I am reading a paper/lecture notes by C.Hipp about Stochastic Control in Insurance and I encountered this: Given $g(s)=\mathbb{E}[V(s-X)]$, where $X$ has a continuous pdf and $V$ admits nonnegative values only.
The claim was this: If $X\sim \exp(mean=1)$, then $g$ satisfies the DE $g'(s)=V(s)-g(s)$. I had a hard time arriving at the result, I only have this $g'(s)=V(s)$. Here's my calculation.
$g(s)=\mathbb{E}[V(s-X)]=\int_0^s V(s-x)e^{-x}dx = \int_0^s V(t)e^{-(s-t)}dt$, if $t=s-x$. Then by the Fundamental Theorem of Calculus, I only got this: $g'(s)=V(s)e^{-(s-s)}=V(s)$. I don't know where the additional term $-g(s)$ came from. Where did I go wrong? Thanks for the help.
$s$ occurs in two places. You've correctly taken into account the occurrence in the integration limit. You also need to differentiate the factor $\mathrm e^{-s}$ in the integrand; that yields a term $-g(s)$. Perhaps this is easier to see if you first pull the factor outside the integral.