Welcome to Lecture five on option pricing via transform techniques. In the last lecture, we actually ended with, I'm borrowing as usual, this is my habit on borrowing the very last slide of the previous lecture where we ended up was call option while its inverse Fourier transform. If you remember, I said once you have this, which was the Fourier transform or modified called Fourier Transform of modified call, we said we can evaluate the call option via this integration. If you remember, as you remember, I would say, the Fourier transform modified call which is this, we have it with respect to the characteristic function of the log of stock price process. That means once you have this, once you have it, the characteristic function of the log of the processes stock price process, and we evaluate it at this point and multiply it by this quantity. What we would get is we would get the Fourier transform or modified call. That means we have it in an integrated form, and to get the call, we simply do the evaluation of this integral. But how would you evaluate this integral? Then what I will do in the next three slides, we will discuss integral evaluation. Now, we have this now knowing that this is known analytically as we did before in lecture three, we do not like to deal with indefinite integral, we want to be definite. That means what we are saying is that suppose you are going to infinity, we will go to B. You might wonder what happened from minus infinity to infinity, which I have one over two pi. I move to one over pi zero to infinity. The reason for this because we have an even function. The integrant was an even function, I wanted to leave this one as a short quiz problem. For the even functions, what we can do is if you have it minus infinity to infinity, it would become, as you see as one over two pi becomes one over pi, and minus infinity to infinity becomes zero to infinity, that's the first thing. Now, we want to make it the definite integral. We simply go to something large enough, is large enough such that by ignoring the rest of the integral, I'm not going to lose much, is within some threshold. Then the B, to B determined actually and is zero to B, and the same integral except is not definite. Now, what we are trying to do we are trying to evaluate that integral. We've seen or the integral evaluation, then I'm using exact same method. Again, as we did it before, we are familiar with this, the set N equal subintervals of length eta, I am trying to be consistent with what we had before that makes it easier that way, and what we are doing is, we are saying eta is B divided by N, and that becomes your grids, the grid points because now we are going from zero to B. We said that 2N equal sub-intervals, and what we're doing is because this is over nu, the first one is nu one. This comes two mu n and plus one, and this are your grid points and do not forget this length is eta. That's how I'm setting up the grid points. This is just the way it is easier to write it the way I'm writing it. You could have said something like this is equal to J eta, and instead of one, you could observe from zero. These are different conventions up to you, but this make it easier as you will see later on to write it the way I'm writing it. Now, then what I'm doing is it goes to n sub-integrals, and each sub-integral looks like this. This is very consistent to the way I did it for the case that we have the conditional probability density function. Now we go to each sub-integrals using trapezoidal rule exactly as before. Again, just to remind you one more time, the way it works is, if we have this as one of the sub-interval, and let's assume we have something like this, I turn it into a trapezoid and I do the area of this trapezoid, and what I would missing is because this is not exactly the area underneath, this is how much I will be missing. But what I'm saying is, if it is eta small enough because this length is eta, this would be a very tight approximation. That's what we are doing here. One more time again, and that would be simply this height plus this height, multiplied by this length, divided by two. That's the that part and then of course this height means this integral evaluated at nu j, this simply means is at this height which means this integrand evaluated that this point, and at that's as I said the very consistent to what we had before. Now, I'm simply rewriting what I had for the sub-integrals. Now, I'm substituting for this, but I'm getting up here which would come down here, and as I said before, now we have a sum, but inside sum, I have two quantities, I can turn a into one by just simply expanding it. Then I'm expanding it, I can't write it as one. The only differences because one more time as I said that before, because when I'm writing the trapezoids, the very first one and the very last one would then share this, but the one in between would share the height twice. That means the one in the beginning and at the end would have eta over two, but the one in the middle would have just eta because I'm utilizing them twice. You might say I'm seeing the first one, what happens to the last one I dropped it. That means, the sum as you see does not go to N plus one. I dropped the last one because I'm saying I can ignore it, and you will see later on that would be very helpful for us to do. Now, as we did it before, we always look at computational cost. If we call each of these as one unit of computation, how many units would you use? You would use N over. That means it would be order of ON, or capital N. Now, I want to do a comparison. Now what I mean by comparison, in a case of having the characteristic function, this would be a function of characteristic function, you already seen this. That would give you the option price. I'm borrowing from lecture three, I'm having the conditional probability density function, conditional probability density function versus these two computational Voice Looks like the both are ON. This is very explicit and you see it because that's simply the payoff, that's simply the conditional PDF, and actually these weights are exactly the same as before, the cost looks the same. But as you see this is pretty actually implicit, because you don't see any payoff, you do see any PDF, you don't see anything like that, but it turns out actually this becomes more generic than the other one. The reason is because for most processes, we can actually, we do have characteristic function of the longest stock price as you will see in the next lecture, and is easy to calculate. We have an integrated form, but when it comes to the PDF, that's not the case. That is why we switched but and when you compare this is fully explicit by explicit, you can see the terms, this is pretty implicit. You can actually see that it's actually looks like a call price. Having said this, the derivation comes from it and you've seen actually the derivation. Now, what are we doing in the next lecture is that I'm now trying to explain to you that what's the benefit of utilizing this as opposed to the other one aside from the fact that in many cases, we may not have conditional PDF, there are other benefits to it as well. Until next lecture. Thank you.
