Scaling a Symmetric Random Walk Toward Brownian Motion

# Intuition behind Scaling Symmetric Random Walk


# Intuition behind Scaling Symmetric Random Walk












I am reading a section in Shreve (2008) where we are scaling down the step size but speeding up the time a symmetric random walk, so that in the limit, we produce a Brownian motion.

I understand the process, but I want to understand the intuition with the $1 bet story set-up.

Is the following intuition correct?

Consider a $1 bet on coin toss where if Heads you win a dollar otherwise you lose a dollar. The cumulative gain on this random variable is a symmetric random walk. We want to speed up the time and scale down the size such that

$$W^{(n)}(t)=\frac{1}{\sqrt{n}}M_{nt}.$$

For example, consider $t=4,n=100.$ Without the scaling, you would flip 4 times, but with the scaling you are flipping 400 coins within 4 seconds. Similarly, for each flip, your step up or down would be $1 as that is how the bet is defined, but with the scaling, your bet now becomes 10 cents.

Reference: Shreve, Steven E. $\textit{Stochastic Calculus for Finance II : Continuous-Time Models}$. Springer, 2008.

## Answer by Magic is in the chain (score 1, accepted)

https://quant.stackexchange.com/a/49864

It is easier if you interpret t as time in years. So let’s say t=4 years.

And the rest is easier if you recall the end result, we want this scaled random walk to approach the standard brownian, which has mean zero and variance t (seeing it as interval from time 0 to t=4).

We are repeating independent and identical tossing game, where the coin is unbiased. Now to get the desired mean and variance, the bet size has to be related to the number of steps. For one step, meaning step size of 4 years, if we set the bet size equal to 2, then the mean will be zero and variance will be 4, as desired. This 2 is related to the step size: $\Delta t=\frac{4}{1}$, which in general terms, assuming m represents the number of steps is $\Delta t=\frac{t}{m}$. So the bet size is square root of $\Delta t$.

Now if you increase the number of steps to say 100, with same t=4, then the bet size would be: $\sqrt{\Delta t}=\sqrt{\frac{t}{m}} =\sqrt{\frac{4}{100}}=\sqrt{\frac{1}{25}}$. The mean is then zero because each of the games has mean zero, and the variance of the sum of independent and identical games is equal to the sum of the variances, which because of homogeneity is $100*\frac{1}{25}=4$

In Shreve’s settings, each unit of t is subdivided into n steps, so our m=n*t; his t=4 and n=1, is equivalent to m=4, you will toss the coin 4 times, each time setting the bet size equal to $\sqrt{\frac{t}{m}}=1$. For t=4, n=100, you have m=400, and the bet size would be $\sqrt{\frac{t}{m}}=\sqrt{\frac{4}{400}}=0.1$