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Expectation of a function of a continuous random variable

  1. Mar 11, 2009 #1
    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!
     
  2. jcsd
  3. Mar 12, 2009 #2

    jambaugh

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    Gold Member

    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.
     
  4. Mar 12, 2009 #3
    Yes. Consider the simplest case, where g(x) = 1 if x is in [a,b] and 0 otherwise. Then

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

    On the other hand, the formula also gives

    [tex]\int_{-\infty}^\infty g(x)f_X(x)\,dx = \int_a^b f_X(x)\,dx.[/tex]
     
  5. Jun 19, 2009 #4
    Go for summations.....that will solve ur problem and rest all may complicate
     
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