Introduction

Suppose you play a game. We model the outcome of the game with the random variable $$X$$ where the probability of winning or losing is $$p_X(1)=p_X(-1)=1/2$$. So it is a fair game and $$\mathbb{E}(X)=0$$.
Suppose now that we play multiple times. We denote by $$X_i$$ the outcome of the $$i$$th game. Suppose we bet a piece of gold each time we play. So after the first round we have $$X_1$$ goldpieces. After the second one we would have $$X_1+X_2$$.
More generally, after the $$t$$-th round we would have \begin{align*} Y_t=\sum_{i=1}^t X_i \end{align*} pieces of gold.
The expected value of $$Y_i$$ in the $$(i-1)$$th round, i.e., after we know $$Y_1$$, …, $$Y_{i-1}$$ would be $$Y_{i-1}$$. Since if we lose, we would have $$Y_{i-1}-1$$ and if we would win, we would have $$Y_{i-1}+1$$.

In this example, the family of random variables $$Y_i$$ is called a simple random walk. This motivates the following more general definition.

Definition

A family of random variables $$Y_0$$, $$Y_1$$, … is called a discrete martingale if $$\mathbb{E}(Y_i)<\infty$$ for all $$i$$ and for the conditional expected value it holds that $$\mathbb{E}(Y_{n+1}|Y_0,\dots,Y_n)=Y_n$$.

Note that the first assumption basically says that the expected value does exist.

Towards continuity

So what is a continuous martingale? Let us try to generalize our random walk. Let $$Y_i$$ be as before a discrete random walk. Now define $$B_{j/N}=\frac{1}{\sqrt{N}} Y_j$$.
We need a scaling factor, such that it works. This has to do with bounding the variance and i do not want to explain it. Note that if we let $$N\rightarrow\infty$$ then this stuff converges to something which i will denote by $$B_t$$. It is also called Brownian motion, therefore the symbol $$B$$. Please also note that $$t\mapsto B_t$$ is almost surely a continuous function. Furthermore because of the scaling we have that $$B_\lambda-B_\mu \sim\mathcal{N}(0,\lambda-\mu)$$ is normally distributed with mean $$0$$ and variance $$\lambda-\mu$$, the differences between any two increments of Brownian motion are independent and $$B_0=0$$.

Definition

Let $$Y_i$$, $$i\geq 0$$ be a family of random variables, $$\mathcal{F}_t=\{Y_i,\,0\leq i\leq t\}$$ a filtration. Then $$Y_i$$, $$i\geq 0$$ is called a martingale if $$\mathbb{E}(Y_i)<\infty$$ for every $$i\leq 0$$ and furthermore $$\mathbb{E}(Y_i|\mathcal{F}_t)=Y_t$$ for every $$i>t$$.

Integrals

Let us integrate this Brownian motion stuff. I mean, integrate as in Riemann integration. You know, we just take a discrete integral of the random walk and then do this limit stuff as explained above and then get an integral of Brownian motion.
Since Brownian motion is a stochastic process, the integral will also be a random variable. Therefore we can characterize it by computing its distribution. Obviously $$I=\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{i=1}^N B_{i/N}$$ should be the integral $$I=\int_0^1 B_t dt$$, if it exists and if the last part of that notation even makes sense. We have to hack around a bit, since only Brownian increments are independent.
\begin{align*} \sum_{i=1}^N Y_i=NX_1+(N-1)X_2+...+X_N=\sum_{i=1}^N (N-i+1)X_i\sim\mathcal{N}\left(0,\frac{1}{N}\sum_{i=1}^N (N-i+1)^2\right) \end{align*}
Where the last part follows, since the $$X_i$$ are independent. Moreover $$\frac{1}{N}\sum_{i=1}^N (N-i+1)^2=\frac{(N+1)(2N+1)}{6}$$ as one can show by induction or a search in your favorite collection of mystic formulae.
\begin{align*} I_n=\frac{1}{N}\sum_{i=1}^N B_{i/N}\sim\mathcal{N}\left(0,\frac{(N+1)(2N+1)}{6N^2}\right)=\mathcal{N}\left(0,\frac{1}{3}+\frac{1}{2N}+\frac{1}{6N^2}\right) \end{align*}
Letting $$N\rightarrow\infty$$ we get $$\int_0^1 B_t dt=I\sim\mathcal{N}(0,\frac{1}{3})$$.
We just integrated Brownian motion, bitches.

25 March 2015

Math