# Homework Help: Indicator/Membership Functions

1. Sep 24, 2008

### Kreizhn

1. The problem statement, all variables and given/known data
Let's say one measures an observation with absolute precision using a set of indicator functions, in a sample space with 3 elements. Then these indicator functions will be the usual discrete indicator functions of the form

$$I_n(j) = \begin{cases} 1 & \text{ if } n = j\\ 0 & \text{ otherwise } \end{cases}$$

Now let's say that your measurement method is flawed, such that 50% of the time your observation is accurate to the event, and the other 50% of the time your observation is random, though uniformly distributed over all three elements in the sample space. What will the membership function be in this case.

2. Relevant equations

The membership function is just a generalized indicator function over "fuzzy" sets. That is, for every member in the sample space, they take on a value between 0 and 1 (though still sum up to 1 over all space).

3. The attempt at a solution

It seems to me we want to say something like $I_n(j)$ will be the `probability' that we observe event n, given that event j happened. This would amount to:

Probability of observing n given j happened + probability of observing n given that j didn't happen.

I boil this down to

P(accurate that j occured)*P(measured n) + P(inaccurate)*P(measured n) + P(accurate)*P(did not measure n) + P(inaccurate in observing j)*P(did not measure n)

2. Sep 24, 2008

### Kreizhn

So this is how I figure it works:

Case 1: n = j

Then we should have P(accurate)*P(measured n) + P(inaccurate)*P(measured n)
which would be $\frac{1}{2} \frac{1}{1} + \frac{1}{2}\frac{1}{3} = \frac{4}{6}$

Case 2: $n \neq j$

Then P(accurate)*P(measured n) + P(inaccurate)*P(measured n) would yeild
$\frac{1}{2} \cdot 0 + \frac{1}{2} \frac{1}{3} = \frac{1}{6}$

Does this seem to make sense?

Last edited: Sep 24, 2008