Chapter 3 - Brownian Motion

Scaled Random Walks

Symmetric Random Walk

Construct a symmetric random walk by repeatedly tossing a fair coin, (i.e., $p$ is probability of heads, and $q = 1 - p$ is probability of tails). Successive outcomes of the tosses,

$\omega$ is the infinite sequence of coin tosses and $\omega_n$ is outcome of $n$-th toss,

By default, $M_0 = 0$, thus, we define,

$M_k$ is the process that is what we call a symmetric random walk. Each toss makes it steps up one unit or down one unit and each of these two possibilities is equally likely.

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Properties of Random Walks

1. Independent increments, for $0 = k_0 < k_1 < \dots < k_m$,

1. Each of the random increments is a random variable that is,

which is an increment of the random walk, and is the change in position of the random walk between times $k_i$ and $k_{i + 1}$.

1. Non-overlapping time intervals are independent because they depend on different coin tosses
2. Each increment $M_{k_{i + 1}} - M_{k i}$ has expected value 0 and variance $k{i + 1} - k_i

1. Variance of a symmetric random walk accumulates are rate one per unit time so that variance of the increment over any time interval $k$ to $l$ for non-negative integers $k < l$ is $l - k$
2. A symmetric random walk is a martingale,

Line one results from separating, line 2 results from linearity of conditional expectations, line 3 is because $M_k$ only depends on $\mathcal{F}_k$ i.e., the first $k$ coin tosses by information and conditioning, and line 4 results from independence.

Quadratic Variation of Symmetric Random Walks Quadratic variation up to time $k$ is defined as,

and is computed path-by-path, by taking the sum of all the squares of all the one-step increments $M_j - M_{j - 1}$, which are either $1$ or $-1$, thus making $\sum_{j = 1}^k 1 = k$.

Note:

• $Var(M_k)$ is computed by taking the average over all paths, considering probabilities
• $[M, M]_k$ is computed along a single path, so up, down probabilities do not enter the computation
• i.e., $Var(M_k)$ is only computed theoretically whereas $[M, M]_k$ does not depend on the particular path chosen so can be computed along realized path explicitly

Scaled Symmetric Random Walk Approximate Brownian motion by speeding up time and scaling down step size of a symmetric random walk, by fixing $n \in \mathbb{Z}^+$, defining the scaled symmetric random walk as,

when $nt$ is an integer, otherwise, the above would be a linear interpolation between the values of the nearest two points (one to the left, one to the right). Like a random walk, a scaled random walk has independent increments for $0 = t_0 < t_1 < \dots < t_m$ and $nt_j$ is an integer, we can say that:

are independent increments. For $0 \le s \le t$, for $ns, nt$ are integers, we can say that:

since the increment is the sum of $n(t - s)$ independent random variables each with expected value 0 and variance $\frac{1}{n}$.

e.g., We have the following increment,

i.e., $n = 100$ and $(0.70 - 0.20) \times 100 = 50$ so we have 50 independent random variables each that takes value of $\pm \frac{1}{\sqrt{100}} = \pm \frac{1}{10}$ each of which has expected value 0 and variance $\frac{1}{100}$ so variance of the increment written above is,

Note: If $s, t$ are chosen so that $ns, nt$ are integers, then the first term on the RHS is independent of $\mathcal{F}(s)$ and $W^{(n)}(s)$ is $\mathcal{F}(s)$-measurable, depending only on the first $ns$ coin tosses, it proves,

for $0 \le s \le t$ such that $ns$ and $nt$ are integers.

Quadratic Variation of Scaled Random Walk For time $t \ge 0$, such that $nt$ is an integer, we can consider the quadratic variation of the scaled random walk, along the path, we evaluate the increment over each time step and square these increments before summing to get the length of the time interval over which we are doing the computation,

NOT an average over all possible paths, but obtains the same answer $t$ along all paths.

Limiting Distribution of Scaled Random Walk Fix time $t$, consider set of all possible paths at the time $t$ for a scaled random walk, and think about the scaled random walk corresponding to different values of $\omega$, sequence of tosses.

e.g., $t = 0.25$ and consider set of all possible values of $W^{(100)}(0.25) = \frac{1}{10} M_{25}$. Thus, $n = 100$ and we have $100 \times 0.25 = 25$ coin tosses, because $M_{25}$ can take any value of odd integer between $-25$ and $25$. So $W^{(100)}(0.25)$ can take any value:

For $W^{(100)}(0.25) = 0.1$ that is to equal 0.1, we must get 13 heads and 12 tails in 25 tosses,

From a plot, we can see that this information nearly is normal, with expected value zero and variance 0.25. Thus, we can get a good approximation for the function $g(x)$ that is continuous and models this distribution by multiplying $g(x)$ by the normal density and integrating,

Theorem: Central Limit $t \ge 0$, and as $n \rightarrow \infty$, the distribution of the scaled random walk $W^{(n)}(t)$ evaluated at $t$ converges to the normal distribution with mean zero and variance $t$.

Proof:

Log-Normal Distribution as Limit of Binomial Model Build a model for a stock price on the time interval $[0, t]$, by choosing $n \in \mathbb{Z}$ and creating a binomial for stock price that takes $n$ steps per unit time.

where $u_n$ and $d_n$ are the up and down factor, respectively. $\sigma$ is a positive constant, which is also the volatility of the limiting stock price process.

$nt$ is the sum of $H_{nt}$ and $T_{nt}$ the number of heads and tails in the first $nt$ coin tosses, respectively. We can write the following, given random walk $M_{nt}$,

the same can be done with the opposite sign for $T_{nt}$. For the model with $u_n$ and $d_n$ the stock price at $t$ is,

Theorem: $n \rightarrow \infty$ the distribution of $S_n(t)$,

converges to the distribution of,

where $W(t)$ is a normal random variable with mean 0 and variance $t$, $S(t)$ is log-normal. Generally, $ce^X$ for any random variable $X$ is normally distributed. For the above, $X = \sigma W(t) - \frac{1}{2} \sigma^2 t$ with mean $-\frac{1}{2} \sigma^2 t$ and variance $\sigma^2 t$.

The proof follows by using the Taylor expansion of $f(x) = \log (1 + x)$.

The distribution of $W^{(n)}(t) = \frac{1}{\sqrt{n}} M_{nt}$ converges to the distribution of a normal random variable with mean zero and variance $t$ a random variable we call $W(t)$. $W^{(n)}(t)$ is multiplied by a term that has $n$ in the denominator and this will have limit zero.

Brownian Motion Take the limit of scaled random walks $W^{(n)}(t)$ as $n \rightarrow \infty$. From this we can get the Brownian motion which inherits properties from the scaled random walk,

Definition: Brownian Motion For a probability space, for each $\omega \in \Omega$, suppose there is a continuous function $W(t)$ of $t \ge 0$ that satisfies $W(0) = 0$ and depends on $\omega$. Then $W(t)$ for $t \ge 0$ is a Brownian motion if for all $0 = t_0 < t_1 < \dots < t_m$ the increments,

are independent and each of these increments is normally distributed and has expected value 0 and variance $t_{i + 1} - t_i$.

• Brownian motion, unlike a scaled random walk, has no linear pieces
• Also, BMs are exactly normal for each $t$ as a consequence of the Central Limit theorem.
• Increments $W(t) - W(s)$ is normally distributed for all $0 \le s < t$.

Intuition

1. A random experiment is performed and the outcome is the path of the BM, so $W(t)$ is the value of the path at time $t$ and this value depends on which path resulted from the random experiment
2. $\omega$ is akin to the outcome of a sequence of coin tosses although now the coin is being toss “infinitely fast”; once the coin tosses are performed and the result $\omega$ is being obtained, then the path of the BM can be drawn. If tossed again, a different $\omega$ is obtained and a different path will be drawn For $\mathcal{F}$ is the $\sigma$-algebra of subsets of $\Omega$ whose probabilities are defined, $\mathbb{P}$ is the probability measure for which distributional statements are made about the Brownian Motion.

e.g., let ${\omega: 0 \le W^{(100)} (0.25) \le 0.2}$ be the set we are working in, Define $M_{25} = 10 W^{(100)} (0.25)$ must fall between 0 and 2 after 25 tosses. because

Since $M_{25}$ can only be an odd number, it falls between 0 and 2, it can only be equal to 1,

Thus, we must get 13 heads and 12 tails in the first 25 tosses. Thus, we can describe the probability of this set as the following,

Distribution of Brownian Motion As we said before, the increments of a Brownian motions are independent and normally distributed. Thus, the Brownian motion at specific time-steps are jointly normally distributed.

Since each $W(t_i)$ has mean 0, then the covariance for $0 \le s < t$, $W(s), W(t)$ is,

We define the $m$-dimensional covariance matrix for the brownian motion,

To get a moment-generating function for a zero-mean normal random variable with variance $t$ and the independence of the increments as specified previously,

Thus, the moment-generating function for a Brownian motion (i.e., the $m$-dimensional random vector $(W(t_1), W(t_2), \dots, W(t_m))$ is given by,

The distribution of the brownian increments can be specified by the specifying the joint-density or the joint-moment-generating function of the random variables $W(t_1), W(t_2), \dots, W(t_n)$.

Theorem: Characterizations of Brownian Motions We have a continuous function $W(t)$ for $W(0) = 0$, that depends on $\omega$, thus the following three properties are considered equivalent,

1. For all $0 = t_0 < t_1 < \dots < t_m$, the increments,

are independent and each is normally distributed with mean $0$ and variance $t_i - t_{i + 1}$

1. For all $0 = t_0 < t_1 < \dots < t_m$ the random variables $W(t_1), W(t_2), \dots, W(t_m)$ are jointly normally distributed with means equal to 0 and covariance matrix given previously
2. For all “”, the random variables “”, have jointly moment-generating function,

Filtration for Brownian Motion For a filtration $\mathcal{F}(t)$ for the Brownian Motion is a collection of $\sigma$-algebras $\mathcal{F}(t)$ satisfying,

1. Every set in $\mathcal{F}(s)$ is also in $\mathcal{F}(t)$ for $s < t$, thus there is at least as much information available at time later time $\mathcal{F}(t)$ as there is at the earlier time $\mathcal{F}(s)$.
2. Information available at time $t$ is sufficient to evaluate the Brownian motion at that time; for each $t \ge 0$, the BM $W(t)$ is $\mathcal{F}(t)$-measurable
3. The increment $W(u) - W(t)$ is independent of $\mathcal{F}(t)$, any increment of the BM after $t$ is independent of the information available at time $t$ $\Delta(t)$ is a stochastic process, and is adapted to $\mathcal{F}(t)$ if for each $t \ge 0$ the r.v. $\Delta(t)$ is $\mathcal{F}(t)$-measurable.

Two Possibilities of $\mathcal{F}(t)$ for BM

1. contains only info obtained by observing BM up to $t$
2. contains info obtained by observing BM and one or more other processes

Martingale Property for Brownian Motion Brownian motion is a martingale.

Proof: For $0 \le s < t$ then we have the following,

Quadratic Variation

We showed that the quadratic variation of a scaled random walk $W^{(n)}(t)$ up to time $T$ is $T$. For a BM, there is no natural step size, so if we are given $T > 0$ then we could choose a step size $\frac{T}{n}$ for some large $n$, and compute the quadratic variation up to time $T$ with this step size.

We want to evaluate this quantity but for small step sizes, so we take the limit of the above from $n \rightarrow \infty$. Thus, we also get $T$ as the quadratic variation. Paths of BM are unusual in that their quadratic variation is not zero making stochastic calculus different from ordinary.

First Order Variation Goal: Compute amount of up and down oscillation undergone by $f(t)$ between $[0, T]$, with down moves adding and up moves subtracting (counterintuitive, but this will be useful in the future).

For $f$,

where the middle term $-[f(t_2) - f(t_1)] = f(t_1) - f(t_2)$ guarantees that the magnitude of the down move of the function $f(t)$ between $t_1, t_2$ is added to rather than subtracted from total.

Take partition $\Pi = {t_0, t_1, \dots t_n}$ of $[0, T]$ which is a set of times, not necessarily equally spaces:

with max step size of the partition denoted:

Thus, first order variation, we take the limit as the number of $n$ partition points goes to infinity, and the longest subinterval $t_{j + 1} - t_j$ goes to zero.

Using MVT, we can show,

on interval $[t_j, t_{j + 1}]$ there is such a point $t_j^*$. Thus, somewhere between these two points, the tangent line is parallel to the chord. Thus, we can get:

Definition: Quadratic Variation The quadratic variation of $f$ up to time $T$ is given by,

where $\Pi$ and the partitions are defined like before.

Suppose $f$ has a continuous derivative, then

• The quadratic variation measures the accumulation of squared increments over finer and finer partitions of an interval
• If $f$ is continuously differentiable then its quadratic variation disappears, meaning it does not exhibit erratic or jumpy behaviour
• Thus, as we can see from the derivation above, the quadratic variation for a continuous-derivative function, is zero.
• Thus, we never consider quadratic variation in ordinary calculus;
• paths of brownian motion can not be differentiated with respect to the time variable

Theorem: $W$ is a Brownian Motion, then $W, W = T$ for all $T \ge 0$ almost surely.

Proof: For sampled quadratic variation corresponding to this partition, we have,

We show that this is a random variable and converges to $T$ as $\left\lVert \Pi\right\rVert \rightarrow 0$. Moreover, it has expected value $T$ and variance that converges to 0. Hence, it converges to expected value $T$ regardless of the path along which we are doing the computation.

The sampled quadratic variation is the sum of independent random variables, therefore its mean and variance are the sums of the means and variances of these random variables.

Which gives us the implication that,

Moreover,

Above, we saw that $\mathbb{E}[(W(t_{j + 1}) - W(t_j))^2] = t_{j + 1} - t_j$ and $Var[(W(t_{j + 1}) - W(t_j))^2] = 2(t_{j + 1} - t_j)^2$ so when $t_{j + 1} - t_j$ is small, the square of it is very small. Therefore, we can write,

Probabilistically, given the approximate equality above, we can write,

but is not in fact near one, no matter how small $t_{j + 1} - t_j$ is, so it is the square of the standard normal random variable given by,

and its distribution is the same no matter how small we make $t_{j + 1} - t_j$. If we choose large $n$ and $t_j = \frac{jT}{n}$ then for $t_{j + 1} - t_j = \frac{T}{n}$ for all $j$, and

Since $Y_1, Y_2, \dots, Y_n$ are iid, then by SLLN, we can see that,

thus, we can say that,

Thus we can write that,

meaning that on an interval $[0, T]$ the BM accumulates $T$ units of quadratic variation. Additionally, let’s take $0 < T_1 < T_2$, then on an interval $[T_1, T_2]$ the sum of the squared increments of BM for each of the subintervals in the partition gives:

Thus, the BM accumulates $T_2 - T_1$ units of QV over the interval $[T_1, T_2]$, and more generally, we can say that BM accumulates quadratic variation at rate one per unit time.

Cross Variation We can compute the cross variation of $W(t)$ with $t$ and the QV of $t$ with itself,

In the first line, we can see that the limit is 0 because,

Thus, we can capture the essence of this by writing,

Volatility of Geometric Brownian Motion For constants $\alpha, \sigma > 0$, we can define the Geometric Brownian Motion,

i.e., the asset-price model used in the Black-Scholes-Merton option-pricing formula. We show how to use the quadratic variation of BM to identify the volatility $\sigma$ from a path of this process.

Using $0 \le T_1 < T_2$ and defined $T_1 = t_0 < t_2 < \dots < t_m = T_2$ and observe the log returns,

over each subinterval $[t_j, t_{j + 1}]$. The sum of squares of the log returns is also called the realized volatility and is given by,

1. Term 1: is approximately equal to its limit which is $\sigma^2$ times the amount of QV accumulated by Brownian Motion on the interval $[T_1, T_2]$ which is $T_2 - T_1$.
2. Term 2: is the $\left(\alpha - \frac{1}{2}\sigma^2\right)^2$ times the QV of $t$ which is 0
3. Term 3: is $2 \sigma \left(\alpha - \frac{1}{2}\sigma^2 \right)$ times the cross variation of $W(t)$ and $t$ which is zero

When the maximum step size is small, then the RHS of the above is approx. equal to $\sigma^2(T_2 - T_1)$.

If $S(t)$ is a GBM with constant volatility $\sigma$ then $\sigma$ is identified from price observations by computing the LHS of the above and taking the root.

Markov Property Let $W(t)$ be a Brownian motion and let $\mathcal{F}(t)$ be a filtration for the Brownian motion, then $W(t)$ is a Markov process.

Proof $0 \le s \le t$ and let $f$ be a Borel-measurable function, then we define another Borel-measurable function $g$ such that,

where $W(t) - W(s)$ is independent of $\mathcal{F}(s)$ and the random variable $W(s)$ is $\mathcal{F}(s)$-measurable. Thus we compute the expectation on the RHS by replacing $W(s)$ with $x$ and hold it constant and take the unconditional expectation of the remaining random variable.

$W(t) - W(s)$ is normally distributed with mean 0 and variance $t - s$ meaning we can write,

Thus, we can now take $g(x)$ and replace $x$ with r.v. $W(s)$, let $\tau = t - s$ and $y = w + x$,

Thus, we define the transition density for Brownian Motion $p(\tau, x, y)$ to be,

so we can decompose and re-arrange this into,

• Conditioned on the information in $\mathcal{F}(s)$, the conditional density of $W(t)$ is $p(\tau, W(s), y)$
• aka, the density in variable $y$
• Density is normal with mean $W(s)$ and variance $\tau = t - s$.
• Information from $\mathcal{F}(s)$ is the only info relevant to value of $W(s)$
• $W(s)$ is relevant is essence of the Markov property

Theorem: Exponential Martingale Exponential martingale corresponding to $\sigma$, is given by

where $W(t)$ is a BM with filtration $\mathcal{F}(t)$, the process $Z(t)$ is a martingale.

Proof Let $0 \le s \le t$, thus we have that,

Given by the fact that we can take out what is known and independence to write the last line.

First Passage Time Define the first passage time to level $m$ as,

which is the first time the BM $W$ reaches the level $m$. If the BM never reaches level $m$, then we set $\tau_m = \infty$. A martingale stopped at a stopping time is still a martingale and thus must have constant expectation.

where $t \wedge \tau_m$ represents the minimum of $t$ and $\tau_m$. Thus, we assume that $\sigma > 0$ and $m > 0$, the BM is always at or below the level $m$ for $\tau \le \tau_m$ so $0 \le \exp {\sigma W(t \wedge \tau_m} \le e^{\sigma m}$. The extensive proof of the following theorem is redacted.

Theorem The first passage time of a BM to level $m$ is finite almost surely, and the LaPlace transform of its distribution is given by the following,

Reflection Principle Fix positive level $m$ and positive time $t$, and want to count the BM paths that reach $m$ at or before $t$. There are two such paths,

1. reach level $m$ prior to $t$ but at time $t$ are at some level $w$ below $m$
2. exceed level $m$ at time $t$ Other cases include ones that are exactly at level $m$ but is unlikely in the case of the Brownian motion and thus has probability 0.

We construct a reflect by switching the up and down moves of the Brownian motion from time $\tau_m$ onwards. The probability that a Brownian Motion ends at exactly $w$ or at exactly $2m - w$ is zero. Consider the paths that reach level $m$ prior to $t$ and are at or below level $w$ at time $t$. This leads to the key reflection equality given by,

Theorem: Random Variable $\tau_m$ The random variable $\tau_m$ has cumulative distribution function given by,

and density given by,

Maximum Date of Brownian Motion The maximum to date for BM to be:

which is used in pricing barrier options. For the value of $t$ the random variable $M(t)$ is indicated. For positive $m$ we have that $M(t) \ge m$ if and only if $\tau_m \le t$.

which can be written as the joint distribution of $W(t)$ and $M(t)$.

Theorem For $t > 0$ the joint density of $(M(t), W(t))$ is given by the following,