If a 1D random walk, along the x-axis has has a closed for expression (Due to the central limit theorem)

$P(N)$ = $\frac{1}{\sqrt{2\pi N}}*e^{\frac{-n^2}{2N}}$, where $N=2n$.

This is a normal distribution with variance $N$, and standard deviation $N^2$.

I used this expression to generate a closed form for a 2D symmetric random walk on a lattice with unit step length.

I think that the x-component $P_x(n)$ will be the same as the 1D probability closed form, except with variance $\frac{N}{2}$ instead of $N$. This makes the x and y components of the equation,

$P_x(N)$ = $\frac{1}{\sqrt{\pi N}}*e^{\frac{-x^2}{N}}$,

$P_y(N)$ = $\frac{1}{\sqrt{\pi N}}*e^{\frac{-y^2}{N}}$,

and hence $P(r)=P_x(n)*P_y(n)$, which gives an expression

$P(e)$ = $\frac{1}{{\pi N}}*e^{\frac{-r^2}{N}}$, where $r=[x,y]$ <-- r is a vector with components

Is this logic correct, because when I try small values of N, the expression has a very large error. Or, what is the correction that I have to make for the expression to make sense?


First a correction, if you have the pdf written correctly, the standard deviation should be $\sqrt{N}$ and the variance is $N$.

You are correct in your assertion that in each direction you will still have the same Normal form. So the pair $\begin{bmatrix}{X\\Y}\end{bmatrix}$ will be a Bivariate normal distribution with covariance matrix:

\begin{equation} \Sigma = \begin{bmatrix} N & 0 \\ 0 & N \end{bmatrix} \end{equation}

If you would like to model the exact location after some steps, you should look into the Bivariate normal pdf.

If you only care about distance from the origin, then you may do what you were doing with $r$. That is

\begin{equation} R = +\sqrt{X^2 + Y^2} \end{equation}

Unfortunately, this may not have a particularly nice form. Here are some facts that may help. If $X \sim Normal$ then $X^2$ will have a (possibly non-central) Chi-Square distribution. Also, the sum of Chi-Square rv's is itself chi square. Finally, the square root of a Chi-Square rv has a "Half Normal" distribution. I'm not familiar enough with the distribution to know the details, but I imagine it could be done.

  • 2
    $\begingroup$ As a potentially non-constructive note, if $X$ and $Y$ were standard then $R^2 \sim \text{Exp}(1/2)$. I'm not sure if we could leverage that in some way. $\endgroup$ – David Feb 24 '17 at 15:31

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