Suppose $X$ is uniformly distributed on $[0, 2\pi]$. Let $Y = \sin X$ and $Z = \cos X$. Show that the correlation between $Y$ and $Z$ is zero.
It seems I would need to know the standard deviation of the sine and cosine, and their covariance. How can I calculate these?
I think I need to assume $X$ has uniform distribution, and the look at the transformed variables $Y=\sin(X)$ and $Z=\cos(X)$. Then the law of the unconscious statistician would give the expected value
$$E[Y] = \frac{1}{b-a}\int_{-\infty}^{\infty} \sin(x)dx$$ and $$E[Z] = \frac{1}{b-a}\int_{-\infty}^{\infty} \cos(x)dx$$
(the density is constant since it is a uniform distribution, and can thus be moved out of the integral).
However, those integrals are not defined (but have Cauchy principal values of zero I think).
How could I solve this problem? I think I know the solution (correlation is zero because sine and cosine have opposite phases) but I cannot find how to derive it.
Best Answer
Since
$$\begin{align} \operatorname{Cov}(Y, Z) &= E[(Y - E[Y])(Z - E[Z])] \\ &= E[(Y - {\textstyle \int}_0^{2\pi} \sin x \;dx)(Z - {\textstyle \int}_0^{2\pi} \cos x \;dx)] \\ &= E[(Y - 0)(Z - 0)] \\ &= E[YZ] \\ &= \int_0^{2\pi} \sin x \cos x \;dx \\ &= 0 , \end{align}$$
the correlation must also be 0.