What is the expected value for Y in the given limiter?

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Homework Help Overview

The discussion revolves around finding the expected value for Y, defined as g(X), in the context of a limiter involving a probability density function f(x). Participants are exploring the integration limits and the implications for calculating expected values.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formulation of expected value using integrals and the challenges of integrating over specific limits. There is a focus on the conditional expected value when the domain is restricted, and questions arise regarding the notation used for different segments of the expected value.

Discussion Status

The conversation is ongoing, with participants attempting to clarify their understanding of the expected value calculations and the implications of the integration limits. Some guidance has been offered regarding notation and the nature of the random variables involved, but no consensus has been reached.

Contextual Notes

There is mention of a specific form for the probability density function f(x) being necessary for further simplification of the expected value calculations. The discussion reflects constraints related to the limits of integration and the definitions of the random variables involved.

magnifik
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For the limiter shown below, find the expected value for Y = g(X)
dh6b6s.png


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
 
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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.
 
magnifik said:
For the limiter shown below, find the expected value for Y = g(X)
dh6b6s.png


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

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
 
hmm..ok
 

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