# Conditional stochastic integration

Let's say we have two functions $h(s)$ and $g(s)$. We can easily simulate a stochastic integral, e.g. $$t \mapsto \int_0^t h(s) dB(s) \sim \mathcal{N}\bigg(0, \int_0^t h(s)^2 ds \bigg).$$ What is the conditional distribution of stochastic integral of $g(s)$ with respect to $B(s)$ then? $$t \mapsto \int_0^t g(s) dB(s) \bigg | \int_0^t h(s) dB(s) \sim ?$$ I have an intuition that these two integrals might have a joint normal distribution with a covariance equal to $\int_0^t h(s)g(s)ds$, but I need a rigorous derivation. Generally, I want to find a way to simulate a bunch of $n$ integrals with respect to the same Brownian motion: $$t \mapsto \bigg( \int_0^t h_1(s) dB(s), \int_0^t h_2(s) dB(s) , \dots , \int_0^t h_n(s) dB(s) \bigg)^T$$

The joint normal distribution follows from the way the stochastic integral is defined. However, if you are already convinced that $$$$I_t(h):=\int_0^t h(s) dB(s) \sim N\bigg(0, \int_0^t h(s)^2 ds \bigg)$$$$ for all $$h\in L^2([0,t])$$, then you have $$$$\sum_1^n r_i I_t(g_i)=I_t\Big(\sum_1^n r_i g_i\Big) \sim N\bigg(0, \int_0^t \Big(\sum_1^n r_i g_i(s)\Big)^2 ds \bigg) \tag{*}$$$$ for all real $$r_i$$'s and all functions $$g_i\in L^2([0,t])$$. So, all linear combinations of the $$I_t(g_i)$$'s are Gaussian and hence the $$I_t(g_i)$$'s are jointly Gaussian (think, e.g., of the joint characteristic function). Formula (*) for $$n=1,2$$ also yields the covariances: \begin{align*} Var(I_t(g_1)+I_t(g_2))&=\int_0^t (g_1(s)+g_2(s))^2 ds \\ &=\int_0^t g_1(s)^2 ds+\int_0^t g_2(s)^2 ds +2\int_0^t g_1(s)g_2(s)\,ds \\ &=Var(I_t(g_1))+Var(I_t(g_2))+2\int_0^t g_1(s)g_2(s)\,ds, \end{align*} so that $$Cov(I_t(g_1),I_t(g_2))=\int_0^t g_1(s)g_2(s)\,ds$$.
Added in response to a comment by Dr_Zaszuś: Thus, for any $$g,h$$ in $$L^2([0,t])$$, the pair $$(I_t(g),I_t(h))$$ has the bivariate normal distribution $$N(\mu_g,\mu_h,\sigma^2_g,\sigma^2_g,\rho)$$, where $$\mu_g=\mu_h=0$$, $$\sigma_g=\|g\|$$, $$\sigma_h=\|h\|$$, and $$\rho=\frac{g\cdot h}{\|g\|\|h\|}$$, where $$g\cdot h:=\int_0^t g(s)h(s)\,ds$$ and $$\|f\|:=\sqrt{f\cdot f}$$. So, the conditional distribution of $$I_t(g)$$ given $$I_t(h)=y$$ is the normal distribution $$N(\mu_{g;y},(1-\rho^2)\sigma_g^2)$$ for any real $$y$$, where $$\mu_{g;y}:=\rho\sigma_g y/\sigma_h$$.
• I understand it has been a year since the answer, but I am not convinced this answers the original question. The calculated variance and covariance are not the conditional variances that are requested, but they are averaged over the realizations of both $I_t(g_1)$ and $I_t(g_2)$. However, what is asked here is to find the conditional variance (and mean) of $I_t(g_2)$ given one specific value of $I_t(g_1)$. I have asked a similar question as I haven't found this one: mathoverflow.net/questions/315781/… – Dr_Zaszuś Nov 20 '18 at 16:17