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.