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Define

$$X_t := \left( \begin{matrix} \cos W_t \\ \sin W_t \end{matrix} \right).$$

where $W = \left( W_t,\mathcal F_t \right) _{t\ge0}$ is a standard Wiener process. Find the Ito differential of X and show that

$$\mathbb EX_t = \left( \begin{matrix} \exp \left( -\frac t2 \right) \\ 0 \end{matrix} \right).$$

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  • $\begingroup$ What did you try, what's the difficulty in solving it? Did you apply Ito's lemma to find the differential of $X$? $\endgroup$ – Ilya May 30 '13 at 14:59
  • $\begingroup$ I'm stuck at the beginning... :( I started with the two-dimensional Ito eq. $$df(t,\mathbf{X}_t) = \frac{\partial f}{\partial t} dt + (\boldsymbol{\nabla}_\mathbf{X}^{\mathsf T} f) d\mathbf{X}_t + \tfrac{1}{2} (d\mathbf{X}_t^\mathsf{T}) (\nabla_\mathbf{X}^2 f) d\mathbf{X}_t$$ but I have difficults to derive the integral form $\endgroup$ – user79133 May 30 '13 at 15:46
  • $\begingroup$ @user79133 This approach requires that you know about the Itô differential $dX_t$. But you don't - you want to determine it! Apply the Itô Formula to the Brownian motion instead... $\endgroup$ – saz May 30 '13 at 19:05
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By applying Itô's Formula $$f(W_t)-f(W_0) = \int_0^t f'(W_s) \, dW_s + \frac{1}{2} \int_0^t f''(W_s) \, ds$$

to $f(x) := \cos x$ we obtain

$$\cos(W_t)-1 = -\int_0^t \sin(W_s) \, dW_s - \frac{1}{2} \int_0^t \cos(W_s) \, ds$$

The mapping $$(t,w) \mapsto M(t,w) := \int_0^t \sin(W_s) \, dW_s$$ is a martingale since it's a stochastic integral with respect to a Brownian motion. This implies $\mathbb{E}M_t=0$. Thus,

$$\mathbb{E}(\cos(W_t))-1 = - \frac{1}{2} \int_0^t \mathbb{E}(\cos(W_s)) \, ds$$

Define $\varphi(s) := \mathbb{E}(\cos(W_s))$. The last equation is equivalent to

$$\varphi(t)-1 = - \frac{1}{2} \int_0^t \varphi(s) \, ds$$

Solve this differential equation and you are done with the first component of $X_t$. Similar argumentation works for $\sin(W_t)$.

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  • $\begingroup$ one issue: if I do as you suggest, I find $$\phi'+\frac 12\phi=\frac 12, \qquad \phi(0)=1$$ Now, the solution is $C \exp (-t/2) +1$ and $\phi(0)=C+1=1$ that means $C=0$ ?!?! Maybe I'm wrong? $\endgroup$ – user79133 Jun 10 '13 at 9:16
  • $\begingroup$ @user79133 How do you arrive at this differential equation? By differentiating the last equation, I obtain $$\varphi' = - \frac{1}{2} \varphi \qquad \varphi(0)=1.$$ $\endgroup$ – saz Jun 10 '13 at 12:51
  • $\begingroup$ yeah, I was wrong with the integral theorem. yeah I agree with your answer. thanks again $\endgroup$ – user79133 Jun 12 '13 at 7:07

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