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Let me now outline a proof of the central
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limit theorem. This proof starts with the
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evolution equation of the probability distribution
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itself after a single step, so let me write
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Pn(x) - what is the probability that after
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n-steps I'm at position x - how can this happen?
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In order to be at position x at the nth step,
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I had to be somewhere else at the n-minus-
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first step - so Pn-1(x'), and then in a single
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step I went from x-prime to x, so little p(x'
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going to x) dx'. So again, this equation which
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is known as the ChapmanāKolmogorov equation
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expresses that in order to be at position x
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at the nth step, I was somewhere else at
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the n-minus-first step and then I made a
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hop from the somewhere else to x and
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then I have to integrate over all these intermediate
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somewhere elses in order to get to x. Now, in
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order to simplify this, I'm going to make use of
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Fourier Transform technology and note that
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this object is known as a convolution because
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we go from x-prime to x by this single step. And so
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if I Fourier Transform this equation, convolution
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in real space turns out to be multiplication
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in Fourier space. So after the Fourier transform,
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we basically have the much simpler statement
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that Pn(k), and again, the k is to emphasize
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that we now have the Fourier Transform function,
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so Pn(k) is not Pn(x) with x replaced by k, but it's a
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different function all together - the Fourier
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Transform - and it's equal to Pn-1(k)*little p(k).
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And so I can now take this recursion formula
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and just recurse it all the way down to the
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start of the walk and infer that...well let's
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just do it one step at a time. Pn-1, well
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that's nothing more than Pn-2(k) times
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little p(k)-squared. And I just continue this
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process until the very end and what I'm
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going to get then is that this is equal to
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P0(k) times little p to the power n. But again,
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if we imagine that a particle starts at the origin,
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then that means P0 at x is a delta function of x,
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it's Fourier Transform is just 1, and so this
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just reduces to little p(k) to the power n.
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And now, to find the probability distribution
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as a function of x, we just have to inverse
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Fourier Transform Pn(k). So let's write that out.
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So Pn(x) is equal to 1 over 2pi, the integral minus over
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all space and so it's going to be p(k) to the
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power n, e to the ikx. Let me do this in
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1-dimension because in higher dimensions
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there is no conceptual difference, it's
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just more clunky, so let's do this as an
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integral from minus infinity to plus infinity
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dk, 1-dimensional integration. But now
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I'm going to make use of the fact that the
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single step distribution has this feature that
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the mean square displacements are finite.
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So let's now return to little p(k) which
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was a Fourier Transform of little p(x), so
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p(x) e to the ikx dx. And let's just expand
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this in a Taylor series. So we have p(x),
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I have one plus ikx minus k-squared
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x-squared over two plus higher-ordered
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terms dx. And let's identify each one of
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these terms. So the first term, p(x) 1 dx,
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this is the probability that I hop anywhere
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in a single step, and by construction, this
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is equal to one. The next term, well the ik
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comes out in front and then I have the
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integral of p(x) xdx and this is nothing
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more than the mean displacement after
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a single step. So this is ik x-average. And
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similarly for the second term, we have
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k-squared over 2 and then I have the integral
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of p(x) x-squared dx which is nothing more
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than the mean squared displacement after
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a single step plus higher-ordered terms.
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To make the rest of this outline of the proof
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simpler, let me just restrict myself to the
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symmetrical case and I'll leave as an
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exercise for the student to worry about the
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asymmetrical case. So I'm going to just forget
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about this term assuming that the
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random walk is symmetrical. So now I'm
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going to plug in this p(k) back into this
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integral from Pn(x), so Pn(x) is equal to
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one over 2pi, the integral of quantity one
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minus k-squared x-squared average over two.
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And it's raised to the power n e to the minus ikx dk.
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And now I make use of the fact that I'm
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interested in the limit of large n and
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small k. Small k corresponds to large x in
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the Fourier domain, and so I can approximate
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this object here by the exponentials. This
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is asymptotically going to one over 2pi e
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to the minus nk-squared x-squared average
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over two times minus ikx dx. And now I
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make use of the fact that the Fourier Transform
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of a Guassian function in k is a Gusassian
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function and so I finally end up...again,
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I'm skipping all of the simple algebraic steps
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which I'm leaving as an exercise for the student
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to do, is one over the square root of 2pi n
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x-squared average e to the minus x-squared
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divided by two n x-squared average. So this is
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just the central limit theorem for the case of a
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symmetrical walk in which the mean displacement
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net in a single step is 0. So again, the thing that I
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want to emphasize is that this distribution is
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Guassian and furthermore it is independent
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of almost all details of the individual steps.
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So this universal behavior is a powerful, unifying
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theme in many problems in statistical physics
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and it played an important role in the
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development of the theory of random walks.