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“All about X, part 1” is really a fancy way of saying a little bit about X.

How do you eat an elephant?
One bite at a time ¹

So let’s eat stochastic difference equations one bite at a time. Let’s consider a first-order linear equation driven by a Gaussian random variable.

$$\begin{align} x_{n+1}&=ax_n+w_n \\ w_n &\sim N(0, \sigma^2) \\ |a| & < 1. \\ \end{align}$$

Let’s dissect this one bit at a time. $x_n$ is our system’s state, or value, at time $n$. Our time variable $n$ takes on integer values, -2, -1, 0, 1, 2, and so on. $w_n$ is our noise term, and it is a normally distributed Gaussian variable with mean 0 and standard deviation $\sigma$ at each time $n$. The values at different times are uncorrelated. We require $|a|<1$ so that our equation is stable, or does not blow up.
What does this series look like? Here is a sample run of 5000 steps for a=0.9967 and $\sigma=$0.082. The significance of these numbers will be revealed later.

Let’s derive some basic properties of our stochastic series $x_n$. Recall that $E()$ is the expectation operator, meaning it computes the expected value of its random variable argument.

$\newcommand{\Var}{\mathrm{Var}}$
$\newcommand{\Cov}{\mathrm{Cov}}$
$\newcommand{\Expect}{{\rm I\kern-.3em E}}$

What is the average (expected) value of $x_n$? Since our process is stationary, that is, a is fixed, and $w_n$ does not varying it’s statistics over time, $\Expect(x_n)$ is constant, let’s call it $\Expect(x)$. Then

$$\begin{align} \Expect(x_{n+1})&=a\Expect(x_n)+\Expect(w_n) \\ \Expect(x)(1-a)&=\Expect(w_n) \\ \Expect(x)&=0 \end{align}$$
where we used $E(w_n)=0$ for the final step.

What about the variance of x?
$$\begin{align} \Var(x_{n+1})&=a^2\Var(x_n)+\Var(w_n) \\ \Var(x)(1-a^2)&=\sigma^2 \\ \Var(x)&=\frac{\sigma^2}{1-a^2} \end{align}$$
$$\begin{align} \Cov(x_{n+1},x_n)&=\Expect[(x_{n+1}-\Expect(x_{n+1}))(x_{n}-\Expect(x_{n}))]\\ &=\Expect[x_{n+1}x_{n}]\\ &=\Expect[(ax_n+w_n)x_n]\\ &=\Expect[ax_n^2]\\ &=\frac{a\sigma^2}{1-a^2} \end{align}$$

Let’s look at the increments:
$$\begin{align} \Delta x_{n+1}&=x_{n+1}-x_n\\ &=(a-1)x_n+w_n\\ \Var(\Delta x)&=(a-1)^2\Var(x)+\sigma^2\\ \Var(\Delta x)&=\sigma^2[\frac{(1-a)^2}{1-a^2}+1]\\ &=\sigma^2\frac{2}{1+a}. \end{align}$$
Note If a is close to 1, then the variance of the increments is close to $\sigma^2$.

Now, let’s get a little more involved and compute the variance over k steps. From some basic system theory (or inspection):
$$\begin{align} x_k=x_0 a^k + \sum_{s=0}^{k-1}a^{k-s-1}w_s \end{align}$$
To make sure we have the right limits, let’s check for k=1:
$$x_1=a x_0 + w_0$$
That looks right, let’s continue on.
$$\begin{align} \Var(x_k-x_0)=(a^k-1)^2\Var(x)+\sigma^2\sum_{s=0}^{k-1}(a^{k-s-1})^2 \end{align}$$
Now, using the Sum of a Geometric Series, we have
$$\begin{align} \Var(x_k-x_0)&=\frac{(a^k-1)^2}{1-a^2}\sigma^2 + \frac{1-a^{2k}}{1-a^2}\\ &=\sigma^2 [ \frac{(a^k-1)^2 + 1 – a^{2k}}{1-a^2}\\ &=2 \sigma^2 \frac{1-a^k}{1-a^2}. \end{align}$$
Checking when $k=1$ and we see it matches our earlier result.

In Part 2 we will look more closely at some simulations and trading results on these series.

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What is the sum of the following geometric series?
$$\sum_{i=0}^{k}a^i=1+a+a^2+…+a^k$$
We will frequently need a simple formula for this finite series. It is called a geometric series because each term is related by a multiple to the previous one. If each term was related by a fixed difference, it would be called an arithmetic series; but that’s for another day.
Let us define the sum as $S$. Then writing the equation for $S$ and $aS$ with some clever alignment:

$$\begin{array}{rrrccc} S=&1&+a&+a^2&+…&+a^k \\ aS=&&a&+a^2&+…&+a^k&+a^{k+1} \end{array}$$
Subtract the equations
$$S(1-a)=1-a^{k+1}$$
or
$$\begin{align} S= \begin{cases} \frac{1-a^{k+1}}{1-a} &a \neq 1 \\ k+1 &\text{otherwise} \end{cases} \end{align}$$
where we have to be careful to divide by $1-a$ only if $a\neq 1$, and the answer for $a=1$ is determined by inspection.
Q.E.D.

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You may have seen the following equation:
$$\begin{align} \lim_{n \rightarrow \infty} \left( 1 +\frac{1}{n} \right)^n = \mathrm{e}. \end{align}$$

which expresses the fundamental constant e as a limit. Wondering how that can be proven?

Consider this graph:

One key fact we will use is

$$\begin{align} \int_1^{x} \, \frac{1}{t} \mathrm{d}t = \ln x \end{align}$$

Looking at the graph, you can see that the area under the $\frac{1}{t}$ curve on the interval $(1,1+\frac{1}{n})$ is bounded by

$$\begin{align} \frac{1}{n}\left( \frac{1}{1+\frac{1}{n}} \right) \leq \int_1^{1+\frac{1}{n}} \, \frac{1}{t} \mathrm{d}t \leq \frac{1}{n} \left( 1 \right) \end{align}$$

Using the definition of $\ln x$ above, we have

$$\begin{align} \frac{1}{1+n} \leq \ln \left( 1+\frac{1}{n} \right) \leq \frac{1}{n} \end{align}$$

Raising $\mathrm{e}$ to the power of each term,

$$\begin{align} \mathrm{e}^{(\frac{1}{1+n})} \leq 1+\frac{1}{n} \leq \mathrm{e}^{(\frac{1}{n})} \end{align}$$

Raising each term to power $n$,
$$\begin{align} \mathrm{e}^{(\frac{n}{1+n})} \leq \left( 1+\frac{1}{n} \right)^n \leq \mathrm{e} \end{align}$$

As n approaches infinity, the left hand term approaches $\mathrm{e}$, therefore
$$\begin{align} \lim_{n \rightarrow \infty} \left( 1 +\frac{1}{n} \right)^n = \mathrm{e}. \end{align}$$

QED.