Distribution of an Exponentially Weighted Wiener Integral
Summary
The answer treats the yield-curve model term as a Wiener integral with a deterministic integrand. Such an integral is normally distributed with mean zero, and its variance is the integral of the squared integrand over time. For an exponentially decaying kernel, that variance can be evaluated by ordinary integration, giving a direct way to draw the term in a simulation as a standard normal variable scaled by its standard deviation.
The explanation also notes that the integrand must satisfy square-integrability conditions. This is useful for understanding the marginal distribution and simulation of the integral at a fixed time. However, it does not provide a full discretization scheme for simulating an entire correlated path, and the supplied variance expression appears to contain an algebraic error: integrating the stated kernel yields a different closed form than the one printed in the answer. The general variance principle remains the key idea.
Key ideas
- A Wiener integral with a deterministic square-integrable integrand is Gaussian with mean zero.
- Its variance is the time integral of the squared integrand.
- A fixed-time value can be simulated by scaling a standard normal draw by the standard deviation.
- The exponential kernel’s variance follows by integrating its square over the interval.
- A single-time distribution does not specify the joint path simulation, and the displayed closed form should be checked.
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# Help with integrating stochastic calculus expression from yield curve model
# 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.
## Answer by Paul (score 2, accepted)
https://quant.stackexchange.com/a/15465
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.