Expectation of a function of a continuous random variable

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Discussion Overview

The discussion revolves around the expectation of a function of a continuous random variable, specifically examining the implications when the resulting variable is discrete. Participants explore whether the integration approach remains valid in such cases and the potential differences between discrete and continuous variables.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that for a continuous random variable X, the expectation of W=g(X) can be expressed as an integral involving the probability density function fX(x).
  • Another participant questions whether the integration approach holds if W is a discrete random variable, suggesting that it might be simpler to use summation instead of integration.
  • A different participant agrees that the integration can still apply in the discrete case but provides an example where g(x) is defined in a specific interval, showing how the expectation can be calculated using probabilities.
  • Another participant recommends using summations to avoid complications, implying that this may be a more straightforward approach.

Areas of Agreement / Disagreement

Participants express differing views on whether to use integration or summation for discrete random variables, indicating a lack of consensus on the best approach in this context.

Contextual Notes

Some assumptions about the continuity of g(X) and the definitions of discrete versus continuous variables are not fully explored, leaving open questions about the applicability of the discussed methods.

kingwinner
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If W=g(X) is a function of continuous random variable X, then E(W)=E[g(X)]=

∫g(x) [fX(x)] dx
-∞
============================
Even though X is continuous, g(X) might not be continuous.

If W happens to be a discrete random variable, does the above still hold? Do we still integrate ∫ (instead of sum ∑)?

Does it matter whether W itself is discrete or continuous?

Thanks!
 
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If the variable is discrete you can still express in terms of integrals using dirac delta functions though it is simpler to replace the integral with a sum over values.
 
kingwinner said:
If W=g(X) is a function of continuous random variable X, then E(W)=E[g(X)]=

∫g(x) [fX(x)] dx
-∞
============================
Even though X is continuous, g(X) might not be continuous.

If W happens to be a discrete random variable, does the above still hold?
Yes. Consider the simplest case, where g(x) = 1 if x is in [a,b] and 0 otherwise. Then

\begin{align*}<br /> E[W] &amp;= 1\cdot P(W = 1) + 0\cdot P(W = 0)\\<br /> &amp;= P(g(X) = 1)\\<br /> &amp;= P(X\in [a,b])\\<br /> &amp;= \int_a^b f_X(x)\,dx.<br /> \end{align*}

On the other hand, the formula also gives

\int_{-\infty}^\infty g(x)f_X(x)\,dx = \int_a^b f_X(x)\,dx.
 
Go for summations...that will solve ur problem and rest all may complicate
 

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