Can someone very simply explain to me how to compute the expected distance from the origin for a random walk in $1D, 2D$, and $3D$? I've seen several sources online stating that the expected distance is just $\sqrt{N}$ where $N$ is the number of time steps, but others say that the expected distance is $\sqrt{\frac{2N}{\pi}}$. Which one is it and is it the same regardless of dimension?



The expected value of the square of the absolute distance from the origin is $N$ (you are adding together $N$ independent random variables with mean $0$ and absolute magnitude $1$), and this is true in any dimension.

So those sources which are telling you $$\sqrt N$$ are giving you this as in some sense the "root mean square" distance from the starting point. It is not the expected value of the distance.

For a one dimensional random walk the expected absolute distance from the origin after $N$ steps is not easy to state explicitly, but as $N$ increases it becomes close to $$\sqrt{\dfrac{2N}{\pi}}.$$ So the sources which give you that are in a sense talking about a limit.

This changes for higher dimensions: if there are $d$ dimensions then the expected absolute distance from the origin after $N$ steps becomes close to $$\sqrt{\dfrac{2N}{d}} \dfrac{\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2})}$$ where $\Gamma$ is the Gamma function. As the number of dimensions increases, this get close to but is still below $\sqrt N$.

This is related to the means of the chi- and chi-squared distributions.

  • $\begingroup$ There is a related question, where the OP asks for some reference for the formula from your post. As he does not have a privilege to post comments, I am pinging you instead of him. $\endgroup$ – Martin Sleziak Sep 1 '13 at 17:50

In an arbitrary dimension d:

Let $\vec{R}$ be the end-to-end distance vector of a random walk of fixed step length $|\vec{r}_i| = l$. $\vec{R}$ can then be expressed as $\displaystyle \vec{R} = \sum_{i=1}^N \vec{r}_i$, where $\vec{r}_i$ is the vector of the $i$-th step. The Root-Mean-Square End-to-End Distance is given by $\textrm{RMS}=\sqrt { \langle R^2 \rangle }$. Since the steps are mutually independent, the covariance of two steps $\vec{r}_i$ and $\vec{r}_j$ is zero if $i\neq j$ and $\textrm{Cov}(\vec{r}_i, \ \vec{r}_j)= \textrm{Var}(\vec{r}_i)$ if $i=j$. The variance of $ \vec{r}_i$ can be expressed as $ \textrm{Var}(\vec{r}_i)= \langle \vec{r}_i \cdot \vec{r}_i \rangle - \langle \vec{r}_i \rangle^2$. Due to symmetry $\langle \vec{r}_i \rangle=\vec{0}$ and therefore the variance of of $ \vec{r}_i$ is simply $ \textrm{Var}(\vec{r}_i)= \langle \vec{r}_i \cdot \vec{r}_i \rangle = |\vec{r}_i|^2 = l^2$. Altogether, the covariance of $\vec{r}_i$ and $\vec{r}_j$ equals $\textrm{Cov}(\vec{r}_i, \ \vec{r}_j)=\delta_{ij}l^2$. The covariance of $\vec{r}_i$ and $\vec{r}_j$ can also be expressed as $\textrm{Cov}(\vec{r}_i, \ \vec{r}_j) = \langle \vec{r}_i \cdot \vec{r}_j \rangle - \langle \vec{r}_i \rangle \cdot \langle \vec{r}_j \rangle$. Combining the two different expressions for the covariance and using that $\langle \vec{r}_i \rangle=0$, results in $\langle \vec{r}_i \cdot \vec{r}_j \rangle =\delta_{ij}l^2$. This result can be used to determine the RMS:

$$\textrm{RMS}=\sqrt { \langle R^2 \rangle } = \sqrt { \langle \vec{R} \cdot \vec{R} \rangle } =\sqrt { \big\langle \sum_{i=1}^N \vec{r}_i \cdot \sum_{j=1}^N \vec{r}_j \big\rangle } =\sqrt { \sum_{i=1}^N \sum_{j=1}^N \langle \vec{r}_i \cdot \vec{r}_j \rangle }= $$ $$=\sqrt { \sum_{i=1}^N \sum_{j=1}^N l^2 \delta_{ij} + 0^2}=\sqrt { \sum_{i=1}^N l^2}=\sqrt { N l^2}=l\sqrt { N }$$

Let $Z_i$ denote the $i$-th coordinate of the end-to-end distance vector $\vec{R}$ after $N$ steps, and let $X_i$ and $Y_i$ denote the number of steps taken in the $i$-th dimension in the positive and negative direction respectively. Then the set of random variables $\{X_i, Y_i\}_{i=1}^d$ follows a multinomial distribution with parameters $N$ and $\displaystyle p_i=\frac{N}{2d}$. For sufficiently large values of $N$, $\{X_i, Y_i\}_{i=1}^d$ are approximately iid (independent and identically distributed) Poisson random variables with parameters $\displaystyle \lambda_i = \frac{N}{2d}$. For $\lambda > 20$, i.e. $N>40d$, $\textrm{Po}(\lambda) \sim \textrm{N}(\lambda, \lambda)$. $ Z_i = l(X_i - Y_i)$ and therefore $\displaystyle Z_i \sim \textrm{N}(l(\lambda - \lambda), l^2(\lambda+\lambda))=\textrm{N}(0, 2l\lambda)=\textrm{N}\left(0, \frac{l^2N}{d}\right)$.

$\displaystyle \langle R \rangle = \langle \sqrt{R^2} \rangle = \left\langle \sqrt{ \sum_{i=1}^d Z_i^2} \right\rangle$. The square root of a sum of $k$ independent $\textrm{N}(0, 1)$-distributed random variables is distributed according to the chi distribution, $\chi_k$. Therefore $\displaystyle \sqrt{ \sum_{i=1}^d \frac{dZ_i^2}{l^2N}}$ is approximately $\chi_d$-distributed for large values of $N$. The expected value of a $\chi_k$-distributed random variable is $\displaystyle \sqrt{2} \frac{ \Gamma \left(\frac{k+1}{2}\right) }{\Gamma \left( \frac{k}{2}\right)}$.

Hence $\displaystyle \langle R \rangle =\left\langle\sqrt{ \sum_{i=1}^d Z_i^2}\right\rangle =\left\langle l \sqrt{\frac{N}{d}} \sqrt{ \sum_{i=1}^d \frac{dZ_i^2}{l^2N} }\right\rangle = l \sqrt{\frac{2N}{d} }\frac{ \Gamma \left(\frac{d+1}{2}\right) }{\Gamma \left( \frac{d}{2}\right)}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.