I am trying to compute the infinitesimal generator of standard Brownian bridge. But I am not sure whether I am correct or not.
This is something I've tried.
Let Z(t) denote standard Brownian bridge, Q the generator.
Then Qf = $ \lim _{t \to 0} \frac{E[f(Z)] -f(x)}{t}$
$E[f(Z)] = E[f(x) + (Z - x)f'(x) + \frac{1}{2}(Z - x)^2f''(x)]\\ = f(x) + E[Z - x]f'(x) + \frac{1}{2}E[(Z - x)^2]f''(x)\\ = f(x) + \frac{1}{2}(t-st)f''(x)$
suppose t<s
Then plug back,
Qf = $\lim_{t \to 0} \frac{ \frac{1}{2} (t-st)f''(x)}{t}\\ = \frac{1}{2} f''(x) (1-s)$
Thank you so much
If made mistakes, please point it out.