probabilityprobability distributionsprobability theorystatisticsuniform distribution
This exercise (partially) comes from Rice 3.9
Suppose that (X, Y ) is uniformly distributed over the region defined
by $0 ≤ y ≤ 1 − x^2$ and $−1 ≤ x ≤ 1$.Find the marginal densities of X and Y.
Find the joint density.
Find the two conditional densities.
$X \sim U[-1,1]$ and $Y∼U[0, 1-x^2]$
then $F_x(x)=x+1$ for $−1 ≤ x ≤ 1$ and $F_Y(y)=\frac{y}{1-x^2}$ for $0 ≤ y ≤ 1 − x^2$
I am not sure if this is correct? I am confused with how to find the joint distribution also.
In general, there is no way of determining the joint density $f_{X,Y}(x,y)$ from knowledge of the marginal densities $f_X(x)$ and $f_Y(y)$ and nothing else. In special cases such as when $X$ and $Y$ are independent, the joint density is, of course, the product of the marginal densities.
As an example, consider random variables $X$ and $Y$ that are uniformly distributed on $[0,1]$. If you are also given that they are independent, then $(X,Y)$ is uniformly distributed on the unit square. But consider that $X$ and $Y$ might have joint density $(2x+2y-4xy)\mathbf 1_{\{(x,y)\colon 0 \leq x, y \leq 1\}}$ on the unit square. Thus, in general, the marginal densities do not determine the joint density.
Read about copulas and Sklar's theorem for further information.
The joint density looks like this:
So, the marginal distribution of $Y$ is given by the following integral
$$\int_y^1 2 \ dx=2(1-y)$$
if $0\leq y\leq 1$.
Best Answer
$X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=\frac12(x+1)$ on that interval. But this would not be the marginal density
Hints for the question:
Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k\, I[-1 \le x \le 1, 0 \le y \le 1-x^2$. You can find the value of $k$ from $\int\limits_{x=-\infty}^\infty\int\limits_{y=-\infty}^{\infty} f_{X,Y}(x,y) \, dy \, dx = \int\limits_{x=-1}^1\int\limits_{y=0}^{1-x^2} k \, dy \, dx = 1$
You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density
For the marginal density for $X$ you have $f_X(x)=\int\limits_{y=0}^{1-x^2} k \, dy$ using the $k$ you found earlier
For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 \le x \le 1, 0 \le y \le 1-x^2$ as $0 \le y \le 1, -\sqrt{1-y} \le x \le \sqrt{1-y}$ and you then want $f_Y(y)=\int\limits_{x=-\sqrt{1-y}}^{\sqrt{1-y}} k \, dx$ again using the $k$ you found earlier