Why Squared Wiener Increments Accumulate into Elapsed Time
Summary
The document explains the stochastic-calculus identity that the square of a Wiener-process increment behaves like a time increment in differential notation. Its argument defines the accumulated squared increments over an interval as the limit of sums across progressively finer partitions. Each increment is represented as the square root of the partition width multiplied by an independent standard normal draw.
The response shows that the sum has an expectation equal to the interval length and a variance that shrinks to zero as the partition is refined. This supports the conclusion that the limiting sum equals elapsed time, giving the quadratic variation of the Wiener process. The identity is a rule of stochastic calculus, not a claim that every individual random increment squared is deterministically equal to a time step. The explanation is an introductory derivation and assumes the standard Wiener-process properties, including independent Gaussian increments.
Key ideas
- A Wiener-process increment over a short interval has variance equal to the interval length.
- Define the accumulated squared increments through sums over finer partitions.
- The sums have an expected value equal to the total interval length.
- Their variance vanishes as the partition is refined, so the limit is elapsed time.
- The differential identity describes quadratic variation and does not equate each individual random square with a deterministic increment.
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# Stochastic Differential
# Stochastic Differential
Let $W_t$ be a Wiener process. It is clear to me that $dW_t$ is of size $\sqrt{dt}$. This can be seen because $$ \mathrm{Var}(W_{t+\Delta} - W_{t})=\Delta. $$ But am I allowed to actually write $(dW_t)^2 = dt$? It looks a bit silly... you have the square of a random variable on the left hand side, and a deterministic variable on the right hand side.
Please can you clarify whether this is right or wrong and perhaps give an explanation for this peculiar identity.
## Answer by Amiro (score 8, accepted)
https://quant.stackexchange.com/a/15327
I can clarify 100% that $(dw)^2$= $dt$ and recommend you to accept it as a fact.
Like any other differential, this differential is defined in terms of its integral: $$ \int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\sum_{k=0}^{n-1}[W(t_{k+1})-W(t_{k})]^{2} $$ Where $t_{k}=t_{0}+k(t_{1}-t_{0})/n$. Since $$ W(t_{k+1})-W(t_{k})=\sqrt{t_{k+1}-t_{k}}\xi_{k}=\sqrt{\frac{t_{1}-t_{0}}{n}}\xi_{k} $$ We have $$ \int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}\frac{t_{1}-t_{0}}{n}\sum_{k=0}^{n-1}\xi_{k}^{2} $$ where $\xi_{0}, \xi_{1},$ . . $\xi_{n-1}$ are independent $N(0,1)$ variables. Clearly the mean ofthe sum is $$ E[\frac{t_{1}-t_{0}}{n}\sum_{k=0}^{n-1}\xi_{k}^{2}]=\frac{t_{1}-t_{0}}{n}\sum_{k=0}^{n-1}E[\xi_{k}^{2}]=t_{1}-t_{0} $$ Since the $\xi$'s are independent, the variance ofthe sum is $$ Var[\frac{t_{1}-t_{0}}{n}\sum_{k=0}^{n-1}\xi_{k}^{2}]=\frac{(t_{1}-t_{0})^{2}}{n^{2}}\sum_{k=0}^{n-1}Var[\xi_{k}^{2}]=\frac{(t_{1}-t_{0})^{2}}{n^{2}}\sum_{k=0}^{n-1}E[(\xi_{k}^{2}-1)^{2}] $$ For unit Gaussian variables, $E[(\xi_{k}^{2}-1)^{2}]=2$, so the variance ofthe sum works out to $$ Var[\frac{t_{1}-t_{0}}{n}\sum_{k=0}^{n-1}\xi_{k}^{2}]=\frac{2}{n}(t_{1}-t_{0})^{2} $$ Thus $$ \int_{t_{0}}^{t_{1}}(dW)^{2}\equiv\lim_{n\rightarrow\infty}S_{n} $$ where the sum $S_{n}$ has mean $t_{1}-t_{0}$ and variance $O(1/n)$ . We conclude that in the limit
$ n\rightarrow\infty$, this integral is $t_{1}-t_{0}$ with certainty. Thus $$ \int_{t_{0}}^{t_{1}}(dW)^{2}=t_{1}-t_{0} $$
For any $t_0$ and $t_1$. Since differentials are defined only in terms of their integral, we can rewrite it as
$(dw)^2 = dt$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.