# Expected values of a limiter

1. Oct 22, 2011

### magnifik

For the limiter shown below, find the expected value for Y = g(X)

attempt at solution:
E[Y] = ∫g(x)f(x)dx, where f(x) is the probability density function with respect to x
so...
E[Y] = E[Y1] + E[Y2] + E[Y3]
E[Y1] = ∫-af(x)dx where the limits of integration are from -∞ to -a
so E[Y1] = -aFx(-a), where Fx is the cumulative distribution function
E[Y2] = ∫xf(x)dx where the limits of integration are -a to a
this is the part i'm having trouble with
E[Y3] = ∫af(x)dx where limits of integration are a to ∞
so E[Y3] = a(1-Fx(a))

i'm having trouble simplifying E[Y2] into one expression because of the limits of integration. i know it would be just E[Y2] if the limits were -∞ to ∞, but that is not the case here

2. Oct 23, 2011

### canis89

I think that when the domain of integration is limited over a certain set A, the expected value becomes the conditional expected value of X given that X lies on A, times the probability that X lies on A.

3. Oct 23, 2011

### Ray Vickson

Unless you have a specific form for f(x) you have gone as far as you can: E(Y2) does not simplify further in any useful way. By the way: I would argue against your notation E(Y1), E(Y2), etc. You do not have three separate random variables Y1, Y2 and Y3; you just have three separate "pieces" of the same random variable Y.

RGV

4. Oct 23, 2011

### magnifik

hmm..ok

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook