When Deterministic Brownian Integrals Preserve Joint Normality
Summary
The document considers integrals of two correlated Brownian motions against a deterministic function. For this setting, each stochastic integral has zero mean, and its variance is the integral of the squared function over time. The covariance between the two integrals is the Brownian correlation multiplied by that same variance term. Because deterministic integrands preserve joint Gaussianity, the pair remains jointly normal with the same correlation and rescaled covariance.
It cautions against extending this result to arbitrary dependence structures or distributions. The answer notes that sums of normal variables remain normal, whereas this closure property does not hold for every distribution, and gives the lognormal case as a contrast. The explanation is limited to deterministic integrands and correlated Brownian motion; it does not derive conditions for random integrands or show that general copulas are preserved under stochastic integration.
Key ideas
- Integrating correlated Brownian motions against a deterministic function yields jointly normal integrals.
- Each integral has zero mean and variance equal to the time integral of the squared integrand.
- The covariance scales with the Brownian correlation and the same integrated squared function.
- The result does not extend automatically to arbitrary marginal distributions or copulas.
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Full text
# Ito integrals and copulas
# Ito integrals and copulas
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 by Bruno (score 2)
https://quant.stackexchange.com/a/14564
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.