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Definite integral over random interval

  1. Oct 7, 2009 #1
    Hi, all,

    assuming a and b are random variables and their pdf f(a) and f(b) are known. then, how do I solve for the definite integral given as [tex]v=\int\limits_{a}^{b} g(x) dx[/tex], where g(x) is a function of x? or, how do I solve the pdf of v?

    Thanks a lot..
     
  2. jcsd
  3. Oct 7, 2009 #2

    EnumaElish

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    I think the answer is "use the fundamental theorem of calculus." Operationally, suppose each of a and b is Bernoulli with probabilities p and q: Pr[a = a1] = 1 - Pr[a = a2] = p, Pr[b = b1] = 1 - Pr[b = b2] = q.

    Then:
    v = G[b1] - G[a1] with prob = pq,
    v = G[b2] - G[a1] with prob. p(1-q)
    v = G[b1] - G[a2] with prob. (1-p)q
    v = G[b2] - G[a2] with prob (1-p)(1-q)

    where "the integral of g from a to b" = G - G[a].
     
  4. Oct 8, 2009 #3
    Another approach is to first work out the cdf P[v<=y]. To do this (assuming a and b are independent) it might be helpful to sketch the 2d region of points (a,b) where G-G[a]<=y.
     
  5. Oct 9, 2009 #4
    If we define that
    [tex]y(a,b)=\int\limits_{b=-\infty}^{\infty} \int\limits_{a}^{b} g(x) dx[/tex]
    we can state the pdf over the joint space of variables a, b and v as
    [tex]f(a,b,v) = \delta\left(v-y(a,b)\right)f(a)f(b)[/tex]
    where [itex]\delta[/itex] is the Dirac delta function.

    The pdf [itex]f(v)[/itex] of v is then the marginal of [itex]f(a,b,v)[/itex] with respect to a and b
    [tex]f(v) = \int\limits_{a=-\infty}^{\infty} \int\limits_{b=-\infty}^{\infty} \delta\left(v-y(a,b)\right) f(a)f(b) db da[/tex]


    The integral of g(x) can as stated above be evaluated using its primitive G(x):
    [tex]\int\limits_{a}^{b} g(x) dx = G(b)-G(a) \quad \texttt{if} \quad b \geq a[/tex],
    [tex]\int\limits_{a}^{b} g(x) dx = G(a)-G(b) \quad\texttt{if} \quad a \geq b[/tex]

    We thus need to account for the two possible cases:
    [tex]
    f(v) =
    \int\limits_{a=-\infty}^{\infty} \int\limits_{b=-\infty}^{a}
    \delta\left(v-G(a)+G(b)\right)
    f(a)f(b) db da
    +\int\limits_{a=-\infty}^{\infty} \int\limits_{b=a}^{\infty}
    \delta\left(v-G(b)+G(a)\right)
    f(a)f(b) db da
    [/tex]
     
    Last edited: Oct 9, 2009
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