3

When looking at an Ito integral are $\int_{0}^t f(s) dW_s $ and $\int_{t}^x f(s) dW_s $ independent under what conditions for $f$. I was trying to solve the exercise where I need to prove that: $\int_{0}^t f(s) dWs$ is a Gaussian random variable if we look at this as a stochastic process by varying $t$.

If we have independence of the time increments i can construct any collection of $\int_{0}^{t_k} f(s) dWs$ as a linear transformation of a vector with components $\int^{t_{k+1}}_{t_k} f(s) dWs$ and a linear transformation of a normally distributed vector is normally distributed.

It is clear for me, that if $f$ is a deterministic elementary function that we have independece of time increments for the ito integral. I also know that I can write my Ito integral of the deterministic function as an ${L}^2$ limit of the ito integral of elementary functions. My proof would be complete, if I knew that independence of random variables carries over via $L^2$ limits. This feels wrong, but I am not sure. Thanks for the help!

Jose Avilez
  • 12,710
Dude1662
  • 101
  • 1
    Every $L^2$-limit has a subsequence that converges almost surely. So we can throw the $L^2$-limits into the bin and pretend that there is a sequence of simple processes (step functions in the time variable) $f_n$ so that $\int_0^tf_n(s),dW_s\to \int_0^tf(s),dW_s$ almost surely. When $f(s)$ is deterministic then $f_n(s)$ is deterministic. This shows very easily that the independence of increments of $W_t$ carries over to the integrals. In the same fashion you should see that for a non deterministic $f_n$ there is almost no hope that its integral has independent increments. – Kurt G. Feb 27 '24 at 13:41
  • 1
    Then $\int_0^tf(s),dW_s$ won't have them either. – Kurt G. Feb 27 '24 at 13:43
  • Why does the independence carry over for P-almost sure convergence? If i understand correctly the statement used is that if we have two independent stochastic processes *(Xn)n and (Yn)n that converge P-almost surely to X or resp. to Y their limits are independent. – Dude1662 Feb 27 '24 at 14:07
  • 1
    Pretty much. That is because the $\sigma$-algebra that is generated by all $\int_a^bf_n(s),dW_s$ is (after adding the $P$-null sets) sufficiently large to ensure that $\int_a^bf(s),dW_s$ is measurable w.r.t. to it. Apply this to two non overlapping intervals $[a,b],[c,d]$ and you get done. If you feel unsure write out all details following my hints. – Kurt G. Feb 27 '24 at 14:13

2 Answers2

2

Denote $X = \int_0^t f(s) dW_s$, $Y = \int_t^x f(s) dW_s$, and $X_n, Y_n$ a sequence of approximating simple stochastic integrals for $X$ and $Y$ respectively. Note that $(X_n, Y_n)$ is jointly Gaussian and uncorrelated for each $n$, hence $X_n$ and $Y_n$ are independent for each $n$.

The preservation of independence in the limit is due simply to convergence in distribution of the vector $(X_n, Y_n)$. Notice that since $(X_n, Y_n)$ converges to $(X,Y)$ in probability, it does so in distribution, so

$$F_{(X,Y)}(x,y) = \lim_n F_{(X,Y)}^{(n)}(x,y) = \lim_n F_X^{(n)}(x)F_Y^{(n)}(y) = F_X(x) F_Y(y)$$ where $F$ denotes a corresponding CDF.

Jose Avilez
  • 12,710
2

Alternatively to @JoseAvilez's answer (+1): you may have seen that the stochastic integral wrt $f \in L^2[0,\infty)$ is adapted and that for any $\xi\in \mathbb{R}$ and $s<t$ we have $$E\bigg[\exp\bigg(i\xi\int_s^tf(u)dW_u\bigg)\bigg|\mathscr{F}_s\bigg]=\exp\bigg(-\frac{\xi^2}{2}\int_s^tf(u)^2du\bigg)$$ or equivalently, that $\int_s^tf(u)dW_u|\mathscr{F}_s\sim \mathcal{N}(0,\int_s^tf(u)^2du)$ (*). Then if you define for $0< t <T$ $$\varphi(\xi_1,\xi_2):=E\bigg[\exp\bigg(i\xi_1\int_0^tf(u)dW_u+i\xi_2\int_t^Tf(u)dW_u\bigg)\bigg]$$ for $\xi_1,\xi_2\in \mathbb{R}$ then an application of the tower rule yields: $$\begin{aligned} \varphi(\xi_1,\xi_2)&=E\bigg[\exp\bigg(i\xi_1\int_0^tf(u)dW_u\bigg)E\bigg[\exp\bigg(i\xi_2\int_t^Tf(u)dW_u\bigg)\bigg|\mathscr{F}_t\bigg]\bigg]\\ &=\varphi(\xi_1,0)\varphi(0,\xi_2) \end{aligned}$$ But this proves independence by Kac's theorem.

(*)

Indeed, given simple stochastic integrals $\int_s^t f_n dW$ approximating $\int_s^t f dW$ in $L^2(P)$ (we can choose $f_n$ simple deterministic approximating $f$ in $L^2[s,t]$), you get - by passing to a subsequence to have convergence a.s. of the stochastic integrals -\begin{align}E\bigg[\exp\bigg(i\xi \int_s^t f_{} dW\bigg)\bigg|\mathscr{F}_s\bigg]&\stackrel{\textrm{DCT}}{=}\lim_{k\to \infty}E\bigg[\exp\bigg(i\xi \int_s^t f_{n_k} dW\bigg)\bigg|\mathscr{F}_s\bigg]\\&\stackrel{W_{t_\ell^{n_k}}-W_{t_{\ell-1}^{n_k}}\perp \mathscr{F}_s}{=}\lim_{k\to \infty}E\bigg[\exp\bigg(i\xi\sum_{1\leq \ell\leq N_{n_k}}a^{n_k}_\ell(W_{t_\ell^{n_k}}-W_{t_{\ell-1}^{n_k}})\bigg)\bigg]\\&=\lim_{k\to \infty}\exp\bigg(-\frac{\xi^2}{2}\sum_{1\leq \ell\leq N_{n_k}}(a^{n_k}_\ell)^2(t_\ell^{n_k}-t_{\ell-1}^{n_k})\bigg)\\&=\lim_{k\to \infty}\exp\bigg(-\frac{\xi^2}{2}\int_s^t(f_{n_k}(u))^2du\bigg)\\&=\exp\bigg(-\frac{\xi^2}{2}\int_s^t(f(u))^2du\bigg)\end{align}

Snoop
  • 15,214