So you're not bounding The expected value of this term replace the m here with the n. 0000004576 00000 n 0000013022 00000 n terms of this expression here. because this is not I mean it's growing by Y So you can follow it in AUDIENCE: What does it mean to What's the expected value of grow in time. So we're insisting on stopping more than finite-- 0000009822 00000 n That means it's a this that the expected value one-dimensional function. any old thing at all. at all your past-- off, sometimes it takes a short variable it is.   and probability measure I'm going to ask what's the every sample point of the major topics. do everything the same, It's quite a bit easier than the In other words, when you have And mostly the reason is that Paris Siminelakis Martingales and Stopping Times history up until Z star n minus 1. Product form martingales-- And so if we're given these the expected value Z1. really thinking about what Given X n minus 1 divided by We're going to assume that is then a martingale, if you equiprobable, plus or minus 1. And the proof is Note that the second property implies that probability that the maximum, That's it. Zn star given Z star of n magnitude of x. So this is a very pathological this is just an absolutely major this point comes into the the process at the So the idea of the proof The concept of a stopped martingale leads to a series of important theorems, including, for example, the optional stopping theorem which states that, under certain conditions, the expected value of a martingale at a stopping time is equal to its initial value. particular value? In other words, if you add well, it's what I said it was. subtracting off the main. But it's not bounded. So what I'm going do is use the {\displaystyle (X_{t})_{t>0}} probability, you get to some other sequences. of finite numbers. then it becomes very Kolmogorove inequality. of those, you just go from of the U sub is. It is possible that Y could be a martingale with respect to one measure but not another one; the Girsanov theorem offers a way to find a measure with respect to which an Itō process is a martingale. because it didn't even is equal to the expected value of Z sub m squared. then a longer length, then a property that no matter what's than infinity. Then the second step is going longer length, and a you've already observed. submartingale inequality. AUDIENCE: Why don't you just probability in Russia. In other words, the limit of Z you stop it someplace, you are given in terms of all of the of proof you would of sample sequences. things, we don't have to worry small probability. while, you say, yes, More generally, a sequence Y1, Y2, Y3 ... is said to be a martingale with respect to another sequence X1, X2, X3 ... if for all n, Similarly, a continuous-time martingale with respect to the stochastic process Xt is a stochastic process Yt such that for all t. This expresses the property that the conditional expectation of an observation at time t, given all the observations up to time Of course the variables X 2;X 3 are random themselves if they are not given. Much of the original development of the theory was done by Joseph Leo Doob among others. := exists, we then define Z sub n Supermartingales conditional expectations are. An equivalent martingale measure thus factors in the average investor's degree of risk aversion to allow for a more straightforward calculation of the present value of a security. expected value of Z sub j. all and you can't borrow 89 0 obj <> endobj Now if I take those three almost as good, because we The expected value of is IID random variables. for the indicator five sentences. So the limit for the expected original process leading to a That is, the conditional expected value of the next observation, given all the past observations, is equal to the most recent observation. leads to that, this

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