Help with integrating stochastic calculus expression from yield curve model
I am very rusty on stochastic calculus, and I am having trouble integrating the following simple term from a yield curve model:
$$ z(t)=\int_0^t\exp(-k(t-s))dW(s) $$ Any suggestions appreciated.
It is a Wiener integral as your integrand is a deterministic function of time.
It is known that the Wiener integral is stationary gaussian process with independent increments. So [Math Processing Error] $ z(t) \sim \mathcal N\left(0, \int_0^te^{-2k(t-s) }~ds\right) $ and $ (z(t)-z(s)) \amalg z(u), \ \forall u,s,t \in \mathbb R_+ \text{ such that }u\leq s, s\leq t $ or alternatively you can just say that $ (z(t)-z(s)) \amalg \mathcal F_s^z, \ \forall s, t \in \mathbb R _+ \text{ with } t\geq s $ where $ \mathcal F_u^z $ is the natural filtration of $ z $ .
Formally you have that $ z(t) \overset{\mathcal L}{=} \int_0^t e^{-2k(t-s)}~ds \frac{1}{\sqrt{t}} W_t $
I suppose you need to use that in a simulation so you can just multiplie a normal random variable by the standard deviation $ \sqrt {\int_0^te^{-2k(t-s) }~ds} $ and you know that $ \int_0^te^{-2k(t-s) }~ds= \frac{1}{2}e^{-2kt}(1-e^{-2kt}) $ (if I made no mistakes).
Actually formal speaking you must ensure that you integrand $ f $ (in your exemple $ f(s) =e^{-k(t-s)} $ satisfies “good” integrability conditions. That means that $ f \in L^2 ( \mathbb R _+,dt) $ (with is the case for your example),where $ dt $ is the Lebesgue measure .
In general terms a process $ I $ defined $ I(t) := \int_0^t f(s) ~ds $ has the properties mentioned above and particularly [Math Processing Error] $ I(t)\sim \mathcal N\left(0, \int_0^tf^s(s)~ds\right) $
I hope that helped you.