I am studying stochastic processes and found the next problem:

Let $A$ and $\Phi $ be two independent random variables such that $E(A) = 0$, $E(A^2) < \infty$, and $\Phi$ is uniformly distributed on $[0, 2\pi]$. Consider the stochastic process $$X_t =A \cos{(\Phi+λt)}, t≥0$$

where $\lambda$ is a fixed real parameter.

Show that the correlation between two different variables of the process is given by $$E(X_tX_s)= \frac{1}{2}E(A^2)\cos {\lambda}(t-s)$$

what i have done:

Using the fact that $Cov (X_s, X_t) = E\{ (X_s - EX_s)(X_t - EX_t) \}$, I compute this and found that $Cov (X_s, X_t) = E(A^2)E\{ \cos(\Phi + \lambda s)\cos (\Phi + \lambda t)- A \cos (\Phi +\lambda s)E(A \cos (\Phi + \lambda t )) - E(A \cos (\Phi+\lambda s))A\cos (\Phi+\lambda s) + E(A)E(A)E(\cos (\Phi+\lambda s) \cos (\Phi + \lambda t)) \}$

Using the hipothesis $E(A) = 0$,

I have that

$Cov (X_s, X_t) = E(A^2)E(\cos (\Phi + \lambda s) \cos (\Phi + \lambda t))$

but I don't know how to calculate $E(\cos (\Phi + \lambda s) \cos (\Phi + \lambda t))$, I know that I need to use the fact that $\Phi \sim U_{[0, 2\pi]}$ but I don't know how to do this

(I think that if I can make this, is possible to use the theorem that says $Cov(X,Y)= E(XY)- EXEY$ to solve the problem)

  • $\begingroup$ Use the fact that $\cos(\Phi+\lambda t)\cos(\Phi+\lambda t) = \frac{1}{2}(\cos (\lambda(t-s))+\cos(2 \Phi+\lambda(t+s)))$. $\endgroup$ – copper.hat Mar 10 '13 at 17:33

Note that $$\cos(x) \cdot \cos(y) = \frac{1}{2}(\cos(x-y) + \cos(x+y))$$

i.e. $$\cos(\Phi+\lambda \cdot s) \cdot \cos(\Phi+\lambda \cdot t) = \frac{1}{2} \cdot( \cos(\lambda \cdot (t-s)) + \cos(2\Phi+\lambda \cdot (s+t))$$

Thus $$\mathbb{E}\big(\cos(\Phi+\lambda \cdot s) \cdot \cos(\Phi+\lambda \cdot t) \big) = \frac{1}{2} \left( \cos(\lambda \cdot (t-s)) + \mathbb{E}\cos(2\Phi+\lambda \cdot (s+t)) \right)$$


$$\mathbb{E}\cos(2\Phi+\lambda \cdot (s+t)) = \frac{1}{2\pi} \int_0^{2\pi} \cos(x+ \lambda \cdot (t+s)) \, dx$$ since $\Phi \sim U(0,2\pi)$. By the periodicity of $\cos$ we conclude $$\mathbb{E}\cos(2\Phi+\lambda \cdot (s+t))=0$$

So we arrive at $$\text{cov}(X_s,X_t) = \frac{1}{2} \mathbb{E}(A^2) \cdot \cos(\lambda \cdot (t-s))$$

By the way, there is a typo in the second line of your computations: Instead of $$\mathbb{E}(A) \cdot \mathbb{E}(A) \cdot \mathbb{E}(\cos(\Phi+\lambda \cdot s) \cdot \cos(\Phi+\lambda \cdot t))$$ it should read $$\mathbb{E}(A) \cdot \mathbb{E}(A) \cdot \mathbb{E}\cos(\Phi+\lambda \cdot s) \cdot \mathbb{E}\cos(\Phi+\lambda \cdot t)$$ (Of course, that doesn't change anything in this case since $\mathbb{E}(A)=0$.)

  • $\begingroup$ Momentary lack of caffeine. Have deleted the comment. $\endgroup$ – copper.hat Mar 10 '13 at 17:42
  • $\begingroup$ thank u, I really appreciate your help $\endgroup$ – LFRC Mar 10 '13 at 17:58

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