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.

  • 1
    $\begingroup$ As stated, your problem is insufficiently defined. Correlation is a concept that applies to random variables, not functions. (Formally, a random variable is a kind of function, namely a measurable function from a probability space to the real numbers equipped with the Borel measure. But just saying "the sine function" doesn't tell you anything about the probability measure in the domain, which is what gets you probabilistic information, including joint distributions.) $\endgroup$ Commented Jul 16, 2016 at 16:16
  • $\begingroup$ If I assume time is a uniform random variable ($X$ in my text), is it not possible to do this? I mean I would be then looking at the correlation of two transformed random variables. $\endgroup$
    – uklady
    Commented Jul 16, 2016 at 16:18
  • 3
    $\begingroup$ So you want $X$ uniformly distributed and then you define $Y = \sin X$ and $Z = \cos X$? That's fine except you also need to specify the support of $X$'s density, since there is no uniform distribution over the whole of $ℝ$, or any other infinitely long interval. $\endgroup$ Commented Jul 16, 2016 at 16:21
  • $\begingroup$ Maybe I could take $[0, 2*pi]$ as the support (I would be assuming that $f=1$, so the interval contains one full cycle). I guess the integration problems will then vanish as well $\endgroup$
    – uklady
    Commented Jul 16, 2016 at 16:23
  • 10
    $\begingroup$ If you do that, then you only need draw a scatterplot--no integration is necessary. That scatterplot is a uniform distribution on the unit circle (obviously). Since the circle is symmetric under any reflection through the origin, the correlation equals its negative, whence it must be zero, QED. $\endgroup$
    – whuber
    Commented Jul 16, 2016 at 18:50

2 Answers 2



$$\begin{align} \operatorname{Cov}(Y, Z) &= E[(Y - E[Y])(Z - E[Z])] \\ &= E[(Y - {\textstyle \int}_0^{2\pi} \sin x \;dx)(Z - {\textstyle \int}_0^{2\pi} \cos x \;dx)] \\ &= E[(Y - 0)(Z - 0)] \\ &= E[YZ] \\ &= \int_0^{2\pi} \sin x \cos x \;dx \\ &= 0 , \end{align}$$

the correlation must also be 0.


I really like @whuber's argument from symmetry and don't want it to be lost as a comment, so here's a bit of elaboration.

Consider the random vector $(X, Y)$, where $X = \cos(U)$, and $Y = \sin(U)$, for $U \sim U(0, 2 \pi)$. Then, because $\theta \mapsto (\cos(\theta), \sin(\theta))$ parameterizes the unit circle by arc length, $(X, Y)$ is distributed uniformly on the unit circle. In particular, the distribution of $(-X, Y)$ is the same as the distribution of $(X, Y)$. But then

$$ - \text{Cov} (X, Y) = \text{Cov} (-X, Y) = \text{Cov} (X, Y) $$

so it must be that $\text{Cov} (X, Y) = 0$.

Just a beautiful geometric argument.


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