How do I relate the std deviation of the step size, to the stdev of the endpoint of a brownian motion, if the step sizes are multiplied by a function

Question

I know that if I take take a brownian motion of, say, 30 steps of standard deviation 1, then the standard deviation of my endpoint will be sqrt(30). But what if the standard deviation of the 30 steps is defined by a function? How do I relate the endpoint standard deviation to that function?

An example in R, first with a standard fixed volatility random walk.

library(xts)
mysd <- 0.1 / sqrt(255) # small sd so that we don't walk too far
genwalk <- function(stepsizes) {i <- 1; for(x in stepsizes) i <- c(i, i[length(i)] * exp(x)); return(i)} # random walk function - geometric
acc <- matrix(0, 1000)
plot(genwalk(rnorm(30, 0, mysd)), type = "l", ylim = c(0.88, 1.12), col = "#0000001A")
for(x in 1:1000) {walk <- genwalk(rnorm(30, 0, mysd)); acc[x] <- last(walk); lines(walk, col = "#0000001A")}
title("1000 30-step random walks with step size 0.1/sqr(255)")

No surprises here:

> sd(acc)
[1] 0.03484051
> mysd * sqrt(30)
[1] 0.03429972
> title("1000 30-step random walks with step size 0.1/sqr(255)")
> library(moments)
> kurtosis(acc)
[1] 3.048236
> mean(acc)
[1] 1.000664

So I know how to get from a standard deviation of x to the standard deviation of a 30-step walk: multiple x by the square root of the number of steps.

But what if x is a function rather than a constant? How do I then relate the function to the endpoint standard deviation? Let's say my function is a decay weighting of length 30 with half life 10:

decay <- exp((-log(2) / 10) * 30:1) # exponential decay, 30 steps

> plot(genwalk(rnorm(30, 0, mysd) * decay), type = "l", ylim = c(0.88, 1.12), col = "#0000001A")
> for(x in 1:1000) {walk <- genwalk(rnorm(30, 0, mysd) * decay); acc[x] <- last(walk); lines(walk, col = "#0000001A")}
> title("Random walks with step size 0.1/sqr(255) * decay weighting")

> sd(acc)
[1] 0.01645629
> mysd
[1] 0.006262243
> kurtosis(acc)
[1] 2.882403
> mean(acc)
[1] 1.000137

So it looks like the endpoints are still nicely normally distributed (mean ~ 0, kurtosis ~ 3), but I have no clue how to relate the standard deviation of the endpoints, that is sd(acc) above = 0.1645629 to mysd which was the step size multiplier for the decay function.

Basically: how do I get from the decay function which was applied to the standard deviation of step sizes, to the standard deviation of the endpoints?

Answer 1

If I understand correctly, we seek the std of the following random variable:

$$ S := \sum_i X_i f(i) = \sum a_iX_i, $$ where $\mathbb{E}[X_i] = 0, \mathbb{E}[X_i^2] = \sigma^2$, where $\{X_i\}_{i=1}^n$ are independent and $a_i$ are non random. Calculate the variance:

$$ var(S) = var(\sum_i a_i X_i) = \sum_i var(a_iXi) = \sum a_i^2 \sigma^2 = \sigma^2\sum a_i^2, $$ where we've relied on the independence of $X_i$ in the second equality (see this). The standard deviation is thus $\sigma\sqrt{\sum_i a_i^2}$.

Please note that you can make the std of $a_iX_i$ be whatever you want by setting $a_i = \frac{\text{wahtever you want}}{\sigma}$.

Answer 2

You'll get what you want by typing this formula:

> mysd * sqrt(sum(decay^2))
[1] 0.01611228

Which is basically the answer @YahirDaon gave you. Let me abuse a little bit of the Brownian Motion definition in this answer.

You should note that this expression only works because, as you put it: "You dont walk too far away". Or, what is the same, your x vector does not differ significantly from zero, since mysd is small.

Because, in this case, your geometric brownian motion can be tought as $e^{\sum_1^nX_i} \approx \sum_1^nX_i$. (Use a Taylor expansion to see this). That is, regular Brownian Motion's formulae are a good approximation. In fact, your formula for variance $\sigma^2 * \sqrt{30}$ is that of a regular Brownian Motion, and only works here because of the same reason: x is a vector with a small norm.

The correct formula (as $\mu = 0$) is: $\sqrt{e^{30\sigma^2}.(e^{30\sigma^2}-1)}$. And you can check that it yields a very similar result:

sqrt(exp(30*mysd^2)*(exp(30*mysd^2)-1))
[1] 0.03433

But if you use a mysd $\gg 0$ both in the iid increments case and the varying ones (which technically isn't even a Brownian Motion because it is defined as having iid increments), then you should check more general Lèvy processes.

btw: Random Walk is not the same as Brownian Motion (the former is a more general concept). You seem to refer to a BM because of the formulae you use, so maybe you should avoid refering to it as Random Walk (which it is, but your formulas are specific to BM, not any RW).

btw2: these notes might help you: http://www.columbia.edu/~ks20/FE-Notes/4700-07-Notes-GBM.pdf