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?

  • $\begingroup$ What is $g(s)$? $\endgroup$ – emcor Aug 29 '14 at 13:56

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.