Suppose $f,g \in L^2[a,b]$. Show that the double Wiener-Ito integral of the function $f(t)g(s)$ is given by $$ I_2(fg) := \int_a^b\int_a^bf(t)g(s)dB(t)dB(s)=I(f)I(g)-\int_a^bf(s)g(s)ds $$ where $B_t$ is the standard Brownian Motion and $I(f)=\int_a^bf(t)dB(t)$ and $I(g)$ is defined similarly.
I feel what I have done leads to the correct answer but I think I am lacking some rigor that I need some help on filling in. Heres what I have so far.
There is a theorem stated in my text that I will use.
$\textbf{Theorem 1:}$ Let $g$ be a continuous function on $\Bbb R$. For each $n \ge 1$, let $\Delta_n=\{t_0,…,t_n\}$ be a partition of $[a,t]$,. Then the sequence $$ \sum_{i=1}^2g(B_{t_{i-1}})\left((B_{t_{i}}-B_{t_{i-1}})^2-(t_i-t_{i-1})\right) $$ converges to $0$ in probability and $\Vert \Delta_n \Vert \to 0$
First, we may my sequences of step functions $\{f_n\}$ and $\{g_n\}$ such that both $f_n \to f$ and $g_n \to g$ in $L^2[a,b]$ and then we define $I(f)= \lim I(f_n)$ and $I(g)=\lim I(g_n)$ where the limit is taken in $L^2(\Omega)$. In particular, if we define $h(t,s)=f(t)g(s)$ and $h_n(t,s)=f_n(t)g_n(s)$ then we can show that $\Vert h-h_n \Vert_{L^2([a,b]^2)} \to 0$ as $\Vert \Delta_n \Vert \to 0$.
Consider the partition $\bigcup_{i\not =j}^n[t_{i-1},t_i)\times[t_{j-1},t_j) \bigcup \bigcup_{i=1}^n[t_{i-1},t_s)^2$ of the square $[a,b]^2$. Note that this second term is the diagonal of the square and the first term contains all the squares off of the diagonal.
Also, $$f_n(t)=\sum\limits_{i=1}^na_i^{(n)}1_{[t_{i-1},t_i)}(t)$$ and $$g_n(t)=\sum\limits_{i=1}^nb_i^{(n)}1_{[t_{i-1},t_i)}(t)$$ so $$h_n(t,s):=f_n(t)g_n(s)=\sum\limits_{i,j = 1}^n a_i^{(n)}b_j^{(n)}1_{[t_{i-1},t_i)}(t)1_{[t_{j-1},t_j)}(s)= \sum\limits_{i \not = j}^na_i^{(n)}b_j^{(n)}1_{[t_{i-1},t_i)}(t)1_{[t_{j-1},t_j)}(s) + \sum\limits_{i=1}^na_i^{(n)}b_i^{(n)}1_{[t_{i-1},t_i)}(t)1_{[t_{i-1},t_i)}(s)$$
Since multiple Ito integrals are defined as the $L^2(\Omega)$ limit of off-diagonal step functions then first we see that $$ I_2(f_ng_n)= \sum\limits_{i,j=1}^na_i^{(n)}b_j^{(n)}(B_{t_{i}}-B_{t_{i-1}})(B_{t_{j}}-B_{t_{j-1}}) – \sum\limits_{i=1}^n a_i^{(n)}b_i^{(n)}(B_{t_{i}}-B_{t_{i-1}})^2 $$
Now this first sum converges in $L^2(\Omega)$ to $I(f)I(g)$ and hence a sub sequence will converge almost surely. Here I guess that this second sum converges to $\int_a^b f(s)g(s)ds $ in probability and again then a sub sequence would converge almost surely and then this would complete the proof.
The convergence of this last last is what I am unsure of, but I believe that we can use $\textbf{Theorem 1}$ in the following way:
$$ \sum\limits_{i=1}^n a_i^{(n)}b_i^{(n)}(B_{t_{i}}-B_{t_{i-1}})^2 = \sum\limits_{i=1}^n a_i^{(n)}b_i^{(n)}\left((B_{t_{i}}-B_{t_{i-1}})^2-(t_i-t_{i-1})\right) + \sum\limits_{i=1}^n a_i^{(n)}b_i^{(n)}(t_i-t_{i-1}) \tag{1}
$$
and so the first sum on the RHS of $(1)$ will converge to $0$ in probability and therefore a subsequence will converge almost surely. I also believe the second sum on the RHS of $(1)$ will converge to $\int_a^bf(s)g(s)ds$ because of the following: $$\int_a^b|f(t)-f_n(t)| dt \le \sqrt{ \int_a^b|f(t)-f_n(t)|^2dt }\cdot \sqrt{\int_a^b1dt} \to 0$$ as $\Vert \Delta_n \Vert \to 0$. The same can be applied to both $g$ and $g_n$.
This implys that $$\lim\limits_{\Vert \Delta_n \Vert \to 0}\int_a^b |f_n(t)g_n(t)-f(t)g(t)|dt = 0$$
which further implys that $$\lim\limits_{\Vert \Delta_n \Vert \to 0}\int_a^b f_n(t)g_n(t) dt= \int_a^bf(t)g(t)dt \tag{2}$$.
Now since $$f_n(t)=\sum\limits_{i=1}^na_i^{(n)}1_{[t_{[i-1},t_i)}(t)$$ and $$g_n(t)=\sum\limits_{i=1}^nb_i^{(n)}1_{[t_{[i-1},t_i)}(t)$$
then this means that $$f_n(t)g_n(t)=\sum\limits_{i=1}^na_i^{(n)}b_i^{(n)}1_{[t_{[i-1},t_i)}(t)$$ and now finally we see that $$ \int_a^b f_n(t)g_n(t) dt = \sum\limits_{i=1}^na_i^{(n)}b_i^{(n)}(t_i-t_{i-1}) \tag{3} $$
We finish by seeing that $(2)$ and $(3)$ implys that the second sum on the RHS of $(1)$ converges to $\int_a^bf(s)g(s)ds$
Is this all correct??
Best Answer
This can be done by using the multidimensional Ito formula.
Now apply this to the function $F(x,y)=x_1x_2$ where $X_t=\int_a^tf(s)dB(s)$ and $Y_t=\int_a^tg(s)dB(s)$ and we get that $$ dX_tY_t=0+ Y_tdX + Y_tdX + d<X,Y>_t $$ and now after integrating and replacing $X$ and $Y$ this yields $$ \begin{align} I(f)I(g) &= \int_a^b\left(\int_a^tg(s)dB(s)\right)f(t)dB(t)+ \int_a^b\left(\int_a^tf(s)dB(s)\right)g(t)dB(t) + \int_a^bf(s)g(s)ds \\ &= \int_a^b\int_a^bf(t)g(s)dB(t)dB(s) + \int_a^bf(s)g(s)ds \end{align}$$