For those of you familiar with the "German tank problem" area of statistics, I have a question set due on problems pertaining to this due at midnight eastern time tonight.

If anyone is particularly good at this, or good at statistics, I can send you a copy of the problems and explain what you have to tell me.

You get to name your price for this after I send you the problems and you decide if you want to do it or not. You will need Microsoft excel or a similar program as you have to compute a decent amount of numbers.

Post quickly please, thanks.

I'll pay you even if you can help me figure out the first couple questions, because then I can take it from there and figure everything out.

Come on guys, I'm sure all of you aren't basic statistics people.

Here's the first question I need help with, this comes after question asking what is the probability that there are 500 tanks if you have a uniform estimate of the number of tanks being between 1-2000. This answer is easily 1/2000

Now, let's imagine we have disabled a tank in battle and observed its serial number to be 100. Using Bayesian reasoning, what is the posterior (after we observe the evidence) probability that there are 500 tanks?

Hints:

Recall that according to Bayes theorem, P(number of tanks is 500 | we observe a serial number of 100) = P(we observe serial number of 100 | number of tanks is 500) * P(number of tanks is 500)/P(we observe a serial number of 100)

The numerator is trivial, so let's focus on the denominator.

For the denominator, to find the total probability of observing serial=100, you will have to consider a lot of different alternatives. You should realize that this probability will be a sum of 2000 terms: P(observe 100 and number of tanks is 1) + P(observe 100 and number of tanks is 2) + ..... + P(observe 100 and number of tanks is 2000).

Recall that P(A and B ) = P(A|B)*P(B).

Save your result for the denominator, P(we observe a serial number of 100) - you'll need it for the next several problems again

You should probably use a spreadsheet for this, or another capable computation software of your choice. Typing thousands of numbers into your calculator is not recommended.





Bayes Theorem says that P(A|B) = P(B|A) * P(A) / P(B)

A = # of tanks = 500
B = Observed serial number is 100

So we are trying to find the probability that there are 500 tanks given we observed serial number 100 (P=(A|B)), we set this equal to...

P=(B|A): (1/500)
P=(A): (1/2000)

(1/500)(1/2000) / P(B) <----(I think this is right)

To find the probability of serial number being 100, we have to add up all of the probabilities of (serial number 100 being observed and number of tanks is N).

We can assume the probability of the first 1-99 is 0%, because if you observe serial number 100 anything below that can't be possible.

So here's where I'm having trouble.

P(A and B ) = The sum of all probabilities (N=100-2000|Serial number 100) * P(Serial number 100)

So I figure, since we know that the answer to the first 1-99 we can use that as a building block to figure out everything above 100.
So P(1 tank|Serial number 100) * P(Serial number 100) = 0%
I assumed that picking serial number 100 has a probability of 1/2000, but I could be wrong.

If that's the case, than P(1 tank|serial number 100) would have to = 0 to make the whole equation equal 0. This seems easy to assume, but if we move that to a problem like P(100 tanks|Serial number 100), I draw blanks on what this is.


Hopefully this is enough information for you to form any help. I can figure out all the other problems if I just figure out this one.

This post was edited by 1messdkid on Oct 27 2013 09:01pm

Let me know if this makes sense for you :

N = total number of tanks
X = serial number of the tank we observe

P ( N=500 given that X=100 ) = P ( N=500 | X=100 ) = P ( N=500 and X=100 ) / P ( X=100 )

P ( N=500 and X=100 ) = ( 1/1901 ) * ( 1/500 )
Here, I discard the possibilities for N=1, N=2 ... N=99, and since N is uniformely distributed between 100 and 2000, we have P(N=500) = 1/1901.

P ( X=100 ) = P(N=100 and X=100) + P(N=101 and X=100) + ... + P(N=2000 and X=100)
P ( X=100 ) = (1/1901)*(1/100) + (1/1901)*(1/101) + ... + (1/1901)*(1/2000)
P ( X=100 ) = (1/1901) * [ 1/100 + 1/101 + ... + 1/2000 ]

In the end :

P ( N=500 given that X=100 ) = (1/500) / [ 1/100 + 1/101 + ... + 1/2000 ] ~ 0.00066645 ~ 1 out of 1500