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\title{Bishop Notes\\February 3, 2006}
\author{Gabriel C. Drummond-Cole}
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So we're going to continue working our way through the book.  Last Monday we made it through the first page.  I want to continue through chapter one, which will be the hardest chapter because I don't have a formal background in probability.  Today I want to talk about integration with respect to Brownian motion, and integration with respect to martingales.  There are some sections on examples and Bessel processes.  These will be sections we need to revisit.  

Later chapters are on complex material that I understand and analytic stuff, and then a synthesis.  So we're shooting for getting through chapter 6, then we might be able to invite some speakers and understand what they're talking about.

\subsection{Integration with respect to Brownian motion}
For Lesbegue integration, you start with simple functions, which only take a finite number of values.  It's a linear combination of characteristic functions of measurable sets.  You take characteristic functions, integrating those is easy.  General functions are limits of simple functions, and measure theory is a bunch of theorems about limits.

Here $H,$ a simple process, is a simple function on the Wiener space, $H_s=\sum_1^n C_j 1_{[t_{j-1},\ldots,t_j)}(s).$  So here $C_j$ is a random variable.  I don't know exactly what that means.  Usually this will be a function, if it were a constant it would be like integrating on an interval.  So that would be like a brownian motion.  I'm reasonably certain that things we stick in here will be Brownian motion and things we derive from it.  

[This will be a sum of Brownian motion, local martingales, random variables.]

$C_j$ is a bounded $F_{t_{j-1}}$ random variable.  I understand that the thing in front should be computable at the beginning of the interval.  You can't wait until you're halfway through the interval to know enough to compute it.  For fixed $s$ this is constant locally.  This is really a function of $s$ and $\omega,$ where $s$ is time.

When you have a general process it should be the limit of such simple processes.  I'm not going to go through it in hopes that it will be analogous to the normal Lebesgue integration.

\subsection{Local Martingales}
Simple martingales, if you have two functions, the constant at zero and a step function with value $\pm k$ before and after $1/2,$ you had $\emptyset, [0,1]$ and for the other also $[0,1/2]$ and $[1/2,1].$  If you choose to integrate $f_1$ over the sigma algebra of $f_0$ they can't be distinguished from one another.  These are discrete, it's hard to imagine this with a continuous variable.  $E[H_t|F_s]=H_s.$  So if you forget that you've done Brownian motion from time $s$ to $t$ you get Brownian motion up to time $s.$

A local martingale, if there are a sequence of times $\tau_1<\tau_2\leadsto\infty$ such that for each $j,\ H_{t\wedge \tau_j}$ is a martingale (stops after time $\tau_j$).  I don't know of martingales that aren't local martingales, all the ones I think of have this property.

A classic example of a stopping time is to run a process in a domain, Brownian motion until you hit the boundary.  It depends on the path you take.  A local martingale is a generalization of a martingale.

One of the most famous discrete martingales is a random walk.  The measure space is infinite sequences of zeros and ones.  So $f_0$ is a constant, you're at the origin.  Next anything that is $.0x$ is to the left, and otherwise to the right.  Then you get the martingale above.  The sequence of these is a martingale.  One way to stop this is at $\pm 10.$  You stop each walk when it hits that.  So $\tau_10(x)$ says ``how long do you have to wait before you hit 10?''  For every $x$ there is a stopping time or there is never a gap of $10.$  For this particular one almost every $x$ will stop in finite time.  In the stopped interval checking that later functions are the same as earlier functions is easy.  All they require is one sequence of stopping times, not every stopping time.  You could also stop, say, at $-1,$ the first time you step left.  This seems to stop faster.  This is actually a very important martingale, analogous to positive harmonic functions.

We might tolk about $f$ holomorphic in the upper half plane, and we cut up the half plane into pieces.  Cut it up in a cantor-type way. and you get some sort of region around the zeros.  You give a rule ``keep doing what you're doing until you meet a criterion, then stop.''  So these are called stopping time regions.  Another common such rule is the square of the gradient, once you've grown enough mass you stop.

So I should have started with thot.  We define $Z_t$ as $\int_0^t H_sdB_s.$  If $H$ is just a characterstic function of $[t_0,t_1]$ multiplied by $c_1,$ this is $0$ for $t\le t_0,$ then $c_1[B_t-B_{t_0}]$ in the interval $(t_0,t_1)$ and then $c_1[B_{t_1}-B_{t_0}]$ after $t_1.$  For a constant you get Brownian motion back.  You just get faster Brownian motion as $c$ increases.  But what if you have $H_s=B_s?$  That would be more interesting.

Later on when we have processes which are limits, the integrals will be limits of these things. 

The things we're going to integrate will be continuous.  We approximate with these simple functions, but this is basically a Riemann integral, so that's pretty easy.  The passage of the limit from simple to continuous that's not very hard.

[You can always approximate a bounded process (in $L^2$) with simple processes.]

Proposition one is linearity of the integral.  Then they talk about quadratic variation.  So this is proposition 1.2 on page 12.  Suppose $H_s$ is simple.  I think this will be true for all processes by limiting later on.  Define $Z_t$ as above.  Define the quadratic variation $\langle Z\rangle_t= \int_0^t H^2_s ds.$  Then they prove something interesting: $Z^2_t-\langle Z\rangle_t$ is a martingale, 
$$
E[Z_t^2]=E[\langle Z\rangle_t]=\int_0^t E[H^2_S]ds.$$
One thing has a square outside the integral sign, the other has one inside one.  You can't pass a square through an integral.  But one is with respect to time, the other is with respect to Wiener space.  So it's not completely wrong.  The one depends on $t$ and $\omega$ while the other depends on, no, it also depends on $\omega.$

So suppose we take a process.  Take the random walk martingale.  If you square it it's no longer a martingale.  You square the second part and you get first step $0$ and second step $1.$  This does not recover this.  If you square it and subtract the quadratic variation, you will recover it.  In particular the original one has mean value zero.  The quadratic variation is giving the correct average, and the new thing will be a martingale again.  It's really easy to prove if you just move the integrals around.

Without getting into this, if you take a process and square it, you get a simple process, and you can define the quadratic variation, and you can show that the expected values of these two things are the same so their difference is a martingale.

I probably can't rely on the discrete approximation for the next proposition.  Keep the notation before, and $\Pi$ is a partition of time into $n$ pieces.  The limit as $n\to \infty,|\Pi|\to 0$ of \sum^n $(z_{t_j}-z_{t_{i-1}})^2$ is the quadratic variation.  The proof is pretty standard, first year measure theory, but I want to relate it to some things.  This brings us back to the interval.  Brownian motion has continuous paths.  In general, does $\int_0^t gdf$ make sense?  Does this symbol make sense?  If $f$ is differentiable we think of $df$ as $f'(x)dx.$  In real analysis we deal with this even if $f$ is not differentiable.  If $\sum |f(t_i)-f(i-1)|$ is bounded below $\infty$ for all partitions, you define the integral as the limit of $g(t_1)(f(t_i)-f(t_{i-1})).$  So the integration with respect to bounded variation functions, it turns out to correspond to positive measures.

Well, Brownian motion looks pretty ragged.  Is it bounded variation?  How far does it travel in $1/n?$  There's a probability that the variation will be about $1/\sqrt{n}.$  There's a probability of about $50\%.$  The total variation should be $n\frac{1}{\sqrt{n}}=\sqrt{n}$ so we don't expect this to be bounded variation.  If you square the difference, that looks more like it's on the order of $1.$

Not only is it finite for Brownian motion, if you run it up to time $t$ you get $t.$  We'll just talk about definitions, we're not going to prove any of this.

In general having the quadratic variation bounded is not enough to make the integral make sense.  If the variation is bounded you get a measure. But you might get a distribution.  It's like the derivative of something you can integrate against.

Almost all of analysis has to do with cancellation.  If you integrate something that integrates to zero with respect to a smooth function, you should get something close to zero.  Integrating with respect to Brownian motion should be like integrating with respect to a symmetric function.  So if the thing you're integrating is smooth you should get something close to zero.  

Part of the reason you can integrate with respect to $dB_s$ it's because of this properties.  You can see that maybe there's some hope for a theory of integration with respect to these martingales.

I'm not going to go into great detail on the next few pages.  They just involve passing to the limit from simple to continuous processes.  That's the content of the next couple of pages.  Then there is linearity of the integral again.

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