Need sum help (information theory)

  • Thread starter Adeimantus
  • Start date
111
1

Main Question or Discussion Point

need sum help plz:) (information theory)

Hello.

I'm working on an information theory-related problem that involves doing a nasty sum. The problem is this: in a widget factory there is a conveyor belt with N widgets on it, and an unknown fraction [itex]\xi = a/N[/itex] of them are defective. You examine a sample of n widgets and find that a fraction [itex] \eta = b/n [/itex]. What is mutual information [itex]I(\eta : \xi)[/itex] between the random variables [itex]\eta[/itex] and [itex] \xi [/itex] ? The idea, i think, is to see how large a sample n you need to take so that the sample defect rate gives you information about the actual defect rate. Let A_a be the event that there are a defective parts in the whole lot and B_b be the event that there are b defective parts in the sample. Then the formula for mutual information is:

[tex] I (\eta : \xi) = \sum_{a=1}^{N} \sum_{b=1}^{n} P(A_{a}) P(B_{b} | A_{a}) \log_{2} { \frac{P(B_{b} | A_{a}) }{P(B_{b})} } [/tex]

which is always nonnegative. Here's what I've got so far: [itex]P(A_{a}) = 1/N [/itex] by principle of insufficient reason (a could be anything from 1 to N with equal probability), and

[tex]
P(B_{b} | A_{a}) = \frac{ \left( ^{a}_{b} \right) \left( ^{N-a}_{n-b} \right) }{ \left(^{N}_{n} \right) }
= \frac{ \left( ^{n}_{b} \right) \left( ^{N-n}_{a-b} \right) }{ \left(^{N}_{a} \right) }
[/tex]
[tex]
P(B_{b}) = \sum_{a=1}^{N} P(A_{a}) P(B_{b} | A_{a}) = \sum_{a=1}^{N} \frac{1}{N} \frac { \left( ^{n}_{b} \right) \left( ^{N-n}_{a-b} \right) }{ \left(^{N}_{a} \right) }
\approx
\int_{0}^{1} \left( ^{n}_{b} \right) x^{b} (1 - x)^{n-b} dx = \frac {\left( ^{n}_{b} \right)}{ \left( ^{n}_{b} \right) (n+1)} = \frac {1}{n+1}
[/tex]

if you pretend it is a Riemann sum and assume that [itex]N \gg n[/itex] and [itex]a \gg b[/itex], which I'm not sure is OK to do. I'm guessing the idea is to get some asymptotic formula for the mutual information as N becomes large, but how do you retain the dependence on N in the sum? For instance, if I apply the "large N" approximation for [itex]P(B_{b} | A_{a}) [/itex], which is [itex]\left( ^{n}_{b} \right) \left(\frac{a}{N}\right)^{b} \left(1 - \frac{a}{N} \right)^{n-b}[/itex], and do the Riemann sum I get an expression that has no dependence on N and apparently diverges to negative infinity (weird because mutual information is nonnegative).

This is not a homework problem, just a "something to think about" problem I came across in an informal book on information theory.

thanks
 

Answers and Replies

Related Threads for: Need sum help (information theory)

  • Last Post
Replies
9
Views
3K
Replies
1
Views
3K
Replies
0
Views
2K
Replies
2
Views
2K
Replies
1
Views
5K
Replies
16
Views
1K
Replies
7
Views
449
Top