# Marginal probability distribution of unit circle random variable

I am given a task stating that there is a bivariate random vector X such that: $$$$p_x(x)=\left\{ \begin{array}{@{}ll@{}} \frac1\pi, & \text{if}\ x^2+y^2 < 1 \\ 0, & \text{otherwise} \end{array}\right.$$$$ and I am asked to find margianl distribution with respect to X and Y. I tried to catch it up by google and found some answers. e.g. $$f(x)_x=\int_\sqrt{1^2-x^2}^\sqrt{1^2+x^2}\frac1\pi dy$$ One thing I am interested in. Where the integration bounds came from. Can anyone explain pls?

• I find different limits: $- \sqrt(1-x^2)$ and $\sqrt(1-x^2)$ – Carlos Campos Jan 9 at 19:35
• Thanks, just realized that I can do in such way – Hillbilly Joe Jan 9 at 19:42

Define the domain $$\mathcal{D} = \{(x_1,x2) \vert x_1^2 + x_2^2 <1\}$$. Given a fix $$x_1$$, if $$(x_1, x_2)$$ belongs to $$\mathcal{D}$$, $$x_2$$ has to satisfy $$- \sqrt{1-x_1^2}< x_2 < \sqrt{1-x_1^2}$$, where we are taken positive roots:
$$$$f_{X_1}(x_1) = \int_{D} f_{X_1,X_2}(x_1,x_2) d x_2 = \left\{ \begin{array}{@{}ll@{}} \int_{- \sqrt{1-x_1^2}}^{ \sqrt{1-x_1^2}} \frac{1}{\pi} d x_2 = \frac{2\sqrt{1-x_1^2}}{\pi} & \text{if}\ x_1^2 < 1 \\ 0, & \text{otherwise} \end{array}\right.$$$$