Calculating probability of a bivariate distribution

Question

Calculate the probability that $$P(X\le 250)$$ Suppose that ( X,Y ) has a bivariate distribution uniformly on the circle $x^2+y^2≤2000^2$.

So far I have integrated the bivariate density from $-/+\sqrt{2000-x^2}$ and now I am considering integrating again except this time I will be integrating from -2000 to 250. Are these bounds for the second integral correct?

Thanks in advance!

Answer 1

Direct approach. Let R=2000, u=250. Then the area of the part of the circle for $0\le x\le 250$ can be the sum of 2 terms, $A=2u\sqrt{R^2-u^2}$ and $B=2\int_\sqrt{R^2-u^2}^R\int_0^{\sqrt{R^2-y^2}}dxdy$. Therefore $P(X\le 250)=\frac{1}{2}+ \frac{A+B}{\pi R^2}$. The one half is $P(X\le 0)$. You do the arithmetic.

Answer 2

Alternative approach - without calculus. R=2000, u=250, v=u/R. $P(X\le u)=\frac{1}{2}+\frac{arcsin(v)+v\sqrt{1-v^2}}{\pi}$.

The first term is $P(X\le 0)$.

The other two terms come from dividing the area of $0\lt x \lt u \ and\ y\gt 0$ into two parts and then multiplying by 2 to include negative y contribution.

The area division is by a radius ending at the point on the circle where x=u. To the right of this radius is a right triangle with area=$u\sqrt{R^2-u^2}/2$. To the left of this radius is a sector with area=$R^2arcsin(v)/2$. To get $P(0\lt X\lt u)$, add these two terms, multiply by 2 and then divide by the area of the circle $\pi R^2$.