# fint joint and marginal distributions of two uniformy distributed variables over a specified region

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.

$$X$$ 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

• thanks. I have obtained $k=\frac34$ and also the marginal distributions: $f_X(x)= \frac34 (a-x^2)$ for $-1\leq x \leq 1$ and $f_Y(y)= \frac32 \sqrt{1-y}$ for $0\leq y \leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=\frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions? Dec 11, 2018 at 15:02
• Inside the region you have the conditionally uniform densities $f(y \mid x) = \frac1{1-x^2}$ for $0 \le y \le 1-x^2$ and $f(x \mid y) = \frac1{2\sqrt{1-y}}$ for $-\sqrt{1-y} \le x \le \sqrt{1-y}$ Dec 11, 2018 at 15:31