3

What is the general expression for the covariance $cov \left[ X_s X_t \right]$ of a stochastic process given by \begin{equation} dX_t = f(X_t,t)dt + g(X_t,t) dW_t \end{equation} for some general (sufficiently well-behaved) functions $f$ and $g$?

TheBridge
  • 5,811
user56643
  • 299
  • 2
  • 8
  • What do you mean by "general expression"? General expression in terms of ... ? – saz Mar 09 '15 at 13:00
  • Formally, the solution to the above is just \begin{equation} X(t) = X_0 + \int_0^t f(X,r) dr + \int_0^t g(X,r) dW_r. \end{equation} for some initial value $X_0$. So now I can just calculate $cov \left[ X(s) X(t) \right]$ by plugging in this formal solution. Of course, I cannot evaluate the integrals but I assume that several of the integrals that will appear in this calculation will drop out due to standard rules from stochastic calculus such as \begin{equation} \mathbb{E} \left[ \int f(X,r) dW_r \right] = 0 \end{equation} and possibly many more. – user56643 Mar 10 '15 at 02:41
  • 1
    Well, yes... so have you tried Itô's formula? – saz Mar 10 '15 at 06:21
  • @ user56643 : You can even use the weaker than ITô's rule, stochastic integration by parts formula (for continuous semimartingale) ? Best regards – TheBridge Mar 12 '15 at 07:36
  • 2
    If $X_0=0$, the covariance of $X_s$ and $X_t$ is $$\int_0^{t\wedge s}E(g(X_u,u)^2)du.$$ – Did Mar 13 '15 at 23:37
  • Thx Did, that's exactly the kind of result that I was looking for. I assume you obtained this by plugging in the above formal solution (with $X_0 = 0$) into the defintion of the correlation and then using that several of the resulting integrals disappear or simplify. Do you know a source which shows the derivation of this? – user56643 Mar 14 '15 at 03:03

2 Answers2

2

Here we simply apply Itô isometry. So as explained Quadratic Variations and the Itô Isometry we have

$$ EX_{t}X_{r}=E[\int^t b ds\int^r b ds]+E[\int^t b ds \int^r \sigma dB_{s}]+E[\int^r b ds \int^t \sigma dB_{s}]+E[\int^{t} \sigma dB_{s}\int^r \sigma dB_{s}]$$

A good reference for quadratic variation computations is Revuz-Yor chapter IV. For Brownian motion, we know that $d\langle B,t\rangle=0$ is zero because generally the cross-variation of a semimartingale and an increasing process is zero Quadratic Variation of Increasing Process?. The quadratic variation for $B_{t}$ is $t$ and so we indeed get by Itô isometry

$$ EX_{t}X_{r}=E[\int^t b ds\int^r b ds]+E[\int^{t} \sigma dB_{s}\int^r \sigma dB_{s}]$$

$$ =E[\int^t b ds\int^r b ds]+\int^{\min(t,r)} E\sigma^{2} ds.$$

Thomas Kojar
  • 7,349
0

Too long for a comment. Temporarily I thought that for non deterministic integrands it is not possible to deduce from the Ito isometry $$\tag{1} \mathbb E\big[\textstyle(\int_0^t\alpha_u\,dW_u)(\int_0^t\beta_u\,dW_u)\big]= \mathbb E\big[\textstyle\int_0^t\alpha_u\beta_u\,du\big] $$ that $$\tag{2} \mathbb E\big[\textstyle(\int_0^t\alpha_u\,dW_u)(\int_0^\color{red}{s}\beta_u\,dW_u)\big]= \textstyle\int_0^{t\wedge s}\mathbb E\big[\alpha_u\beta_u\big]\,du $$ holds. But this is in fact quite simple: The LHS of (2) equals \begin{align} &\mathbb E\big[\textstyle(\int_0^t\alpha_u\,dW_u)(\int_0^t\beta_u1_{\{u\le s\}}\,dW_u)\big]\stackrel{(1)}{=}\mathbb E\big[\int_0^t\alpha_u\beta_u1_{\{u\le s\}}\,ds\big]\\[2mm] &=\textstyle\int_0^t\mathbb E\big[\alpha_u\beta_u\big]1_{\{u\le s\}}\,ds= \int_0^{t\wedge s}\mathbb E\big[\alpha_u\beta_u\big]\,du\,.\tag{3} \end{align}

Kurt G.
  • 17,136