I haven't found any similar question in the forum, so I trust some of you will find this thought-provoking (at the very least). Perhaps you can help me.

Let's consider first the two following stochastic differential equations: $$dX_t = (6+3X_t)dt + (2+X_t)dB_t, X_0 = 1$$ $$dY_t = 3Y_tdt + Y_tdB_t, Y_0 = 1$$

Of course $B_t$ is a standard Brownian Motion started at zero.

Now the question is: how can I compute the quadratic covariation of these two processes, and then obtain its expectation? That is, how can I compute $E\left(\left<X,Y\right>_t\right)$.

Regards and thanks in advance.

  • $\begingroup$ What definition of quadratic (co)variation do you know/use? $\endgroup$ – saz Jan 13 '14 at 15:51
  • $\begingroup$ well, for example, if $X$ and $Y$ are continuous semimartingales, we could define its covariance process with the integration by parts formula: $<X,Y>_t - <X_0,Y_0> = \int_0^t Y_sdX_s + \int_0^t X_sdY_s$ $\endgroup$ – Adam Jan 13 '14 at 16:20
  • $\begingroup$ The covariation (not covariance) process is not what you write. Check WP. $\endgroup$ – Did Jan 13 '14 at 17:18
  • $\begingroup$ yeah sorry in that last comment I meant covariation, not covariance. Anyway I need to compute $E<X,Y>_t$, regardless of how you would like to name it. $\endgroup$ – Adam Jan 13 '14 at 17:22

Define the quadratic variation of a semimartingale $(X_t)_{t \geq 0}$ by

$$[X,X]_t := \mathbb{P}-\lim_{n \to \infty} \sum_{j=0}^n (X_{t_j}-X_{t_{j-1}})^2$$

where $\Pi_n := \{t_0<\ldots<t_n<t\}$ is a sequence of partitions such that $|\Pi_n| \to 0$. Moreover, we set $$[X,Y]_t := \frac{1}{4} ([X+Y]_t-[X-Y]_t).$$

Using this definition it is not difficult to show the following result.

Lemma 1 Let $(M_t)_{t \geq 0}$ a continuous square-integrable martingale and $(A_t)_{t \geq 0}$ a continuous process of bounded variation. Then,

  1. $[M]_t$ is the unique previsible process such that $M_t^2-[M]_t$ is a martingale.
  2. $[M,A]_t=[A,A]_t=0$

Now we are ready to calculate the quadratic covariation. By definition,

$$X_t\pm Y_t= \underbrace{\int_0^t (2+X_s \pm Y_s) \, dB_s}_{=:M_t} + \underbrace{c_{\pm}+ \int_0^t (6+3X_s \pm 3Y_s) \, ds}_{=:A_t}.$$

where $c_+=2$, $c_-=0$. Obviously, $(M_t)_{t \geq 0}$ is a (continuous) martingale and $(A_t)_{t \geq 0}$ of bounded variation. Consequently, we obtain by applying Lemma 1

$$[X \pm Y,X \pm Y]_t = [M,M]_t+2[M,A]_t+[A,A]_t = \int_0^t (2+X_s \pm Y_s)^2 \, ds.$$


$$[X,Y]_t = \frac{1}{4} ([X+Y]_t-[X-Y]_t) = \int_0^t (2+X_s) \cdot Y_s \, ds \tag{1}$$

In order to compute $\mathbb{E}[X,Y]_t$, we have to find $\mathbb{E}Y_t$ and $\mathbb{E}(X_t \cdot Y_t)$. Since stochastic integrals with respect to a Brownian motion are martingales, we find

$$f(t) :=\mathbb{E}(Y_t)=1+\underbrace{\mathbb{E} \left( \int_0^t Y_s \, dB_s \right)}_{0} + 3 \int_0^t \mathbb{E}(Y_s) \, ds$$

i.e. $f$ satisfies the ODE

$$f'(t) = 3f(t) \qquad f(0)=1$$

Obviously, the unique solution is given by $$\mathbb{E}Y_t = f(t)= e^{3t} \tag{2}$$ Similarly, we obtain from Itô's formula that

$$g(t) := \mathbb{E}(X_t \cdot Y_t) =1+\mathbb{E} \left( \int_0^t(7X_s Y_s+8Y_s) \, ds \right) = 8 \int_0^t f(s) \, ds + 7 \int_0^t g(s) \, ds$$

i.e. $$g'(t) = 8f(t)+7g(t) = 8e^{3t}+7g(t) \tag{3}$$

This ODE can be solved explicitely; I leave it to you. Combining $(1)$, $(2)$ and $(3)$ allows us to compute $\mathbb{E}[X,Y]_t$.

  • $\begingroup$ Thanks very much, although in the equation $$X_t\pm Y_t= \underbrace{\int_0^t (2+X_s \pm Y_s) \, dB_s}_{=:M_t} + \underbrace{1+ \int_0^t (6+3X_s \pm 3Y_s) \, ds}_{=:A_t}$$ I do not understand why the bounded variation process begins with the 1. I think it should be 2, since $X_0 + Y_0 = 2$. $\endgroup$ – Adam Jan 13 '14 at 18:02
  • $\begingroup$ @Adam Yes, it should be $0$ or $2$, depending on $\pm$ ... I'll correct it. $\endgroup$ – saz Jan 13 '14 at 18:04
  • $\begingroup$ One last thing, after applying Itô's product rule before the last step I believe we should be left with: $$g(t) := \mathbb{E}(X_t \cdot Y_t) =\mathbb{E} \left( \int_0^t(6X_s Y_s+6Y_s) \, ds \right).$$ $\endgroup$ – Adam Jan 13 '14 at 18:29
  • 2
    $\begingroup$ @Adam No, I don't think so. Note that $$X_t \cdot Y_t-1 = \int_0^t X_s dY_s+ \int_0^t Y_s \, dX_s + \int_0^t d [X,Y]_s$$ I guess, you forgot the last term. $\endgroup$ – saz Jan 13 '14 at 19:09

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