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Question

Consider a random walk on the plane $\mathbb{R}^2$, a point randomly chooses an angle $\theta$ at each step and walks forward one unit length.

Then after $t$ steps, what is the probability that this point is within a unit length from the origin?

My Attemption

The problem seems to be equivalent to solving the following probability distribution:

$$r_t\sim\sqrt{(\sum_{i=1}^t\cos\theta_i)^2 + (\sum_{i=1}^t\sin\theta_i)^2},\quad\mathtt{where}\quad\theta_i\sim U[0,2\pi]$$

I tried some numerical simulations:

$P_1 = 1$, $P_2 = \dfrac13$, $P_3 \stackrel{?}{=} \dfrac14$

Is there any good way to estimate the value of $P_t$?

Update1:

for big enough $t$, $P_t \stackrel{?}{\approx} 1-e^{-1/t}$.

How to prove or disprove these results?

Aster
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  • Related paper: "Three and four step random walk integrals", Borwein, Straub & Wan. http://arminstraub.com/downloads/pub/walks2.pdf Their $W_n(1)$ is the expected value of your $r(n)$. – GEdgar Apr 13 '21 at 14:25
  • @GEdgar, really useful, but knowing any moment is still difficult to calculate this probability, it seems that the probability distribution needs to be derived from the momentum generating function. – Aster Apr 14 '21 at 02:46
  • Just a random thought: We could alternatively define the problem by making the angle $\theta_i = 0$ denote away from the origin for $i \geq 2$, so that $$r_{t+1}^2 = r_t^2 + 2r_t\cos(\theta_{t+1}) + 1 \ r_0 = 0$$ I feel like another solution should be able to come from this re-definition, but I'm not thinking of anything at the moment. – Brian Moehring Apr 17 '21 at 08:45

1 Answers1

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Let $Z_k = \begin{bmatrix}\cos\theta_k \\ \sin\theta_k\end{bmatrix}$ where the $\theta_k$'s are i.i.d. $\text{Uniform}[0,2\pi]$, then $\mathbb{E}[Z_k] = \vec{0}$ and $\mathbb{E}[Z_kZ_k^T] = \dfrac{1}{2}I$.

Then by the central limit theorem, $\dfrac{1}{\sqrt{t}}\displaystyle\sum_{k = 1}^{t}Z_k$ converges to a $\mathcal{N}(\vec{0},\tfrac{1}{2}I)$ distribution.

You can check that the magnitude-squared of a $\mathcal{N}(\vec{0},\tfrac{1}{2}I)$ distribution has an exponential distribution with mean $1$ while the argument of a $\mathcal{N}(\vec{0},\tfrac{1}{2}I)$ distribution has a uniform distribution over $[0,2\pi)$.

Hence, $\mathbb{P}\left\{\left\|\displaystyle\sum_{k = 1}^{t}Z_k\right\| \le 1\right\} = \mathbb{P}\left\{\left\|\dfrac{1}{\sqrt{t}}\displaystyle\sum_{k = 1}^{t}Z_k\right\|^2 \le \dfrac{1}{t}\right\} \approx \mathbb{P}\left\{\text{Exp}(1) \le \dfrac{1}{t}\right\} = 1-e^{-1/t}$, which is the approximation you got.

You can probably justify how accurate this approximation is as $t \to \infty$ by using some concentration inequalities. I'll update this answer if I come up with something interesting.

JimmyK4542
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