Recent content by DCG

I like this approach. Since the only thing that I want to prove is that if g increases then q can only increase or stay the same, it suffices to prove that: \frac{\partial q(g)}{\partial g} \geq 0 Thank you all for the answers.This was really helpful. I think that I am in the right...

Thank you for your answer but I do not understand what happened to f(x) in the integral of (a2(x)-a1(x))f(x)dx >0 this does not imply that integral(a2(x)-a1(x))dx > 0 since f(x) could allocate more probability mass to intervals where a2(x)>a1(x) and less to opposite ones.

So I read a bit about functional derivatives. It seems like q=q(a) is a functional. It maps functions to numbers. So I can try and find its derivative with respect to a. According to what I read: \int \frac{\partial q(a)}{\partial a(x)}h(x)dx = {lim}_{\epsilon \rightarrow 0} \frac{q(a +...

I am sorry, my choice of variable names is terrible. Let: q = \int_{\mu}^{\infty} (1-F(x))a(x)dx In my first post I named it f and it has nothing to do with f(x). g = \int_{\mu}^{\infty} a(x)f(x)dx Here F(x),f(x) are fixed. They are the cdf and pdf of a regular distribution, \mu is...

Oh I see. Any pointers on how I could approach this though? Is there a name for this kind of problem? Anything I could read that could help me? Thanks a lot!

Hi all, I have the following quantity: f = \int_{\mu}^{\infty} (1-F(x))a(x)dx I want to claim that by increasing the following quantity: g = \int_{\mu}^{\infty} a(x)f(x)dx then f can only increase. Can I differentiate f with respect to g ? Is the following correct...