1
$\begingroup$

In Proof of Proposition 1.2.20 in the following lectures notes

http://math.uni-heidelberg.de/studinfo/reiss/sode-lecture.pdf

I found following quote " stochastic integrals with respect to independent Brownian motions are uncorrelated (attention: they may well be dependent!)."

So I was wondering now when this can occur.

Given two independent Wiener processes $W_{1}$ and $W_{2}$. We consider two Ito integrals:

$\int_{0}^{t} F(s)\,dW_{1}(s)$ and $\int_{0}^{t} G(s)\, dW_{2}(s)$.

From general (multi-dimensional) Ito-isometry, the two integrals are uncorrelated. Using approximations of $F$ and $G$ by simple functions, one can show uncorrelation probably pretty straight forward. However, I am wondering when the two integrals can become dependent. My questions therefore are:

  1. Is there an easy example for this?

  2. Assuming that $F$ and $G$ are deterministic, does then follow independence? And if so, how does one show this?

$\endgroup$
0

1 Answer 1

1
$\begingroup$

Note that, by Ito's isometry, \begin{align*} E\left(\int_{0}^{t} F(s)\,dW_{1}(s)\int_{0}^{t} G(s)\, dW_{2}(s)\right) &= \int_0^t E\big(F(s) G(s) \big) d\langle W_1, W_2\rangle_t =0. \end{align*} That is, $\int_{0}^{t} F(s)\,dW_{1}(s)$ and $\int_{0}^{t} G(s)\,dW_2(s)$ are uncorrelated.

If $F(s)$ and $G(s)$ are deterministic, then, for any constants $a$ and $b$ \begin{align*} a\int_{0}^{t} F(s)\,dW_{1}(s) + b\int_{0}^{t} G(s)\,dW_2(s) \end{align*} is the limit, in probability, of the sum \begin{align*} \sum_{i=1}^n \Big(aF(t_{i-1}) \big(W_1(t_i)-W_1(t_{i-1})\big) +bG(t_{i-1}) \big(W_2(t_i)-W_2(t_{i-1})\big)\Big),\tag{1} \end{align*} where $0=t_0 < t_1 < \cdots < t_n = t$ is a partition of $[0, t]$. Note that all terms in $(1)$ are normal and independent, and then their sum is also normal. Since the limit, in probability, of normal random variables is also normal, \begin{align*} a\int_{0}^{t} F(s)\,dW_{1}(s) + b\int_{0}^{t} G(s)\,dW_2(s) \end{align*} is normal. That is, $\int_{0}^{t} F(s)\,dW_{1}(s)$ and $\int_{0}^{t} G(s)\,dW_2(s)$ are jointly normal, and are therefore independent (note that the joint characteristic function is the product of the individual characteristic functions).

However, if $F(s)$ and $G(s)$ are not deterministic, then $\int_{0}^{t} F(s)\,dW_{1}(s)$ and $\int_{0}^{t} G(s)\,dW_2(s)$ are not necessarily independent. For example, consider $X_t=\int_{0}^{t} W_2(s)\,dW_{1}(s)$ and $Y_t=\int_{0}^{t} W_1(s)\,dW_2(s)$. Then $X_t$ and $Y_t$ is uncorrelated, but are not independent.

$X_t$ and $Y_t$ are uncorrelated but are not independent.

Note that \begin{align*} dX_t^2 = 2X_tdX_t + W_2^2(t) dt, \end{align*} and \begin{align*} dY_t^2 = 2Y_t dY_t + W_1^2(t) dt. \end{align*} Then, \begin{align*} X_t^2 Y_t^2 = 2\int_0^t Y_s^2 X_s dX_s+2\int_0^t X_s^2 Y_s dY_s + \int_0^t \left(W_1^2(s)X_s^2 + W_2^2(s) Y_s^2 \right) ds. \end{align*} By symmetry, \begin{align*} E\left( X_t^2 Y_t^2\right) = 2\int_0^t E\left(W_1^2(s)X_s^2\right) ds.\tag{2} \end{align*} Note that \begin{align*} dW_1^2(t) = 2W_1(t) dW_1(t) + dt. \end{align*} Then \begin{align*} W_1^2(t)X_t^2 &= 2\int_0^t X_s W_1(s)^2 dX_s + 2\int_0^t W_1(s) X_s^2 dW_1(s)\\ &\qquad +\int_0^t \left(W_1(s)^2 W_2(s)^2 + X_s^2 + 4X_sW_1(s)W_2(s)\right)ds.\tag{3} \end{align*} Furthermore, note that \begin{align*} W_1(t)W_2(t) = X_t + Y_t. \end{align*} Then, \begin{align*} E\left( X_tW_1(t)W_2(t)\right) &= E\left(X_t^2 + X_t Y_t\right)\\ &=\int_0^t s ds=\frac{1}{2}t^2. \end{align*} Therefore, from $(3)$, \begin{align*} E\left(W_1^2(t)X_t^2\right) &=\int_0^t \left(s^2 + \frac{1}{2}s^2 + 2s^2 \right)ds = \frac{7}{6}t^3, \end{align*} and, from $(2)$, \begin{align*} E\left( X_t^2 Y_t^2\right) &= 2\int_0^t E\left(W_1^2(s)X_s^2\right) ds\\ &=\frac{7}{3} \int_0^t s^3ds = \frac{7}{12} t^4. \end{align*} However, \begin{align*} E\left( X_t^2 \right)E\left(Y_t^2\right) = \frac{1}{4}t^4. \end{align*} That is, $X_t$ and $Y_t$ are not independent.

$\endgroup$
2
  • $\begingroup$ Great answer. Thanks alot. The convergence in (1) also holds in $L^{2}$, doesn't it!? $\endgroup$
    – Strickland
    Oct 3, 2017 at 18:56
  • $\begingroup$ Yes, it hold in $L^2$, but, more generally, in probability. $\endgroup$
    – Gordon
    Oct 3, 2017 at 18:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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