What is your estimate of the number of tiles in the bag?

A: This and problems like these are called German War Tank problems. During WW-II German tanks were numbered in sequential order when they were manufactured. Allied forces needed an estimate of how many tanks were deployed and they had a handful of captured tanks and their serial numbers painted on them. Using this, statisticians estimated the actual tanks to be far lower than what intelligence estimates had them believe. So how does it work?

Let us assume we draw a sample of size \(k\). The maximum in that sample is \(m\). If we estimate the maximum of the population to be \(m\) then probability of the sample maximum to be \(m\) is
$$
P(\text{Sample Max} = m) = \frac{m-1 \choose k-1}{N \choose k}
$$
The \(-1\) figures because the maximum is already taken out of the sample leaving behind \(m - 1\) to choose \(k -1 \) from. The expected value of the maximum using this strategy is thus
$$
E(\text{Maximum}) = \sum_{m=k}^{m=N}m\frac{m-1 \choose k-1}{N \choose k}
$$
Note, we run the above summation from \(k\) to \(N\) as for \(m < k\) the expectation is \(0\) because the sample maximum has to be at least \(k\). After a series of algebraic manipulations ( ref ) the above simplifies to
$$
E(\text{Maximum}) = M\big( 1 + \frac{1}{k}\big) - 1
$$
which is quite an ingenious and simple way to estimate population size given serial number ordering.

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