Suppose $P$ is the set of functions where $p\in P: R^{+2}\to R^+$ and $p(t,s)$ is differentiable in $t$. $\forall t, p(t,\cdot)$ is a probability distribution on the positive axis $s\in [0,\infty)$, i.e. $s,t\ge 0$, $\int_0^\infty p(t,s)ds=1,\forall t$ and $p(t,s)\ge 0, \forall s,t$. Let $$f(p;t,k) := \int_0^\infty (s-k)_+\frac{\partial p(t,s)}{\partial t}ds,$$ where $x_+:=\max(x,0)$. For given $\forall t,k$,

1) solve for $\displaystyle\max_{p\in P} f(p;t,k)$;

2) solve for $\displaystyle\max_{p\in P} f(p;t,k)$, if in addition $p\in P$ satisfies $t = \int_0^\infty sp(t,s)ds, \forall t\in [0,\infty)$;

3) I conjecture that $f(p;t,k)\le 1, \forall p\in P, (t,k)\in R^{+2}$. What is a proof or counterexample?

I suppose this can be accomplished with calculus of variation, or just linear programming, with Lagrangian multipliers to deal with the constraints, but currently the partial derivative of $t$ gives me pause.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.