Let me now outline a proof of the central

limit theorem. This proof starts with the

evolution equation of the probability distribution

itself after a single step, so let me write

Pn(x) - what is the probability that after

n-steps I'm at position x - how can this happen?

In order to be at position x at the nth step,

I had to be somewhere else at the n-minus-

first step - so Pn-1(x'), and then in a single

step I went from x-prime to x, so little p(x'

going to x) dx'. So again, this equation which

is known as the Chapman–Kolmogorov equation

expresses that in order to be at position x

at the nth step, I was somewhere else at

the n-minus-first step and then I made a

hop from the somewhere else to x and

then I have to integrate over all these intermediate

somewhere elses in order to get to x. Now, in

order to simplify this, I'm going to make use of

Fourier Transform technology and note that

this object is known as a convolution because

we go from x-prime to x by this single step. And so

if I Fourier Transform this equation, convolution

in real space turns out to be multiplication

in Fourier space. So after the Fourier transform,

we basically have the much simpler statement

that Pn(k), and again, the k is to emphasize

that we now have the Fourier Transform function,

so Pn(k) is not Pn(x) with x replaced by k, but it's a

different function all together - the Fourier

Transform - and it's equal to Pn-1(k)*little p(k).

And so I can now take this recursion formula

and just recurse it all the way down to the

start of the walk and infer that...well let's

just do it one step at a time. Pn-1, well

that's nothing more than Pn-2(k) times

little p(k)-squared. And I just continue this

process until the very end and what I'm

going to get then is that this is equal to

P0(k) times little p to the power n. But again,

if we imagine that a particle starts at the origin,

then that means P0 at x is a delta function of x,

it's Fourier Transform is just 1, and so this

just reduces to little p(k) to the power n.

And now, to find the probability distribution

as a function of x, we just have to inverse

Fourier Transform Pn(k). So let's write that out.

So Pn(x) is equal to 1 over 2pi, the integral minus over

all space and so it's going to be p(k) to the

power n, e to the ikx. Let me do this in

1-dimension because in higher dimensions

there is no conceptual difference, it's

just more clunky, so let's do this as an

integral from minus infinity to plus infinity

dk, 1-dimensional integration. But now

I'm going to make use of the fact that the

single step distribution has this feature that

the mean square displacements are finite.

So let's now return to little p(k) which

was a Fourier Transform of little p(x), so

p(x) e to the ikx dx. And let's just expand

this in a Taylor series. So we have p(x),

I have one plus ikx minus k-squared

x-squared over two plus higher-ordered

terms dx. And let's identify each one of

these terms. So the first term, p(x) 1 dx,

this is the probability that I hop anywhere

in a single step, and by construction, this

is equal to one. The next term, well the ik

comes out in front and then I have the

integral of p(x) xdx and this is nothing

more than the mean displacement after

a single step. So this is ik x-average. And

similarly for the second term, we have

k-squared over 2 and then I have the integral

of p(x) x-squared dx which is nothing more

than the mean squared displacement after

a single step plus higher-ordered terms.

To make the rest of this outline of the proof

simpler, let me just restrict myself to the

symmetrical case and I'll leave as an

exercise for the student to worry about the

asymmetrical case. So I'm going to just forget

about this term assuming that the

random walk is symmetrical. So now I'm

going to plug in this p(k) back into this

integral from Pn(x), so Pn(x) is equal to

one over 2pi, the integral of quantity one

minus k-squared x-squared average over two.

And it's raised to the power n e to the minus ikx dk.

And now I make use of the fact that I'm

interested in the limit of large n and

small k. Small k corresponds to large x in

the Fourier domain, and so I can approximate

this object here by the exponentials. This

is asymptotically going to one over 2pi e

to the minus nk-squared x-squared average

over two times minus ikx dx. And now I

make use of the fact that the Fourier Transform

of a Guassian function in k is a Gusassian

function and so I finally end up...again,

I'm skipping all of the simple algebraic steps

which I'm leaving as an exercise for the student

to do, is one over the square root of 2pi n

x-squared average e to the minus x-squared

divided by two n x-squared average. So this is

just the central limit theorem for the case of a

symmetrical walk in which the mean displacement

net in a single step is 0. So again, the thing that I

want to emphasize is that this distribution is

Guassian and furthermore it is independent

of almost all details of the individual steps.

So this universal behavior is a powerful, unifying

theme in many problems in statistical physics

and it played an important role in the

development of the theory of random walks.