9
votes
Calculate value of Integral of Wiener process $\int_{0}^t e^{\lambda u } dZ_u$
This is just a Wiener integral (stochastic integral with respect to a Brownian motion and deterministic integrand), hence a centered Gaussian random variable with variance
$$
\int_0^t{e^{2\lambda u}\...
5
votes
Accepted
Why this stochastic integral is calculated with Riemann integral
For any semi martingale $X$ (in particular for $X_t=W_t$ or for $X_t=t$) we have
$$\tag{1}
\int_0^t dX_s=X_t-X_0\,.
$$
You are correct that the Ito integral uses the limit procedure
$$\tag{2}
\int_0^...
4
votes
Integral of brownian motion wrt. time over [t;T]
The last integral is correct as
$$\int_t^T W_s ds = \int_t^T (T-s) dW_s \sim N\left(0, \int_t^T(T-s)^2ds\right) = N\left(0,\frac{1}{3}(T-t)^3\right).$$
Ref. Arbitrage Theory in Continuos Time (Björk, ...
2
votes
Accepted
Calculate value of Integral of Wiener process $\int_{0}^t e^{\lambda u } dZ_u$
I'll try to use Ito's Lemma to come up with a solution. Ito's lemma states that:
$$F(Z_t,t)=\int_0^t\left(\frac{\partial F}{\partial u}+\frac{\partial F}{\partial Z}a+0.5\frac{\partial^2 F}{\partial Z^...
2
votes
Accepted
What does it mean to "compute" an Itô integral?
Note that SDE (4) does have a "closed-form" representation.
Let $X$ be
$$X := S^p, $$
so (4) is a geometric Brownian motion SDE
$$dX = (p\alpha + 2^{-1}p(p-1) \sigma^2) X dt + p \sigma X dW,...
1
vote
Parametric Stochastic Integral
I am having trouble to understand your notation
$$
\int_t^Td\xi(t,s)g(s)\,ds\,.
$$
What is the meaning of this when you switch from the differential form $dF_t$ to the integral form
$$
F_t=F_0-\int_0^...
1
vote
Accepted
Regression of stochastic integral on Wiener process
By definition,
$$
{\mathbb Cov}(M_t,W_T) = {\mathbb E}[M_t W_T] - {\mathbb E}[M_t] {\mathbb E}[W_T] = {\mathbb E}[M_t W_T]
$$
since ${\mathbb E}[M_t] = {\mathbb E}[W_T] = 0 $. We now consider the ...
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