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Let $X$ and $Y$ be independent with uniform $(-1,1)$ distribution. Please help in finding:

a) $P(X^2+Y^2 \leq r^2)$
b) The CDF of $R^2 = X^2 + Y^2$
c) The density of $R^2$

All I tried was breaking it down into cases, for part a we have 2 cases, where the circle is inscribed in the square and vice versa.

I need help espcecially with part b, then I can try c. Can someone please explain?

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It may be convenient to use the fact that $X^2$ and $Y^2$ are distributed as $U_1 ^2$ and $U_2 ^2$, respectively, where $U_1$ and $U_2$ are independent uniform$(0,1)$ variables. –  Shai Covo Apr 14 '11 at 6:58

2 Answers 2

up vote 2 down vote accepted

Solving the problem hinges on finding the answer to part (a). A geometric approach will be described.

There are 3 cases. If $0 \le r \le 1$, the probability is just $1/4$ the area of a circle of radius $r$, so it is $\pi r^2/4$.

If $1 <r < \sqrt{2}$ things are messy. Draw a picture! We have a circle, part of which lies inside the square. We want to find the area of the part of the circle that is inside the square, and then divide by $4$.

The area of the part of the circle inside the square is the area of the circle, minus the sum of the areas of the $4$ parts that stick out of the square.

Look at one of those parts. It is a sector of a circle, with angle $2\arccos(1/r)$, minus a triangle with base $2\sqrt{r^2-1}$ and height $1$.

So the area of the $4$ parts that stick out is $$4\left[r^2\arccos(1/r)-\sqrt{r^2-1}\right]$$ Subtract from $\pi r^2$, divide by $4$. We get $$\pi r^2/4 + \sqrt{r^2-1} -r^2\arccos(1/r)$$ Finally, the easiest case: if $r>\sqrt{2}$, our probability is $1$.

To find the cdf of the random variable $Z=X^2+Y^2$, we want to find an expression for the probability that $Z \le z$. We get this by setting $r=\sqrt{z}$ in the answers to part (a).

After all this work, the pdf is (sort of) easy. Differentiate the cdf of $Z$.

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You can use the following formula: $$ F_R(r) = P(X^2+Y^2\leq r^2) = P(X^2\leq r^2 - Y^2) = \int\limits_{-\infty}^\infty P(X^2\leq r^2 - y^2)f_Y(y)$\,dy $$ where $f_Y(y)$ is a density function of $Y$, i.e. $$ f_Y(y) = \frac{1}{2}I_{[-1,1]}(y), $$ where $I$ is an indicator function which is equal to $1$ on $y\in[-1,1]$ and $0$ otherwise. So $$ F_R(r) = \frac{1}{2}\int\limits_{-1}^1 P(X^2\leq r^2 - y^2)\,dy. $$ The function in the last integral I you can find easily. With this you have answer to 1) and also $F_R(r)$ is a CDF for the question 2). PDF you can find as its derivative.

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