Uniformly distributed variable

Question

I've stumbled upon this correlation between two different variables thread and I can't understand this equation:

$$\mathbb{E}\cos(2\Phi+\lambda \cdot (s+t)) = \frac{1}{2\pi} \int_0^{2\pi} \cos(2x+ \lambda \cdot (t+s)) \, dx\text{ since }\Phi \sim U(0,2\pi)$$

Could anyone expand it?

I'm not sure if I'm right, but we might also expand it with trigonometric identities: $\mathbb{E}\cos(2\Phi+\lambda \cdot (s+t))=\mathbb{E}\cos(2\Phi)+\mathbb{E}\cos(\lambda \cdot (s+t))-\mathbb{E}\sin(\lambda(t+s))\mathbb{E}\sin(2\Phi) $

Does it mean that $\mathbb{E}\cos(2\Phi)=\int^{2\pi}_{0}\frac{1}{2\pi} \cos(2x) \, dx$ ?

is there a $2$ missing inside the cosinus that is the integrand? — Jean-Sébastien, Apr 29 '13 at 21:25
Then, as Did already answered, it is just the definition of the expected value of a function of $\Phi$. — Jean-Sébastien, Apr 29 '13 at 23:09

score 1 · Answer 1 · answered Apr 29 '13 at 21:42

1

For every (say, measurable and bounded) function $u$, $E[u(\Phi)]=\frac1{2\pi}\int\limits_0^{2\pi}u(x)\mathrm dx$. This is what $\Phi$ being uniformly distributed on $(0,2\pi)$ means, actually.

answered Apr 29 '13 at 21:42

Did

279,727

Trivial. Thank you all for help! – atomoutside Apr 30 '13 at 07:52

Uniformly distributed variable

1 Answers1