Applying the Central Limit Theorem to a Scaled Random Walk

Summary

The document asks how the Central Limit Theorem can establish that a scaled symmetric random walk converges at a fixed time to a normal distribution with mean zero and variance equal to that time. The concern is that the sequence of random-walk values across scaling levels is not independent, even though each value has the same distributional role in the limit statement.

The answer resolves this by focusing on the construction within each walk: at a fixed time, the value is a sum of symmetric Bernoulli increments scaled by the square root of the number of steps. Those increments are independent and identically distributed, so the classical central limit theorem applies to the sum as the number of steps grows. The relevant independence is among the increments making up each walk, not among the sequence of walk values across different scaling levels. The exchange provides a brief explanation rather than a full derivation, and the stated convergence concerns a fixed time, not a proof of convergence of the entire process.

Key ideas

• At fixed time, the scaled random walk can be expressed as a sum of scaled symmetric increments.
• The increments within each walk are independent and identically distributed.
• The classical central limit theorem applies to that sum as the step count increases.
• Independence across different scaling levels is not required for this argument.
• The explanation addresses convergence at a fixed time, not full process convergence.

Tags

# How to apply CLT on scaled symmetric random walk--Shreve unclear


# How to apply CLT on scaled symmetric random walk--Shreve unclear












"Theorem 3.2.1 (Central limit)" in the book "Stochastic Calculus for Finance II Continuous-Time Models" by Steven Shreve says:

> Theorem. Fix $t\geq0$. As $n\to \infty$, the distribution of the scaled random walk $W^{(n)}(t)$ evaluated at time $t$ converges to the normal distribution with mean zero and variance $t$.

A "proof" is given there, without using the general Central Limit Theorem.

However, my question is the following: Suppose we wish to simply apply the CLT, how can we even do that here?

For a fixed $t$, the sequence $W^{(n)}(t) \quad (n= 1,2,3, \ldots)$ is not i.i.d. (in fact, they are of course identically distributed, but they are not independent!). Therefore I can't see how we can deduce this using the CLT.

## Answer by Alex Lapanowski (score 1)

https://quant.stackexchange.com/a/77864

If I recall correctly, Shreve defines $W^{(n)}(t)$ as constructed from increments: $$ W^{(n)}(t):=\sum_{i=1}^{tn}\frac{1}{\sqrt{n}} S_{i} $$ where $S_{i}$ are symmetric Bernoulli random variables.

When expressed as a scaled sum of i.i.d. increments, the CLT applies and we can deduce the distribution of Brownian Motion as the limit of $W^{(n)}(t)$.

## Answer by Rylan (score 0)

https://quant.stackexchange.com/a/76288

Using the phrasing of the central limit from Wolfram, you can think of it as the central limit theorem being applied to the (scaled) $X_i$ "coin tosses" that make up $W^n(t)$.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.