Uncorrelated Brownian Drivers Do Not Ensure Uncorrelated Processes
Summary
The document addresses whether zero instantaneous covariance between the Brownian drivers of two stochastic differential equations guarantees zero covariance between the resulting processes at a fixed time. The answer is no in general, and it distinguishes the covariance of process values from the correlation of their local increments. It also cautions that zero covariance does not generally imply independence.
An example uses one process as a geometric diffusion and makes the second process’s dynamics depend on the first. Even with uncorrelated Brownian increments, the evolving levels can become correlated through their dynamics. The answer attributes this effect to the drift terms, though the example also makes the second process depend directly on the first, so the general lesson is that shared state dependence can matter as well. A second response sketches an Itô product calculation: zero cross-variation removes one term, but does not by itself show that the covariance vanishes. The example is illustrative rather than a general proof or classification of conditions for independence.
Key ideas
- Zero instantaneous covariance between Brownian drivers does not generally imply zero covariance between process values.
- Uncorrelated random variables are not necessarily independent.
- Dependence between process dynamics can generate correlation even when the driving Brownian increments are uncorrelated.
- Zero cross-variation removes a term in the Itô product rule but does not settle the covariance by itself.
- The example illustrates the issue without specifying general sufficient conditions for independence.
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Full text
# Are two stochastic processes independent if the Wiener processes inside are uncorrelated
# Are two stochastic processes independent if the Wiener processes inside are uncorrelated
Assume there are two stochastic processes: $dx_t = \alpha_1(x_t,t)dt + \beta_1(x_t,t)dW^1_t$ and $dy_t = \alpha_2(y_t,t)dt + \beta_2(y_t,t)dW^2_t$.
Does $dW^1_t\times{dW^2_t} = 0$ imply that $\operatorname{cov}(x_t, y_t) = 0$?
If it does, please give me a proof.
## Answer by Quantuple (score 3)
https://quant.stackexchange.com/a/26398
[Edit]
My "answer" below is not a really an answer for I have completely misinterpreted your original question. I thought you asked about the covariance of 2 processes over a given time horizon (i.e. for a fixed $\omega$) and not the covariance of two random variables (fixed $t$). Also note that $\text{cov}(x,y)=0$ does not mean that $x$ and $y$ are independent (except special case where they have an elliptic distribution), it just means they are uncorrelated (Pearson linear correlation assumed).
No this is not true in general.
Take the example below where I've assumed \begin{gather} dX_t = X_t( r_X dt + \sigma_X dW_t^X ),\ \ X(0) = X_0 \\ dY_t = X_t( r_Y dt + \sigma_Y dW_t^Y ),\ \ Y(0) = Y_0 \\ d\langle W^X, W^Y \rangle_t = \rho_{XY} \end{gather} with \begin{gather} X_0 = 1, Y_0 = 2 \\ r_X = 50\%, r_Y = -50\%\\ \sigma_X = 50\%, \sigma_Y = 25\% \end{gather}
Now take $\rho_{XY} = 0\%$ and you get the figure below: although the Brownian increments are uncorrelated (here represented through the log-returns, bottom subplot), the processes $X_t$ and $Y_t$ clearly exhibit significant correlation (see top subplot with a significant sample Pearson correlation).
This is essentially due to the drift terms in the SDE.
Note that if you take $\rho_{XY}=99\%$ you won't see that correlation (recall it is a correlation between local increments) transpire at the global level either, see below
## Answer by user16651 (score 1)
https://quant.stackexchange.com/a/26391
Hint:
By Integration, we have $$x_t=x_{0}+\int_{0}^{t} \alpha_1(x_s,s)ds+\int_{0}^{t} \beta_1(x_s,s)dW_1(s)$$ $$y_t=y_{0}+\int_{0}^{t} \alpha_2(y_s,s)ds+\int_{0}^{t} \beta_2(y_s,s)dW_2(s)$$ then $$E[x_t]=x_0+E\left[\int_{0}^{t} \alpha_1(x_s,s)ds\right]$$ $$E[y_t]=y_0+E\left[\int_{0}^{t} \alpha_2(y_s,s)ds\right]$$ Now we apply Ito's lemma $$d(x_ty_t)=x_tdy_t+y_tdx_t+\underbrace{d[x_t,y_t]}_{0}$$ as a result $$x_ty_t=x_0y_0+\int_{0}^{t}x_sdy_s+\int_{0}^{t}y_sdx_s$$ on the other hand $$Cov(x_t,y_t)=E[x_ty_t]-E[x_t]E[y_t]$$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.