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Consider two unit $\mathbb R^2$ vectors $v$ and $w$. Then $v+w$ lies within a (closed) circle with radius 2, that is, in the region $x^2+y^2\leq4$.

Intuitively, the probability of $v+w$ lying close to the center is higher.

What is the exact probability density?

I tried to solve a problem in a naive way, but I get strange infinite quantities.

That is, i get $\frac{1}{2\pi} \delta(|r-1|-1)$.

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You have to know the probability density on $v$ and $w$ to find the probability density of their sum. You haven't specified this, so there's no way to answer the question. –  Jonathan Christensen Jan 2 '13 at 19:53
    
The two vectors are chosen randomly on the unit circle. Sorry if I hadn't been clear enough. –  user54987 Jan 2 '13 at 19:56
    
You probably mean that the variables are chosen "uniformly" on the unit circle. A "uniform" random variable is one that treats any two equally-sized rectangles as equally likely to contain a realization of the random variable ("random" simply means that the value is uncertain). Formally, a uniform random variable is one that uses a scalar multiple of Lebesgue measure as the underlying probability measure. –  GMB Jan 2 '13 at 20:09
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If $v_1=\mathrm e^{\mathrm i\theta_1}$ and $v_2=\mathrm e^{\mathrm i\theta_2}$, then $v=v_1+v_2$ is $v=r\mathrm e^{\mathrm i\theta}$ with $r=2\cos(\frac12(\theta_1-\theta_2))$ and $\theta=\frac12(\theta_1+\theta_2)$. If $\theta_1$ and $\theta_2$ are i.i.d. uniform on the circle $S^1=\mathbb R/2\pi\mathbb Z$, then $r$ and $\theta$ are independent and $\theta$ is uniformly distributed on $S^1$. Furthermore, $|r|=2\cos(\alpha)$ where $\alpha$ is uniform on $(0,\frac\pi2)$ hence $\mathbb P(|r|\geqslant2x)=\mathbb P(\alpha\leqslant\arccos(x))=\frac2\pi\arccos(x)$ for every $x$ in $(0,1)$. Thus, the density $f_r$ of $r$ is $$ f_r(x)=\frac{\mathbf 1_{|x|\lt 2}}{\pi\sqrt{4-x^2}}. $$

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Close to what I found. I found it to be: $$\frac{1}{(2\pi)^2r\sqrt{1-\frac{r^2}{4}}}.$$ What is the problem: it is not renormalizable. –  user54987 Jan 3 '13 at 14:54
    
Indeed. In more standard (at least to probabilists) terms, neither the function you suggest nor any of its multiples can be the density of a random variable since its integral is infinite. –  Did Jan 3 '13 at 18:02
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