- #1
magnifik
- 360
- 0
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
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