2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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

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

Hence,

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

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 $\mathbb{E}[X,Y]_t$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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