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Instructor (Brad Osgood):I wonder how long I can keep this up. All right. So the first

thing – any questions? Any comments? All right. The first thing I want to do is fix a little

mistake I made at the end of last time. I was upset about it for days afterwards, so I just

want to correct it. This was [inaudible] central limit theorem, which I hope you will agree

was a real treat. I mean, it’s just an amazing fact, and the way that it brings in

convolution, the [inaudible] transform of the galaxy and so on is really, I think, a real

highlight of the course to see those things emerge and to – the whole spooky nature of

repeated convolutions approaching the galaxy, I think, is just an amazing thing.

But I did screw it up at the end, so I just wanted to correct that. Everything was fine

except for one final formula that I wrote, so I’m not going to redo it. I just want to point

out where I wrote the wrong thing. It’s correct in the notes. I just wrote the wrong thing

on the board. This was the setup – so X1 through XN are, as they say in the biz,

independent and identically distributed random variables. You don’t have to write this

down. I’m just recalling what the notation was. And P of little X is the distribution that

goes along with all of them because they have the same distribution.

There’s a single function, P of X, which describes how each one of them is distributed.

So it’s a distribution for each. And then I formed P of N of X was the distribution for the

sum scaled by square root of N. So it’s the average – excuse me. There was some

assumption we made on the Xs on normalation, that is. We assume they had mean zero

and we assume they had standard deviation or variance one, and then if you form the

sum, the mean of the sum is zero but the standard deviation or the variant center

deviation of the sum is the square root of N, so it’s scaled by the square root of N.

You divide by square root of N to have this sum, SN, I called it, have mean zero and

standard deviation one, and P of N of X was the distribution for this. What we found was

that the Fourier transform – here we go. The Fourier Transform of P of N of S was the

Fourier transform of P and S over the square root of N to the Nth power. And then the

trick of the analysis – the way the proof worked was to compute the Fourier transform of

P and S over the square root of N just using the definition of Fourier transform that uses a

very sneaky thing of looking at the Taylor series of expansion with the complex

exponential integrating the terms and so on, and it was really quite clever.

What I found was that the Fourier transform of P at S over the square root of N was

approximately if N is large was one minus two Pi squared S squared over N. That was

fine. That was fine. And then for some reason, I raised this to the Nth power, so the

Fourier transform P at S over the square root of N to the N was then approximately this

thing raised to the Nth one minus two Pi squared S squared over N to the N and then

inexplicably I had the wrong approximation for this in terms of an exponential in terms of

the power of E.

That is this one minus two Pi squared S squared over N is approximately E to the minus

two Pi squared S squared. I’m going to look that up again to make sure I have it right this

time. Sure enough. All right. Then from there, and for some reason I wrote two Pi

squared over S squared last time, but it’s two Pi squared times S squared.

thing – any questions? Any comments? All right. The first thing I want to do is fix a little

mistake I made at the end of last time. I was upset about it for days afterwards, so I just

want to correct it. This was [inaudible] central limit theorem, which I hope you will agree

was a real treat. I mean, it’s just an amazing fact, and the way that it brings in

convolution, the [inaudible] transform of the galaxy and so on is really, I think, a real

highlight of the course to see those things emerge and to – the whole spooky nature of

repeated convolutions approaching the galaxy, I think, is just an amazing thing.

But I did screw it up at the end, so I just wanted to correct that. Everything was fine

except for one final formula that I wrote, so I’m not going to redo it. I just want to point

out where I wrote the wrong thing. It’s correct in the notes. I just wrote the wrong thing

on the board. This was the setup – so X1 through XN are, as they say in the biz,

independent and identically distributed random variables. You don’t have to write this

down. I’m just recalling what the notation was. And P of little X is the distribution that

goes along with all of them because they have the same distribution.

There’s a single function, P of X, which describes how each one of them is distributed.

So it’s a distribution for each. And then I formed P of N of X was the distribution for the

sum scaled by square root of N. So it’s the average – excuse me. There was some

assumption we made on the Xs on normalation, that is. We assume they had mean zero

and we assume they had standard deviation or variance one, and then if you form the

sum, the mean of the sum is zero but the standard deviation or the variant center

deviation of the sum is the square root of N, so it’s scaled by the square root of N.

You divide by square root of N to have this sum, SN, I called it, have mean zero and

standard deviation one, and P of N of X was the distribution for this. What we found was

that the Fourier transform – here we go. The Fourier Transform of P of N of S was the

Fourier transform of P and S over the square root of N to the Nth power. And then the

trick of the analysis – the way the proof worked was to compute the Fourier transform of

P and S over the square root of N just using the definition of Fourier transform that uses a

very sneaky thing of looking at the Taylor series of expansion with the complex

exponential integrating the terms and so on, and it was really quite clever.

What I found was that the Fourier transform of P at S over the square root of N was

approximately if N is large was one minus two Pi squared S squared over N. That was

fine. That was fine. And then for some reason, I raised this to the Nth power, so the

Fourier transform P at S over the square root of N to the N was then approximately this

thing raised to the Nth one minus two Pi squared S squared over N to the N and then

inexplicably I had the wrong approximation for this in terms of an exponential in terms of

the power of E.

That is this one minus two Pi squared S squared over N is approximately E to the minus

two Pi squared S squared. I’m going to look that up again to make sure I have it right this

time. Sure enough. All right. Then from there, and for some reason I wrote two Pi

squared over S squared last time, but it’s two Pi squared times S squared.

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