Random walk on $\mathbb{R}^2$ 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?
 A: 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.
