1. ## 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.

All the best,
Bob

2. Is the following integral correct? (I wasn't certain what was meant by $dF(y)$.)

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

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

and, I think the answer is:

$\int\nolimits_{-\infty }^{\infty }\left[ [1-G(p-y)]-yg(p-y)\right] f(y)dy
$

4. 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

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