As a part of a project, I would like to know what the distribution is of the absolute distance people traveled on a particular moment of the day (in comparison to their home). I think it would be best modeled with a simple random walk $(S_n)_{n \in \mathbb{N}}$ in $\mathbb{Z}^2$. The central limit theorem gives $n^{-1/2}S_n \to Z \sim \mathcal{N}(0,(1/2)I_2)$ with $I_2$ the $2\times 2$ identity matrix. So, basically, my question is (using the Euclidean norm):
What is the distribution of $\| Z \|$, where $Z \sim \mathcal{N}(0,(1/2)I_2)$?
I tried to calculate this distribution by integration along a circle with radius $r$. I get the density function
\begin{align*} f(r) = 2r\exp(-r^2). \end{align*}
This gives $f(0)=0$. However, my intuition says otherwise: the most likely place a random walk will be after a some time, is at the origin. What am I doing wrong?
Here is my calculation for the density function. The multivariate normal distribution $Z \sim \mathcal{N}(0,2^{-1}I_2)$ has the density function $f: \mathbb{R}^2 \to \mathbb{R}$ given by
\begin{align*}
f(x,y) = \frac{1}{\pi} \exp(-(x^2+y^2)).
\end{align*}
Consider an $s > 0$. I used the parametrization function $s: [0,2\pi] \to \mathbb{R}^2$ given by $\theta \mapsto (r\cos\theta, r\sin \theta)$. The integral along the circle with radius $r > 0$ then becomes
\begin{align*}
\int_0^{2\pi} f(r\cos\theta,r\sin\theta) \| s'(\theta) \| \ \text{d}\theta = \frac{1}{\pi}\int_0^{2\pi} r\exp(-r^2) \ \text{d}\theta = 2re^{-r^2}.
\end{align*}
I found one question similar to this one (The distance distribution from the mean for an n-dimensional normal(Gaussian) distribution), but it is unanswered.
Best Answer
\begin{align} & \Pr\left(\sqrt{X^2+Y^2}\le w\right) \\[10pt] = {} & \iint\limits_{\begin{smallmatrix} \text{disk of} \\[2pt] \text{radius } w\end{smallmatrix}} \frac 1 {2\pi} e^{-(x^2+y^2)/2} \,d(x,y) \\[10pt] = {} & \int_0^{2\pi} \int_0^w \frac 1 {2\pi} e^{-r^2/2} \Big( r\,dr\,d\theta\Big) \\[10pt] = {} & 2\pi \frac 1 {2\pi} \int_0^w e^{-r^2/2} r\,dr = \int_0^{w^2/2} e^{-u}\,du \\[10pt] = {} & 1 - e^{-w^2/2}. \end{align}
The density is then the derivative of that, which is $w\mapsto we^{-w^2/2}$ for $w\ge0$.