Marginal density of $X = (X_1, X_2)$ distributed on region $A$

Question

Consider the region $A = \{(s,t) \in \mathbb{R}^2; s \geq 0, t \geq 0, s^2+t^2 \leq 1 \}$. Be $X = (X_1, X_2)$ a random vector that is uniformly distributed on $A$.

Now, how can I calculate the marginal density $f_{X_1}$?

My first idea regarding the joint pdf is that since the vector has a uniform distribution on $A$, the density function is constant for all $s,t$. Thus, $f_{X_1,X_2}(s,t) = \frac{1}{A}$.

Answer 1

$A$ is a set, not a number, so $\frac{1}{A}$ does not make sense. You are almost correct that the density function is constant, but it is only constant for all $(s,t)$ that lie in $A$. Outside $A$, the density is zero. Can you find the correct joint density now?

$$f_{X_1, X_2}(s,t) = \begin{cases}\frac{1}{\text{area}(A)} & (s,t) \in A \\ 0 & (s,t) \notin A\end{cases}$$

Then, to compute the marginal density of $X_1$, you do the usual thing: integrate the joint density with respect to its second argument. $$f_{X_1}(s) = \int_{-\infty}^\infty f_{X_1, X_2}(s,t) \, dt.$$

The tricky part is to figure out which values of $t$ make the joint density nonzero, and which values of $t$ make the joint density zero.

Answer 2

The joint density $f(x,y)$ is $\frac 4 {\pi}$ for $(x,y) \in A$ and the marginal density is just its integral w.r.t. $y$. So $f_{X_1}(x)=\frac 4 {\pi}\sqrt {1-x^{2}}$ for $0\leq x \leq 1$ (the integral is from $0$ to $\sqrt {1-x^{2}})$.

Answer 3

$f_{X_1}(x_1)= \int_{0}^{\sqrt{1-x_{1}^2}}\,\,4/\pi \, \mathrm{d}x_2$

since area of region $A$ is $\pi/4$