# 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.

## 2 Answers

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

• It appears to me that $cov(X_t, Y_t)=0$, if $\rho_{X,Y}=0$. – Gordon Jun 2 '16 at 15:01
• Well it doesn't seem to me: neither when starting from BehrouzMaleki's hint nor from my simulations. Could you explain your reasoning? – Quantuple Jun 2 '16 at 15:56
• What I meant was from a formulaic perspective based on your dynamics, since they are log-normal, the co-variance can be computed explicitly. – Gordon Jun 2 '16 at 15:58
• I realise that my answer has nothing to do with the initial question. I was talking about the cross-sectional covariance (fixed realisation) while you talk about the covariance between two random variables. My bad! – Quantuple Jun 2 '16 at 16:13

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 $$Cov(x_t,y_t)=E[x_ty_t]-E[x_t]E[y_t]$$

• I also got that hint by myself but I got stuck at that stage. So, please give me more detail for example whether $\operatorname{cov}$ is zero or not. – imp Jun 1 '16 at 23:09