Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. It only takes a minute to sign up.
Sign up to join this community
This may be a trivial question but I want to consult with you all.
Let U be a continuous random variable taking values int he interval [0,2pi]. Let X = cos(U), Y = sin(U). Determine the Pearson correlation between X and Y.
How to calculate the E[XY]?
Using the identity $\sin(2U) = 2\sin(U)\cos(U)$, we get:
$\displaystyle\mathbb{E}[XY] = \mathbb{E}[\sin(U)\cos(U)] = \mathbb{E}\left[\frac{\sin(2U)}{2}\right] = \frac{1}{2}\mathbb{E}\left[\sin(2U)\right] = \frac{1}{2}\int_0^{2\pi}\frac{1}{2\pi}\sin(2\theta)d\theta = \frac{1}{8\pi}\int_0^{4\pi}\sin(x)dx = 0 $
This argument requires some geometric intuition and knowing that the Pearson correlation is proportional to the linear regression coefficient of $Y$ against $X$.
As $U$ ranges from $0$ to $2\pi$, $(X,Y)$ describes a circle, covering the upper half of the circle in the same way it covers the lower half. Thus the conditional distribution of $Y$ given $X$ is symmetric about $0$, forcing the regression line to be $Y=0$, implying the correlation is zero, whence the covariance is zero, too. Since (by circular symmetry) both $X$ and $Y$ have zero expectations, the covariance is merely $\mathbb{E}(XY)$ and we're done.
That had a hand-waving intuitive flavor to it. Maybe we can do a little better. Here's a more elaborate version of the same idea. It doesn't require knowing about regression.
Changing $U$ to $-U$ does not change the distribution of $(X,Y)$: it only causes $(X,Y)$ to traverse the circle clockwise rather than counterclockwise and the circle is still covered uniformly because the distribution of $-U$ is uniform between $-2\pi$ and $0$. But changing $U$ to $-U$ leaves $X$ alone and converts $Y$ to $-Y$. Because the distribution is unchanged, the expectations remain the same, too: $$\mathbb{E}(XY)=\mathbb{E}(X(-Y)) = -\mathbb{E}(XY).$$ Since $|XY|$ is bounded, the expectation of $XY$ exists and is finite, whence the only possible value of its expectation is zero.
OK, maybe that doesn't seem sufficiently rigorous. Here's another approach.
Cosine and sine depend only on $U$ up to a multiple of $2\pi$. Therefore you may allow $U$ to range uniformly over any interval of length $2\pi$, such as from $\pi/2$ to $\pi/2+2\pi$. That's equivalent to adding the start of the interval, $\pi/2$, to $U$. In light of this, notice that
$$X = \cos(U)\text{ becomes } \cos(U+\pi/2) = \sin(U) = -Y$$
and
$$Y = \sin(U)\text{ becomes } \sin(U+\pi/2) = \cos(U) = X$$
The expectation of $(X+Y)^2$ thereby becomes
$$\eqalign{ \mathbb{E}(X^2+Y^2) + \mathbb{E}(2XY) &= \mathbb{E}((X+Y)^2) \\ &= \mathbb{E}((-Y+X)^2) = \mathbb{E}(X^2+Y^2) + \mathbb{E}(-2XY). } $$
(These equations are the consequences of straightforward algebraic relationships $(X\pm Y)^2 = X^2+Y^2\pm 2XY$ and the linearity of expectation.)
Subtracting the right from the left and dividing that by $4$ yields
$$\mathbb{E}(XY) = 0,$$
QED.
