Ito integrals and copulas

Question

Let $X_{t}$ and $Y_{t}$ be two brownian motions and let their joint distribution be given by $F$. So in regularly correlated BM's where $dX_{t}dY_{t}=\rho dt$, we have a bivariate normal distribution for $X$ and $Y$.

Does this mean that $\int_{0}^{t}g(s)dX_{s}$ and $\int_{0}^{t}g(s)dY_{s}$ have the same bivariate distribution? So again in the case of regularly correlated BM's, this would imply a bivariate normal distribution but with different mean, and covariance matrices?

Does this work for all distributions? Lets say $X_{t}$ and $Y_{t}$ are distributed with copula $C$, does this mean that $\int_{0}^{t}g(s)dX_{s}$ and $\int_{0}^{t}g(s)dY_{s}$ are also distributed with copula $C$?

And if so, why?

Answer 1

0/ Let's me use more common notations to avoid misunderstanding. We will consider $B_t^x$ and $B_t^y$ - two correlated Brownian motions, e.g. $<dB_t^x,dB_t^y>=\rho dt$.

Just to recall, Ito's process: $$X_t = X_0 + \int_0^t \mu(s,\omega) ds + \int_0^t \sigma(s,\omega) dB_s^x\\ dX_t=\mu(t,\omega) dt + \sigma(t,\omega) dB_t^x$$

1/ Single BMs: $$\mathbb{E}(B_t) = 0\\ \mathbb{E}((B_t)^2) = t \\\mathbb{Cov}(B_t^x,B_t^y) = \rho t$$

2/ Integrals: $$I(t) = \int_0^t g(s) dB_s $$ $I(t)$ is Ito process with zero drift => $\mathbb{E}(I(t)) = 0$.

From your notation it seems like $g(t)$ is deterministic, hence
$$\mathbb{E}(I(t)^2) = \int_0^t g^2(s)ds \\ \mathbb{Cov}(I_x(t),I_y(t)) = \rho \int_0^t g^2(s)ds.$$

So we may say that in both cases you have multivariate normal distribution with the same correlation matrix but different scaling factor.

3/ Does this work for all distributions? => No, just consider lognormal distribution. The trick is that the sum of normal distributions is a normal distribution, which is not the case for any distribution.