# APPENDIX A

# Central Limit Theorem-Plausibility Argument

If a path begins at a point *x*_{0} then the initial distribution over *x* is the Dirac delta distribution, which means merely that the path has probability 1 of being located exactly at the starting point, expressed as

*p*(*x*; *t* = 0) = *δ*(*x* − *x*_{0})*.*

Then use a *square shock*, meaning that the distribution is convoluted with a square of width 2*α* and height 1*/*(2*α*),

and the new distribution is then a square. Then convolute again and the new distribution is a triangle,

and so on. Now, in more general terms, let the shock be some function *f*(*x*), not necessarily a square. Fourier transform the function to get *f*(*p*), and expand it as follows:

*f*(*p*) = *a*_{0} exp(−*a*_{2}(*p* − *p*_{0})^{2} + *a*_{3}(*p* − *p*_{0})^{3} + · · ·)

about its maximum, at *p*_{0}. This expansion is meaningful as long as the derivative of *f*(*p*) is zero and continuous at *p*_{0} and the second derivative of *f*(*p*) is continuous and negative (i.e., *a*_{2} *>* 0). Then take the *N*-th power of this, to get the Fourier transform of the operator that gives us the *N*-step distribution:

Now the argument is that because the first term looks like the Fourier transform of a Gaussian of standard deviation

and for a particular standard deviation, ...

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