Let's go through the setup. As usual, we assume S cap T is T-time price. That means the price at time T which is unknown of course. Now as before, we assume that we have the density of ST. As you remember you say, but you said you don't have it. The assumption is we say that assume that we have it and then you'll see that we actually would then need to have it. Now when we write the density of ST, what we typically work in the log space. Then when it comes to density of log of ST, which we call it as the smallest t, is what I call Q. Then I go from F to Q. Now also I'm working the log space for the strike as well. I call it small k which is a log of capital K. This one as we said that earlier is a characteristic function of the log of the stock price process, fine. Now I'm kind of abusing the notation here. Previously what we call we write it like this. If you remember in the previous lectures. Now here, a bit for simplicity and also because we work in the log space, I'm writing it as CT(k). This is exactly the same thing, except I'm having this one over here as opposed to capital K. I'm having the small k which is a log of K and I'm dropping that 0 is known, that is today's price. This is just a notation. I'm sorry that I'm abusing it, but this is what we typically intend to do. The CT(k) is a price of a T-maturity called with this strike capital K, which is e to the power small k. Now, what's the formulation? As before, you've seen this before, the discounted, you remember this expectation, which that expectation was exactly this. You remember when we dropped the positive part, it turned to this. That's exactly what I'm borrowing from the previous slides. Now I'm turning everything from capital S, K2, its so called log space. Then instead of ST, I'm writing it at e to power smallest tk. I'm writing it into e to the power k. This one as I said would become q and that one of course I'm changing. Remember when you go from here to this, you do the change of a variable. That means you go from S to the log of S which is S. And that is what I will be calling that one. This is just simply a starting. What we have, do the change of variable, do everything in the log space. Good, now just for you to see the relationship between f and q, you've already seen this for the log normal distribution, you've seen that before in the previous lecture. In this lecture, when I'm going from f to q, it looks like this. You may say this is exactly the same because this is what is exactly that. This is exactly what is exactly this. The only difference is, as you see, there is no ST in the log in the denominator. The reason is because you do the changeover variable, that would be gone. I'm going to give this one as an exercise for you actually. When we set up the exercise of the quiz, I'm going to ask you to do this, which is a simple change of a variable just in one line. But it's a good exercise for you to do it. Now, as I said, the goal for us is to link the characteristic function to options pricing. Then the next few slides, I'm going to do a step by step, just some derivation. And after result and as you'll see, at the end, I'll come up with a generic form of pricing, which would be a nice simple model-free set up. But I have to go through some derivation just to see how we're going to link it. Few things I'm going to do. The first thing is this, I already started from CT(k) but I wouldn't work with CT(k). I would work with a small ct(k), which I call it the so called modified call, which the modify called means the original call multiplied by this. You may see later where this is coming from. At this point, just take it for granted. This is just e alpha to k, which tk is the same k. And alpha is some positive number. Now, what I will do is I'm gona look at the Fourier transform of this modified call. Then the Fourier transform of modify call, which I'm going to call it psi sub t e new, and I'm using the definition. Then the definition of Fourier transform was simply this, multiply whatever you want to get, if you remember, for f of x, that was this Fourier transform. Then sub f, I'm substituting this, nothing more. And then, what I'm doing is, of course, itself means that, I just did a simple substitution. Now, let's say you have this, then what I said, if you have this to recover, my apologies. Let's say you have this one is given to you, in order to recover CTk, we just do inverse Fourier transform. I'm just using the definition, and if you remember, it was 1/2Ï€, whatever that one is, but constrain, this was coming out. That's exactly where this is coming, and remember, it becomes the reverse of it. Then it's 1/2Ï€, and of course, this one, assuming this is known, and that's exactly by definition. That's if you go back to the slides that I had the definition of Fourier transform, inverse Fourier transform. That's exactly where it's coming from, assuming you have that. You may say a few things. First of all, I do not have this, that's correct, but you will see, you're going to get it actually analytically, and that means you're going to get that one once you have that one. Good, let's see, then, what I would do is I'm trying to get this one analytically, which it was exactly that, I do simple substitution. The simple substitution is this, remember, we had the e alpha k, and the rest of it comes from the definition of the core, which is exactly what we had in the previous two slides. Now, I want to evaluate this one. In order to evaluate it, you'll see actually, we didn't, we can't go any further than this, but what I'm trying to do is this, I see what I can take out, what I can take out is this. That's this one comes out, by taking out means, no dependency on s or on k. Remember, this becomes a double integral after the substitution. Now, the other thing is, I can make these two into that form. Nothing new here, but what I'm doing is, the integral was going from minus infinity to infinity from k to infinity. DSDK, what I'm trying to do is, I'm trying to do the change of integral which is called Fubini. Fubini will tell us how to do this, then as opposed to going from minus infinity to plus infinity, k to infinity DSDK, this turns out to be minus infinity to infinity. Minus Infinity 2 to S dkds. I'm leaving this one and an exercise for you as well. Again it's just a two liner. You're simply looking at, the way you do integration, you're just turning it around, and then you will see you come to this term. That's exactly what I did. This, coming from here to this, I'm using Fubini. Fubini theorem would help us to do this changeover integration. While this is important for us? Because the minute you do the change of integration, then you see this qs does not depend on k, then you do the integration first with respect to k. That means you can pull it out. That's exactly what I did here, and whatever I'm leaving over here are those that depends on k, the rest, I'm taking it out. And then you will see a candidate's integral analytically, which I'm showing it in the next slide. Then the next slide, what I'm doing is, I'm taking that integral out, and do the integration, I do the substitution, and that's where I'm ending. Now, what you would notice is, if you didn't have this alpha, that means, if you go ahead and set equal to 0, this tube would have been undefined. Again, another exercise, I'm leaving it for you, that if you set alpha equal to 0, you do the substitution at minus infinity, it would be undefined. Having alpha greater than zero, therefore any alpha greater than 0, theoretically, you will see with vanish, it would be 0. That's why it would be evaluated at es, evaluated at es, common denominator, I would get that one at. And as you're noticing, that inner integral, now I've known analytically, I can substitute it back. By substituting back, that's what I would get after the substitution, and I can take the denominator down because it doesn't depend on s. That's denominator, it's out, this one is already out, and then you'll see, I end up with This integral you may say hold on, but I cannot do this one analytical the answer to it is yes at this point, but, you already know the characteristic function of log of the stock price. The characteristic function is this, That's the definition of characteristic function, which is this, you already have this. Once you have this, then you will notice that this is nothing but this characteristic function being evaluated at this point. And then, as you will see, the Fourier Transform a modified call, can be known analytically as long as you have the characteristic function of the log of the stock price process, that's the beauty of this. Now you already got this fun, and as I said earlier, in order to get the call back, it would become nothing but this e- alpha k divided by 2 pi- infinity infinity of e- i nu k. Multiplied by this, which I put it in a box because I known it analytically and do not for get what that one is, this is this. I'm going to just let it be for this lecture in the next lecture what we're trying to do is, now we're trying to evaluate this one. But as you see, the call is link to characteristic function through that one, which I'm going to substitute it in, and explain to you how to do it, see you in the next lecture.