# On the Brownian motion and Law of Large numbers

According to the Law of Large Numbers, if I throw a million coins, I expect to observe half of them face head, and the other half tail. Why doesn’t this apply to (1-dimensional) Brownian motion? A suspended particles undergoes the collision with millions of molecules, at every (infinitesimal) time interval. Therefore, why does it move at all?

• It is actually unlikely that you would get exactly half heads and half tails. May 5 at 13:21
• "undergoes collision with millions of mollecules at every time instant" sounds a bit nonsensical to me. It possibly comes from an incorrect understanding, but it would be easier to judge if you could edit it to make your meaning clearer. May 5 at 13:53
• @AXensen well the mean collision time is $\sim10^{-13}$ seconds, which means that there's $\sim10^{13}$ collisions happening each second. Pretty standard stuff, I thought. May 5 at 14:16
• Brownian motion is tough because of the issues with its power spectral density. See, for example, the answers and question here: dsp.stackexchange.com/a/58811/41790. Good luck with it!
– Ed V
May 5 at 17:35
• This is a somewhat related thread in math stackexchange that has a very interesting mathematical answer: math.stackexchange.com/questions/4533724/… May 5 at 19:41

...I expect to observe half of them face head, and the other half tail.

Well that's part of your problem. Just because you have $$N$$ coin flips, it does not mean that in every attempt you would have precisely $$N/2$$ heads and $$N/2$$ tails. You can have $$N/2-10$$ heads and $$N/2+10$$ tails; or the other way around. This is what makes it a probability distribution, the outcome is not deterministic but relies on chance.

For Brownian motion, it is more proper to consider the scaling limit of the random walk, which is the Weiner process (i.e., instead of unit steps, we use $$\mathrm{d}x$$), rather than the random walk. One of the key properties of this stochastic process is that the increments in time (e.g., from $$t$$ to $$t+\mathrm{d}t$$) are distributed normally with mean 0 and variance $$\sqrt{\mathrm{d}t}$$.

Since the variance is non-zero, then we can reasonably expect that there are outcomes in which the particle suspended in the fluid can drift because there is a probability of being slightly skewed in one direction over the other, even though the mean of the distribution is zero.

I think that you meant the random walk. It is true that the strong law of large numbers (SLLN) says that for $$n\to\infty$$ there are equal amounts of going left and going right almost surely. Let $$r_n$$ the displacement after $$n$$ steps: $$\frac{r_n}{n}\to 0 \quad a.s.$$ However, if you don't divide by $$n$$, you only get: $$r_n = o(n) \quad a.s.$$ You certainly do not have convergence almost surely of $$r_n$$ to $$0$$, and this is seen by estimating the various moments. In particular, taking the mean of the absolute value could (and does) give a sub-linear contribution (though the variance is easier to calculate).

Physically this is diffusion. Mathematically, you need to be more precise in quantifying the fluctuations about the mean. The usual approach is by using central limit theorem (though depending on what you are interested in, other approaches like large deviations could be more relevant). It says: $$\frac{r_n}{\sqrt{n}}\to r \quad \text{(in law)}$$ with $$r$$ a (gaussian) random variable. In particular, you recover that the growth is sub-linear as it is in $$\sqrt n$$, which is consistent with the SLLN.

Hope this helps.

• "It is true that the strong law of large numbers (SLLN) says that for $n\to\infty$ there are equal amounts of going left and going right almost surely." – I don't think this is phrased quite accurately. As you point out, the difference between the number of steps left and the number of steps right does not approach 0, so describing the situation as "equal amounts of going left and going right" simplifies things too much. May 5 at 22:14
• If you write $R_n$ (resp. $L_n$) the number of steps to the right (resp. left) the SLLN says $\frac{R_n}{L_n}\to1$ a.s. This is what I mean by there are equal amounts almost surely.
– LPZ
May 5 at 22:37

When you throw a million heads, you expect roughly $$\sqrt{\text{a million}}$$ excess of either heads or tails (1000 more heads than tails or vice versa). The distribution in the number of heads is described a binomial distribution. The variance of the number of heads (the average squared distance from the mean) is $$N/4$$, so the standard deviation is $$\sqrt{N}/2$$ - roughly speaking, how much you expect the number of heads to deviate from N/2 (wiki for binomial distribution). In the same way, with brownian motion, the distance from the initial point grows proportional to the square root of time.

To more clearly explain diffusion in air, the mean free path for a molecule is about $$10^{-7}\text{ m}$$, and the order of magintude of a molecule's velocity is $$500\text{ m}/\text{s}$$. So there's a collision (re-randomizing velocity) $$5\times10^9$$ times per second. So the order of magnitude of the diffusion of a test molecule in a gas is $$(10^{-7}\text{ m})\sqrt{T*5\times10^9/s}$$, or $$7\text{ mm}\sqrt{T}/\sqrt{s}$$. Certainly observable over reasonable timescales.

For pollen riding the surface of water, the story is quite a bit more complicated, and frankly I don't know how it works exactly. As you may have alluded to in your question, the pollen is in contact with many water molecules at once which are much smaller than it. The end result is still that the pollen's position grows like $$\sqrt{T}$$, but one cannot calculate the constant in such a simple way. The wiki for Brownian motion discusses some models.

• Another way to think of it is that your first sentence is basically saying that for a supposedly unbiased coin, you'll always start drifting towards more H or more T, rather than $N/2$ that the binomial distribution says you should obtain. That seems to not pass the sniff test. May 5 at 17:16
• Excuse me, does the Law of Iterated Logarithm play a role in all this? May 5 at 17:23
• @KyleKanos its 1000 more heads than tails, not 1000 heads, so N/2-500 heads and N/2+500 tails is a plausible outcome. The average distance from the mean value (which as you say is obviously N/2) for any distribution is called the standard deviation (the square root of the variance), which is given in the linked wiki as $\sqrt{npq}$, where $q=p=0.5$. So the number of heads on average differs from N/2 by $\sqrt{N}/2$ May 5 at 17:33
• @ric.san the law of iterated logarithm (i've just now heard of this) is definitely related. It seems to be a slightly stronger statement saying that for any random thing you redo many times and add the results, the sum differs from the mean by less than $\sqrt{2n\log\log n}$, which is very close to $\sqrt{2n}$... so it's a similar rule that applies to more general distributions and has a slightly bigger growth rate. May 5 at 17:43