# Indicator/Membership Functions

## Homework Statement

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.

## Homework 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).

## 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)

Related Calculus and Beyond Homework Help News on Phys.org
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: