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We have a dartboard with radius $1$, the dart will always hit the dartboard. The hitting point of the dart is uniformly distributed, with a stochastic vector $(X,Y)$. Now I want to determine the probability mass and density function.

Say $A=\{(x,y): x^2 + y^2 \leq 1\}$. So $f_{X,Y}(x,y) = 0$ if $(x,y) \notin A$ and $f_{X,Y}(x,y) = c $ if $(x,y) \in A$. Now: \begin{align} \int \int f_{X,Y}(x,y) dx dy = \int\int\limits_A = \text{area } A \cdot c = 1. \end{align} So $1=2\pi \cdot c$ and $c=\frac{1}{2\pi}$. We found that $f_{X,Y}(x,y)=\frac{1}{2\pi}$. Now I am interested in $f_X(x)$ and $F_X(x)$. For the first one we can say that: \begin{align} f_X(x) = \int\limits_{-1}^1 \frac{1}{2\pi} dy = \frac{1}{\pi}. \end{align} But how to find $F_X(x)$? We have to integrate over $x$ and $y$ but how to deal with the restrictions of $A$?

iJup
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1 Answers1

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To find $f_X(x)$, your idea is correct but the limits should be the range of $y$ at any given $x$, so should be from $-\sqrt{1-x^2}$ to $\sqrt{1-x^2}$ As you are integrating along $y$ it just becomes $f_X(x)=\frac 2\pi\sqrt{1-x^2}$ (correcting your $c$). Then for $F_X(x)$ you integrate $\int_{-1}^x f_X(t)dt=\int_{-1}^x\frac 2\pi\sqrt{1-t^2}dt$

Ross Millikan
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