That part is Xk. The largest eigenvalue turns out to be 1. In that case, then expectation of your value at the stopping time, when you've stopped, your balance, if that's what it's modeling, is always equal to the balance at the beginning. So this one-- it's more a intuitive definition, the first one, that it's a collection of random variables indexed by time. This is one of over 2,200 courses on OCW. Then, if it's a Markov chain, what it's saying is, you don't even have know all about this. So we have either-- let's start from 0-- random variables like this, or we have random variables given like this. I was confused. For example, if you know the price of a stock on all past dates, up to today, can you say anything intelligent about the future stock prices-- those type of questions. That means, for all h greater or equal to 0, and t greater than or equal to 0-- h is actually equal to 1-- the distribution of Xt plus h minus Xt is the same as the distribution of X sub h. And again, this easily follows from the definition. And b is what is the long term behavior of the sequence? That's what I'm trying to say here. ยป That means your lambda is equal to 1. I don't see what the problem is right now. There are Markov chains which are not martingales. Here, it was, you can really determine the line. So that's it for today. It is 1/3, actually. PROFESSOR: Yes. So if look at these times, t0, t1, up to tk, then random variables X sub ti plus 1 minus X sub ti are mutually independent. A discrete time stochastic process is a Markov chain if the probability that X at some time, t plus 1, is equal to something, some value, given the whole history up to time n is equal to the probability that Xt plus 1 is equal to that value, given the value X sub n for all n greater than or equal to-- t-- greater than or equal to 0 and all s. This is a mathematical way of writing down this. So happens with 1/2, 1/2. I'm not sure if there is a way to make an argument out of it. But using that, you can also model what's the probability that you jump from i to j in two steps. One of the most important ones is the simple random walk. This part is irrelevant. And if you want to look at the three-step, four-step, all you have to do is just multiply it again and again and again. q will be the probability that it's broken at that time. We stop at either at the time when we win $100 or lose $50. And moreover, from the first part, if these intervals do not overlap, they're independent. Just look at 1 and 2, 1 and 2, i and j. I want to make money. So example, you play a game. He mainly teaches Derivatives, Investments, Behavioral Finance and taught Financial mathematics for more than 20 years at University Paris-Dauphine. It's not a fair game. Springer, New York, Pliska SR (1997) Introduction to mathematical finance. So what you'll have is these two lines going on. If it's heads, he wins. We have one to one correspondence between those two things. And the question, what happens if you start from some state, let's say it was working today, and you go a very, very long time, like a year or 10 years, then the distribution, after 10 years, on that day, is A to the 3,650. And then I say the following. Broken to working is 0.8. There is no 0, 1, here, so it's 1 and 2. And then what it says is expectation of X tau is equal to 0. And they are random variables. And in the future, you don't know. Email: blockj@math.upenn.edu References: 1. You have a machine, and it's broken or working at a given day. There are martingales which are not Markov chains. So when you start at k, I'll define f of k as the probability that you hit this line first before hitting that line. So at the 100 square root of t, you will be inside this interval like 90% of the time. From the practical point of view, you'll have to twist some things slightly and so on. So I think it's easier to understand discrete time processes, that's why we start with it. And the third one is even more interesting. Your expected value is just fixed. What is a simple random walk? Under that assumption, now you can solve what p and q are. You're supposed to lose money. And one more thing we know is, by Perron-Frobenius, there exists an eigenvalue, the largest one, lambda greater than 0, and eigenvector v1, v2, where v1, v2 are positive. Even though the extreme values it can take-- I didn't draw it correctly-- is t and minus t, because all values can be 1 or all values can be minus 1. For example, to describe one stochastic process, this is one way to describe a stochastic process. The distribution is the same. The applications of stochastic processes and martingale methods (see Martingales) in finance and insurance have attracted much attention in recent years. Then Xk is a martingale. And what it's saying is, if all the entries are positive, then it converges. Of course, there are technical conditions that have to be there. This picture looks a little bit more clear. Analysis and Linear Algebra for Finance: Part I, Analysis and Linear Algebra for Finance: Part II, Patrick Roger is a professor of Finance at EM Strasbourg Business School, University of Strasbourg.

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