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I'm reading the notes here, particularly page 33, which gives Einstein's formulation of Brownian motion. One statement made in the construction seems to lack justification.

  1. Suppose that the probability density of the event that an ink particle moves from $x$ to $x+y$ in (small) time $\tau$ is $f(τ, y)$. Then $$u(x,t+\tau) = \int_{-\infty}^\infty u(x-y,t)f(y,\tau)dy = \int_{-\infty}^\infty \left( u-u_xy+\frac12 u_{xx}y^2+\cdots \right)f(y,\tau)dy.$$

Since $f$ is a probability density, $\int f dy=1$. And since $f$ is symmetric, $\int yfdy=0$. Now if we assume that $\int y^2 fdy=D\tau$, for some constant $D>0$, the author correctly concludes that

$$\frac{u(x,t+\tau)-u(x,t)}{\tau} = \frac D2u_{xx}(x,t) + \{\text{higher order terms}\}.$$

He then says that if $\tau\to0$, the higher order terms vanish and we have $u_t = \frac D2 u_{xx}.$

Question: why do the higher order terms vanish?

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    You changed from $\rho$ to $f$ partway through the discussion. – 2017-02-09
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    Anyway, none of this is rigorous at all, so you should really just be fishing for intuition. The intuition is that if the variance of $\rho(\cdot,\tau)$ is linear in $\tau$ then the $2n$th moment "should" be proportional to $\tau^n$ (while the odd moments obviously vanish). Of course it is not at all clear how to *guarantee* that, but that's the idea. – 2017-02-09
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    I changed them all to f's now. That doesn't seem intuitive to me at all. Is this even a reasonable thing to expect for well behaved functions? – 2017-02-09
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    Well, for instance this is true for the normal distribution, which is really what you're going to see fall out of all of this in the end. But it is also true for any "scale invariant" $f$ which has all finite moments: this means that the random variable with density $f(\cdot,\tau)$ has the same distribution as $\tau^{1/2} X$ where $X$ has the density $f(\cdot,1)$. As for why the scaling has to be $\tau^{1/2}$, this is actually a pretty standard phenomenon: you might look up literature explaining why the units of the diffusion constant are what they are. – 2017-02-09
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    I don't want to get in a protracted discussion about this in the comments, but do you know of a resource to explore this more rigorously? Alternatively if you expanded a bit or provided a reference in an answer I would be happy to accept it. (EDITED: I do see how a normal distribution fits the bill) – 2017-02-09
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    It really depends on where you want to start; Brownian motion is rather "universal" in the sense that there is a wide variety of assumptions that you can start from and still arrive at it. The place where I got some really nice intuition is from looking at Brownian motion as the limit of simple symmetric random walks, where you send $\Delta x,\Delta t$ to zero in such a way that $\frac{\Delta x^2}{\Delta t}$ converges to a finite, nonzero number (which is the diffusion constant, or maybe some universal multiple of it, I don't recall). – 2017-02-09
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    This approach shows us that this scaling is "critical": if $\Delta x$ is any smaller than that, then the process obtained in the limit doesn't even move in finite time, whereas if $\Delta x$ is any larger than that, then the variance blows up. – 2017-02-09
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    Thanks, Ian. Do you have a resource you'd recommend that explores this particular approach? The same book does have a discussion of this, but it is handwavy and unsatisfactory to me for similar reasons. – 2017-02-09

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