Variance Accumulation in a Symmetric Random Walk
Summary
The document explains what it means for the variance of a symmetric random walk to accumulate at a rate of one per time step. Each increment represents a fair coin bet paying either one unit or minus one unit, so its expected value is zero and its variance is one. Since the increments are independent, adding a step increases the cumulative process variance by one.
Starting from the first step, this yields variance equal to the number of steps. Over any interval from step k to step l, the increment has variance equal to the number of steps in that interval. Thus, the stated accumulation rate describes the growth of the cumulative process's squared dispersion, not a rate measured in the original payoff units. The explanation assumes the fair, independent, symmetric walk described in the question and does not address more general random walks with drift or dependent increments.
Key ideas
- Each fair symmetric step has mean zero and variance one.
- Independent increments make cumulative variance increase by one per step.
- The cumulative walk has variance equal to the number of steps taken.
- Variance over an interval equals the number of increments in that interval.
- The interpretation relies on the symmetric independent-step setup.
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# Accumulation Rate of Variance in Random Walk
# Accumulation Rate of Variance in Random Walk
I am slightly confused with the terminology Shreve (2008), he states:
> "The variance of the symmetric random walk accumulates at rate one per unit time, so that the variance of the increment over any time interval $k$ to $l$ for nonnegative integerst $k<l$ is $l-k$."
I understand the latter half of the statement, but I don't understand the variance's accumulation rate. This is something not familar to me. The way I thought of variance's unit is that it is squared, so it is usually not easily interpreted.
The context in which this symmetric random walk is defined is think of a random variable that pays you \$1 if a coin comes up heads and -\$1 otherwise. Consider a cumulative gain or loss of this dollar bet at $k$-th bet, and we call this process a symmetric random walk.
So what does he mean here about the rate at which the variance accumulates? The rate at which the dispersion of the cumulative bet accumulates? Confused.
Reference: Shreve, Steven E. \textit{Stochastic Calculus for Finance II : Continuous-Time Models}. Springer, 2008.
## Answer by Cettt (score 1, accepted)
https://quant.stackexchange.com/a/49802
If we denote the random walk with $(X_k)_{k \in \mathbb{N}}$ than for all $k$ the random variable $\Delta X_k := X_{k} - X_{k-1}$ has mean zero and variance one: \begin{align} \mathbb{E}[\Delta X_k] = \frac 12\cdot 1 + \frac 12 \cdot (-1) = 0, \quad \text{Var}(\Delta X_k) = \mathbb{E}[(\Delta X_k)^2] - \bigl(\mathbb{E}[\Delta X_k]\bigr)^2 = 1 \end{align}
This is what is meant by the second part.
The first part deals with the variance of $X_k$ which can be computed like this: \begin{align} \text{Var}(X_k) &= \mathbb{E}\Bigl[\text{Var}(X_k \mid X_{k-1}) \Bigr] + \text{Var}\Bigl(\mathbb{E}[X_k \mid X_{k-1}] \Bigr) = 1 + \text{Var}(X_{k-1}), \\ \text{Var}(X_1) &= \mathbb{E}[(X_1)^2] - \bigl(\mathbb{E}[X_1]\bigr)^2 =1. \end{align}
Therefore $\text{Var}(X_k) = k$ for all $k \in \mathbb{N}$. In particular, the variance of the symmetric random walk accumulates at rate one per unit time.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.