How would you go about solving this? $$p(x,t)=C\exp\left[-x+\int_0^t\int_0^\infty y\,p(y,\tau)\,\mathrm{d}y\,\mathrm{d}\tau\right]$$
Here $p(x,t)$ is the time-dependent probability distribution of a variable $x$, so it should be normalized to 1, and positive everywhere. Also, we have the initial condition: $$p(x,0) = \exp\left[-x\right]$$
I hit across this equation in some work I'm doing, and I'm stuck. Don't know enough about integral equations to know where to begin.
Is there any hope at all to solve this? Where would you start?
Notice that the term $$\int_0^t \int_0^\infty y p(y,\tau) dy d\tau$$ only depends on $t$, so $$p(x,t) = C e^{-x} e^{f(t)}$$ with $$f(t) := \int_0^t \int_0^\infty y p(y,\tau) dy d\tau$$ replacing $p$ in the last equation $$f(t) = C \int_0^t e^{f(\tau)} d\tau \underbrace{\int_0^\infty y e^{-y} dy}_{1} = C\int_0^t e^{f(\tau)} d\tau$$.
$$f'(t) = C e^{f(t)}$$ and you can easily solve this DE.(notice that $f(0) = 0$)