I cannot figure out how to solve the following problem, and I do not know where to look for methods to solve it.
$$\min_{p(t)} \int_0^1 g(p(t))f(t)dt+\int_0^1\int_t^1 p(x)dxF(t)dt$$
Essentially, I am trying to find the function $p(t)$ that minimizes an expected cost, where $f(t)$ is the pdf of $t$, and $F(t)$ is the CDF. I know that $p(0)=0$ and $p(1)=1$.
If I did not have the term $\int_t^1 p(x)dx$, then I would simply minimize pointwise by taking the derivative with respect to $p(t)$. However, I do not understand how to take the derivative of $\int_t^1 p(x)dx$ with respect to $p(t)$. From my very limited understanding, this is a functional (or variational?) derivative.
If on the other hand I had $\int_0^1\int_t^1 p(x)dxf(t)dt$ instead of $\int_0^1\int_t^1 p(x)dxF(t)dt$, I could integrate by parts to get rid of the inside integral.
When I look to articles on this (https://en.wikipedia.org/wiki/Calculus_of_variations#Euler.E2.80.93Lagrange_equation), I cannot figure out how to translate my problem.
Any help that you could give me on solving this or pointing me in the right direction would be greatly appreciated.
Thanks a lot!
EDIT:
Ultimately I want
$$\frac{d}{dp(t)}\int_t^1p(x)dx$$
Define $$P(t):=\int_1^tp(x)dx=-\int_t^1p(x)dx$$
Note that $\dot P(t)=p(t)$
Then you can rewrite your problem as $$\min_{P(t)} \int_0^1 g(\dot P(t))f(t)dt-\int_0^1P(t)F(t)dt$$ This is a standard calculus of variations problem with the Lagrangian: $$L(t,P,\dot P)=g(\dot P(t))f(t)dt-P(t)F(t)$$ You just solve this, and then convert $P$ back to $p$.
By the way, I'm not sure why you want to have the derivative $$\frac{d}{dp(t)}\int_t^1p(x)dx$$ But note that this is equal to $$-\frac{d}{dp(t)}\int_1^tp(x)dx=-\frac 1 {\frac {dp(t)}{dt}}\cdot\frac d {dt} \int_1^tp(x)dx=-\frac {p(t)}{\dot p(t)}$$