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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?

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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.

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  • $\begingroup$ Great answer. Thanks alot. The convergence in (1) also holds in $L^{2}$, doesn't it!? $\endgroup$ – Strickland Oct 3 '17 at 18:56
  • $\begingroup$ Yes, it hold in $L^2$, but, more generally, in probability. $\endgroup$ – Gordon Oct 3 '17 at 18:58

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