I'm very new to this topic. I have a distribution similar to the picture below but with the center at zero.

As I said, I'm very new to this, but if I understand correctly, if there was no hole in the middle, the probability distribution of the magnitude would have a Rayleigh distribution. Now my question is, what type of distribution do you have when there is a hole in the middle? and what would be the formula for the variance?

I have a uniform concentration of solute in a 2D ring (doughut) shape that is diffusing and I would like to know if the probability distribution of the magnitude is somehow related to a gaussian distribution like the rayleigh distribution. Hope this is more clear? I do have an analytical solution to this problem but not on the computer that I'm working on right now.

Really hope someone can help me with this, or can give me some advice on what books to read to get a better understanding of this topic.

Picture is from previous question on this site, this is the link to that question

[Picture is from previous question on this site]

  • $\begingroup$ Your question is somewhat puzzling, because almost every formula for the variance (or other moments) of distributions applies to all distributions. What formula(s) do you know? To be able to make progress, you will need more than a picture: how is your distribution specified mathematically? $\endgroup$ – whuber Aug 5 '15 at 13:05
  • $\begingroup$ OK, let's try this another way. Having a "hole in the middle" is not sufficiently specific to permit any kind of helpful answer. Please tell us what you know about your distribution. Do you have a formula for it? Is it estimated from data (and if so, what are the data like)? $\endgroup$ – whuber Aug 5 '15 at 13:55
  • $\begingroup$ Thanks a lot for taking the time to answer. I'm sorry for not being very clear. I have a uniform concentration of solute in a 2D ring (doughut) shape that is diffusing and I would like to know if the probability distribution of the magnitude is somehow related to a gaussian distribution like the rayleigh distribution. Hope this is more clear? I do have an analytical solution to this problem but not on the computer that I'm working on right now. Thanks a lot. $\endgroup$ – Maria Aug 5 '15 at 14:11
  • 1
    $\begingroup$ It is related to a Gaussian only in a remote (and not terribly useful) sense. A correct answer to your question can be obtained only from the specific analytic solution you have. If you are modeling true diffusion, then at any positive time the density at the origin will be nonzero. Moreover, at some times the density at the origin ought to become large--the opposite of a "hole" in the middle. At any time, the variance of this distribution will (literally) be the second moment of the mass about the origin: are you sure this is what you need to find out? $\endgroup$ – whuber Aug 5 '15 at 14:33
  • $\begingroup$ If you really have a hole in the middle, it's a truncated distribution, but w/ the truncation on the inside, not the outside. So this may (or may not) be a Rayleigh distribution (see Dilip Sarwate's answer below), except that there is some "jacking up" of the density on the non-truncated portion, so that the density integrates to 1. This could be determining the probability P of the truncated portion, then multiplying the density of the non-truncated portion by 1/(1-P); or it could be something else, depending on how the hole is created, and the remaining portion of the distribution affected. $\endgroup$ – Mark L. Stone Aug 5 '15 at 14:40

Diffusion of a single particle is a random walk in two dimensions. As a function of time, $t$, the probability density for its location will therefore be Gaussian, centered at the particle's original location, with a variance $\sigma^2 t$ directly proportional to time. (The constant of proportionality $\sigma^2$ is a quantitative expression of the diffusivity of the medium.)

Because diffusion is usually modeled as an additive process--the motions of the particles are assumed to be independent and non-interfering--a uniform distribution of particles within any region $\mathcal R$ will evolve into a uniform location mixture of such Gaussians. In statistical terminology, the distribution at time $t$ will be the sum of the uniform distribution over $\mathcal R$ and the (two-dimensional) Gaussian distribution centered at $(0,0)$ with covariance matrix

$$\Sigma(t) = \pmatrix{\sigma^2 t & 0 \\ 0 & \sigma^2 t}.$$

Because variances of independent variables add, the variance of the distribution at time $t$ will be the variance of the uniform distribution over $R$, plus $\Sigma(t)$.

It is straightforward to compute the variance of a uniform distribution over an annulus of radii $r_1 \lt r_2$: it is the diagonal matrix with values $(r_1^2 + r_2^2)/4$. Therefore the variance of the diffused particles after time $t$ is given by the matrix

$$\left(\frac{r_1^2 + r_2^2}{4}+ \sigma^2 t\right)\pmatrix{1 & 0 \\ 0 & 1}.$$

As $t$ grows large, the distribution becomes closer and closer to the solution obtained by starting with all the points near the centroid of $\mathcal R$--regardless of what shape $\mathcal R$ might have or of how the particles were distributed within it. Provided $\mathcal R$ is bounded--as in this case--the diffused particles will eventually adopt a (circularly symmetric) Gaussian distribution centered at the centroid of $\mathcal R$. Its variance is given by the preceding formula.

To illustrate, I placed 10,000 points uniformly and randomly within an annulus of radii $7/4$ and $2$ and diffused them via a random walk with Gaussian steps of variance $\sigma^2 = 1/8^2$.

Figure: the original points

Along the way I kept track of the mean variance of the $x$ and $y$ coordinates relative to the theoretical variance as given above: namely, the starting variance plus $\sigma^2 t$. Their ratio will not be constant, due to the random fluctuations: it is a form of a random walk itself. But it will stay close to $1$:

Figure: the variance ratio

The progress of the diffusion can be plotted by showing the (smoothed) density on a 3D plot (whose height depicts the density).

![Figure: densities

Notice that by time $1/\sigma^2 = 64$, the average amount of particle motion is sufficient to place many of them in the very center. The annulus appears to be "filled up", and thereafter the mode of the distribution is in the center of the ring. This filling is apparent in a left-to-right scan of these images, which will take you from step 1 through step 72 in a quadratic progression of intervals. (The image will show detail when magnified.)

  • $\begingroup$ Thank you so much for the clear explanation! This is exactly what I was looking for. Very helpful! $\endgroup$ – Maria Aug 5 '15 at 18:17

If $X$ and $Y$ are random variables whose joint density function $f_{X,Y}(x,y)$ has circular symmetry about the origin, that is, the value of $f_{X,Y}$ at any point $(x,y)$ in the plane depends not on the individual values of $x$ and $y$ but only on $r = \sqrt{x^2+y^2}$, the (Euclidean) distance of the point $(x,y)$ from the origin, then we can write $f_{X,Y}(x,y)$ as a function $g(r)$ of $r$ alone. Note that $g(r) = f_{X,Y}(r,0)=f_{X,Y}(0,r)$. Then, with $R = \sqrt{X^2+Y^2}$ denoting the distance of the random point $(X,Y)$ from the origin, we have that the density of $R$ is given by

$$f_R(r) = 2\pi r g(r), \quad 0 \leq r < \infty.\tag{1}$$

In the special case when $X$ and $Y$ are independent $N(0,\sigma^2)$ random variables, we have that $$f_{X,Y} (x,y) = \frac{1}{2\pi\sigma^2}\exp\left(-\frac{1}{2\sigma^2} \left(x^2+y^2\right)\right) = \frac{1}{2\pi\sigma^2}\exp\left(-\frac{r^2}{2\sigma^2}\right)\tag{2}$$ giving the Rayleigh density $$f_R(r) = \frac{r}{\sigma^2}\exp\left(-\frac{r^2}{2\sigma^2}\right), \quad 0 \leq r < \infty\tag{3}$$ that you mentioned. But you should understand clearly that this is a special case. Indeed, if $X$ and $Y$ have a circularly symmetric density, then they are independent if and only if they are zero-mean normal random variables with the same variance. So, $(3)$ holds only in this special case.

A proof of $(1)$ in the general case is straightforward. We have that for $a \geq 0$, \begin{align}F_R(a) &= P\{R \leq a\}\\ &= P\{\sqrt{X^2+Y^2} \leq a\}\\ &= \iint_{\sqrt{x^2+y^2} \leq a}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy & \scriptstyle{\text{integral over circle of radius }a}\\ &= \int_{r=0}^a \int_{\theta=0}^{2\pi} g(r)\cdot r\,\mathrm d\theta \cdot \mathrm dr & \scriptstyle{\text{change to polar coordinates}}\\ &= \int_{r=0}^a 2\pi r g(r)\, \mathrm dr & \scriptstyle{\text{inner integral has value }2\pi} \end{align} which upon comparison to $\displaystyle F_R(a) = \int_{-\infty}^a f_R(r)\,\mathrm dr = \int_0^a f_R(r)\,\mathrm dr$ leads us to $(1)$.

For more details on this calculation, as well as what happens when $X$ and $Y$ do not enjoy circular symmetry, see Examples 4.6.2 (p. 185) and 4.7.6 (p. 192) in the Course Notes for ECE 313. I will also recommend these Notes for gaining a solid understanding of probability theory at the undergraduate level. (end of commercial).

Finally, in response to your question about the "hole in the middle", the function $g(r)$ would have value $0$ for small values of $r$ (corresponding to the hole). Since $(1)$ gives the density function of the magnitude $R$ of the distance from the origin, you should be able to compute the variance of $R$ by yourself.

  • $\begingroup$ Thanks a lot for taking the time to answer my question. The course notes will be very helpful. $\endgroup$ – Maria Aug 5 '15 at 18:23

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