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Math Help - derivative of integral

  1. #1
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    derivative of integral

    Hi all,

    I am new to the forum.

    I have an integral that I am using, and I am unsure if I am computing the derivative of it correctly. I want to take the derivative of the following expression with respect to p:

    int( p*[1-G(p-y)] + int(z*g(z)dz,-inf,p-y) dF(y) , -inf,inf)
    where G is a CDF of z, and g is its pdf, and F is the CDF of y, -inf and inf are the bounds of the outer integral, and -inf and p-y are the bounds of the inner integral.

    My calculation gives:

    int( 1-G(p-y)-y*g(p-y) dF(y), -inf,inf ) where -inf and inf are the bounds of the integral.

    Thanks for your help.

    All the best,
    Bob
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  2. #2
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    Is the following integral correct? (I wasn't certain what was meant by dF(y).)

    \displaystyle \int_{-\infty}^{\infty} \left ( p(1 - G(p - y)) + \left ( \int_{-\infty}^{p-y} z g(z) \, dz \right ) F(y) \, \right ) dp
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  3. #3
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    Thanks for your reply. Actually, it is the following:

    $\int\nolimits_{-\infty }^{\infty }\left[ p[1-G(p-y)]+\int\nolimits_{-%<br />
\infty }^{p-y}zg(z)dz\right] f(y)dy$

    and, I think the answer is:

    $\int\nolimits_{-\infty }^{\infty }\left[ [1-G(p-y)]-yg(p-y)\right] f(y)dy<br />
$
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  4. #4
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    I suspect I am committing a gross error, but isn't the derivative of a definite integral 0 because a definite integral is a constant the derivative of a constant is 0? If so, then isn't the answer to your question 0 because

    \displaystyle \int_{-\infty}^{\infty} \left ( p(1 - G(p - y)) + \left ( \int_{-\infty}^{p-y} z g(z) \, dz \right ) \right ) F(y) \, dp

    is a definite integral?

    Edit: I didn't notice the outside integral is with respect to y. I knew my solution was too easy! :P

    \displaystyle \int_{-\infty}^{\infty} \left ( p(1 - G(p - y)) + \left ( \int_{-\infty}^{p-y} z g(z) \, dz \right ) \right ) F(y) \, dy
    Last edited by NOX Andrew; April 7th 2011 at 06:36 AM.
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