# Analytic proof of area probability

$$X,Y$$ 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 ?

• Now that you know there is an analytic proof feel free to solve these problems geometrically. – Ethan Bolker Apr 23 at 11:44
• 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. – Paul Sinclair Apr 23 at 16:32

$$4P(X^{2}+Y^{2} \leq 1)=\int_{-1}^{1} \int_{-\sqrt {1-x^{2}}}^{\sqrt {1-x^{2}}} dydx$$ which is $$2\int_{-1}^{1}\sqrt {1-x^{2}}dx$$. Make the substitution $$x =\sin(\theta)$$ to evaluate this.
• Thank you sir. But why is there a 4 before $P(X^2+Y^2\leq 1)$? – Martund Apr 23 at 9:44
• 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. – Kavi Rama Murthy Apr 23 at 9:45
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) $$P(X^2+Y^2\leq1) = \iint_D g(x, y)dxdy = \frac{1}{4}\iint_D dxdy = \frac{\pi}{4}$$
$$\mathbb P\{X^2+Y^2\leq 1\}=\frac{1}{4}\int_{\{(x,y)\mid x^2+y^2\leq 1\}}\boldsymbol 1_{[-1,1]\times [-1,1]}(u,v)\,\mathrm d u\,\mathrm d v=\frac{1}{4}Area(\{(x,y)\mid x^2+y^2\leq 1\})$$