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I am given a task stating that there is a bivariate random vector X such that: $$\begin{equation} p_x(x)=\left\{ \begin{array}{@{}ll@{}} \frac1\pi, & \text{if}\ x^2+y^2 < 1 \\ 0, & \text{otherwise} \end{array}\right. \end{equation} $$ 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?

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

\begin{equation} 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. \end{equation}

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  • $\begingroup$ Thank you for your explanation. $\endgroup$ – Hillbilly Joe Jan 9 at 19:46

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