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 Chapmanā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.