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.


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 \(B_0=0\).


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 \(i>t\).


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