# 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 $z(t) \sim \mathcal N\left(0, \int_0^te^{-2k(t-s) }~ds\right)$ 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 $I(t)\sim \mathcal N\left(0, \int_0^tf^s(s)~ds\right)$

I hope that helped you.

• Many thanks - so for the expectation of z(t) this dW term just drops out (since by construction it has a 0 mean). Very much appreciate your help! – bondprop Nov 15 '14 at 16:20
• You're welcome. You are right, you can translate that formally by $z(t) \overset{\mathcal L}{=} \int_0^t e^{-2k(t-s)}~ds \frac{1}{\sqrt{t}} W_t$. I added that to the answer – Paul Nov 15 '14 at 19:20