Choose a point $(X,Y)$ uniformly in the unit disk. What is $\rho(X,Y)?$

Question

Solution in the book:

The joint pdf is $f(x,y)=1/\pi.$ Denote the unit disk by $D$ to obtian

$$E[XY]=\frac{1}{\pi}\iint_Dxy \ dxdy.$$

which we realize must be $0$ by symmetry. Also, by symmetry, $E[X]=E[Y]=0,$ so we get $\text{Cov}[X,Y]=0$ and hence also $\rho(X,Y)=0.$

Questions:

1. How would one show that $f(x,y)=1/\pi?$

2. How did they get the dobule integral expression for $E(XY)?$ Is it the analouge for having a single random variable, for example: $$E(X)=\int_{-\infty}^{\infty}x\cdot f(x) \ dx, \ \text{so} \ \ E(XY)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xy \ dxdy \ ?$$

3. The reason that $E[X]=E[Y]=0$, is it the following:

The marginal pdfs are

$$f_X(x)=f_Y(x)=\int_{-1}^1\frac{1}{\pi} \ dx=\frac{2}{\pi}.$$

So $$E(X)=E(Y)= \frac{2}{\pi}\int_{-1}^{1}x \ dx = 0 \ ?$$

Answer 1

Let $(X,Y)$ denote the random point on the unit disk.

Convince yourself that dependent random points $(-X, Y)$ and $(X, -Y)$ have the same distribution as $(X,Y)$.

It now follows that $\rho(X, Y)$ and $\rho(X, -Y) = -\rho(X, Y)$ must be equal, which is only possible is $\rho(X,Y)$ assumes a particular value.

Answer 2

When we are dealing with a uniform distribution over a given set (in this case the disk), the probability density is given by one over the (Lebesgue) measure (or volume / area) of that set. In your example, the disk $D$ has total area of $\pi$, and hence the joint density of $X$ and $Y$ is \begin{align} f_{X,Y}(x,y) = \frac{1}{\pi} I_{D}(x,y), \end{align} where $I_D(x,y)$ is the indicator function of the set $D$, i.e., $I_D(x,y) = 1$ if $(x,y) \in D$ and zero otherwise. Just as an aside, if you would deal with the unit ball $B$ in three dimensions you would have for example the density \begin{align} f_{X,Y,Z}(x,y,z) = \frac{3}{4\pi} I_{B}(x,y,z). \end{align}

To compute the expectations you can indeed use the law of the unconscious statistician (https://en.wikipedia.org/wiki/Law_of_the_unconscious_statistician): \begin{align} \mathbb{E}[g(X,Y)] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} g(x,y) f_{X,Y}(x,y) dx dy, \end{align} where $g$ is a suitable (measurable) function, for example $g(X,Y) = XY$. Together with the density this is then enough to derive the results.

To get the marginal density of $X$ you can check: Uniform distribution on unit disk

Combining all of that should help you to derive the results.

Answer 3

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