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$X,Y$ are i.i.d. $unif(-1,1)$ random variables. Prove that $$P(X^2+Y^2\leq 1)=\frac{π}{4}$$

Geometrically, I understand how that happens. $(X,Y)$ is a random point in square having centre at origin and vertices $(-1,-1),(-1,1),(1,-1),(1,1)$. Probability that a random point in this square lies in the unit circle is the ratio of their areas.

But, is there any analytic proof ?

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    $\begingroup$ Now that you know there is an analytic proof feel free to solve these problems geometrically. $\endgroup$ – Ethan Bolker Apr 23 at 11:44
  • $\begingroup$ Definitely have to agree with Ethan here - those "analytic proofs" below are just the geometric argument you gave dressed up in the language of calculus. $\endgroup$ – Paul Sinclair Apr 23 at 16:32
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$4P(X^{2}+Y^{2} \leq 1)=\int_{-1}^{1} \int_{-\sqrt {1-x^{2}}}^{\sqrt {1-x^{2}}} dydx$ which is $2\int_{-1}^{1}\sqrt {1-x^{2}}dx$. Make the substitution $x =\sin(\theta)$ to evaluate this.

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  • $\begingroup$ Thank you sir. But why is there a 4 before $P(X^2+Y^2\leq 1)$? $\endgroup$ – Martund Apr 23 at 9:44
  • $\begingroup$ Well, put $4$ on the left instead of $\frac 1 4$ on the right to make typing simpler. $\frac 1 4$ on the right comes from the density of uniform distribution. $\endgroup$ – Kavi Rama Murthy Apr 23 at 9:45
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Let $D = \{(x, y): x^2+y^2 \leq 1\}$. Let $g(x, y)$ be the density of $(X, Y)$. It exists because $X$ and $Y$ are independent, and equals to multple of their densities. (Fubini theorem) $$P(X^2+Y^2\leq1) = \iint_D g(x, y)dxdy = \frac{1}{4}\iint_D dxdy = \frac{\pi}{4}$$

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$$\mathbb P\{X^2+Y^2\leq 1\}=\frac{1}{4}\int_{\{(x,y)\mid x^2+y^2\leq 1\}}\boldsymbol 1_{[-1,1]\times [-1,1]}(u,v)\,\mathrm d u\,\mathrm d v=\frac{1}{4}Area(\{(x,y)\mid x^2+y^2\leq 1\})$$

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