0
$\begingroup$

Suppose that $\theta_1, \cdots, \theta_n$ are distributed independently and that $\theta_j$ has probability density function (PDF) $f_j = \frac{1}{2\pi}$ ($i.e.$, the uniform distribution) for $j = 1, \cdots, n$. What is the PDF of $R$ given that $R^2 = C^2 + S^2$, with $C = \sum_{j=1}^{n}{cos \theta_j}$ and $S = \sum_{j=1}^{n}{\sin \theta_j}$ for a small $n$, $e.g.$, $n = 2$?

$\endgroup$

2 Answers 2

2
$\begingroup$

Solution for $n=2$: In view of the rotational invariance, the distribution the distance of the random point $(C,S)$ from the origin is the same as that for $(1,0)+(\cos U,\sin U)=(1+\cos U,\sin U)$, where $U$ is uniformly distributed on $[0,2\pi]$. So, $R$ is equal in distribution to $\sqrt{(1+\cos U)^2+\sin^2 U}=2|\cos U/2|$ and hence to $2\cos V$, where $V$ is uniformly distributed on $[0,\pi/2]$. So, the pdf $f_R$ of $R$ is given by $$f_R(r)=\frac2{\pi\sqrt{4-r^2}}1_{0<r<2} $$ for real $r$.


Consider also the case of large $n$. Note that $R=R_n$ is the length of the vector $S_n$ that is the sum of $n$ iid copies of the random vector $X:=(\cos U,\sin U)$, with $U$ as above. The mean of $X$ is $(0,0)$ and its covariance matrix is $\frac12\,I_2$, where $I_2$ is the $2\times2$ identity matrix. So, by the multivariate central limit theorem, the distribution of $\sqrt{\frac 2n}S_n$ converges to the standard bivariate normal distribution. So, the distribution of $\sqrt{\frac1n}R_n$ converges to the Maxwell distribution, with pdf $f_M$ given by $$f_M(r)=2r e^{-r^2}1_{r>0} $$ for real $r$.

$\endgroup$
8
  • $\begingroup$ Dear, thanks for your answer. One question, what does the number "1" mean in the $f_R(r)$ expression, just before the limits of r? It also appears for the case when $n$ is large. Is that a typo?Thanks a lot! $\endgroup$ Commented Nov 29, 2019 at 15:27
  • 1
    $\begingroup$ This is the indicator (or characteristic) function of a set. See en.wikipedia.org/wiki/Indicator_function for clarification. $\endgroup$ Commented Nov 29, 2019 at 17:10
  • $\begingroup$ @MickyboYakari, thanks! $\endgroup$ Commented Nov 29, 2019 at 17:12
  • 1
    $\begingroup$ see mathoverflow.net/q/347317/11260 for a clarification. $\endgroup$ Commented Nov 30, 2019 at 21:59
  • 2
    $\begingroup$ @FelipeAugustodeFigueiredo : I meant the distribution of the distance of the random point from the origin, rather than the distribution of the random point itself. Sorry about that confusion. This is now corrected. $\endgroup$ Commented Nov 30, 2019 at 22:28
2
$\begingroup$

Generalization to arbitrary $n$: The desired distribution $P_n(r)$ of the distance $r$ from the origin after $n$ steps can be expressed as an integral over Bessel functions $J_0$, $$P_n(r)=r\int_0^\infty k [J_0(k)]^n J_0(k r)\,dk,\;\;0<r<n,$$ $$F_n(r)\equiv\int_0^r P_n(r')\,dr'=\int_0^\infty [J_0(k/r)]^nJ_1(k )\,dk,\;\;0<r<n,$$ see theorem 2.1 in arXiv:1508.04729 for a derivation (and the generalization to $d$ dimensions). [A more tutorial exposition is given in A Short Walk can be Beautiful.]

For $n=2$ we recover $P_2(r)=(2/\pi)(4-r^2)^{-1/2}$. The cited paper contains closed form expressions for $n=3,4$, in terms of hypergeometric functions (eqs. 74, 79).

I note a remarkable exact result (first obtained by J.C. Kluyver in 1905) for the probability $p_n$ to return to the unit disc after $n$ steps, $$p_n=F_n(1)=\frac{1}{n+1},\;\;n>1.$$ A purely geometric derivation is given in arXiv:1508.04729.

For $n\gg 1$ we may approximate $J_0(k/r)^n\approx\exp\left(-\frac{nk^2}{4r^2}\right)$, see page 345 of Kluyvers paper, to obtain $$F_n(r)\approx\int_0^\infty \exp\left(-\frac{nk^2}{4r^2}\right)J_1(k)\,dk=1-e^{-r^2/n},$$ which gives the expected Maxwell distribution for $$P_n(r)=dF_n(r)/dr=(2r/n)e^{-r^2/n}.$$

$\endgroup$
3
  • $\begingroup$ Is it possible, based on this integral, to show that it tends to the Maxwell distribution when $n$ is large? $\endgroup$ Commented Dec 1, 2019 at 5:47
  • $\begingroup$ @FelipeAugustodeFigueiredo --- yes, I have added the large-$n$ limit, with a reference to the 1905 paper by J.C. Kluyver where this problem was treated in much detail. $\endgroup$ Commented Dec 1, 2019 at 12:18
  • $\begingroup$ Dear Carlo, dank u wel! $\endgroup$ Commented Dec 1, 2019 at 17:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .