Scaling Brownian Integrals: Equality in Distribution Versus Pathwise Equality
Summary
The document examines whether the time integral of Brownian motion over [0,T] can be obtained by scaling the integral over [0,1]. A change of variables gives an expression involving the process at times sT, but Brownian scaling does not make those values equal to the same sample path at times s multiplied by √T. The distinction is between equality in distribution and equality for each realization.
For T greater than one, the integral depends on Brownian path values between 1 and T, which cannot be inferred from the path on [0,1]. The response therefore rejects the proposed pathwise identity while noting that Brownian scaling gives the two sides the same distribution. The discussion is a conceptual correction rather than a trading method; its relevance is to stochastic modeling, where confusing distributional scaling with pathwise equality can invalidate an argument. It does not develop broader applications or provide empirical evidence.
Key ideas
- Brownian scaling relates distributions of rescaled paths, not necessarily values on the same realization.
- The integral over [0,T] depends on path values throughout that interval.
- The proposed identity can hold in distribution without holding almost surely path by path.
Tags
Full text
# the order of integral of Brownian motion
# the order of integral of Brownian motion
When we want to obtain the order of $\int_{0}^{T} B_{t} d t$, we can use the scale property of Brownian motion.
Let $B$ be a Brownian motion. Is the order of $\int_{0}^{T} B_{t} d t$ correctly calculated:
$$\int_{0}^{T} B_{t} d t=\int_{0}^{1} B_{s T} \cdot T d s=\int_{0}^{1} T^{1/2} \cdot T B_{s} d s=T^{1+1/2} \int_{0}^{1} B_{s} d s$$
## Answer by fes (score 1)
https://quant.stackexchange.com/a/60079
This (in particular the 2nd equality) is incorrect. Let $\omega \in \Omega$ denote the sample realization. You state:
$$\int_{0}^{T} B_{t} (\omega) d t=T^{1+1/2} \int_{0}^{1} B_{s}(\omega) d s$$
For example assume $T>1$. This is claiming that in order to compute the integral you only need to consider realizations of $B_t$ between $[0,1]$ and then scale them up. However, the value of this integral also depends on the random realizations of $B_t$ between $[1,T]$, which cannot be deduced simply from its values between $[0,1]$. The LHS can take a value clearly different from the RHS depending on $\omega$. Intuitively if $W_t$ is e.g. a stock price, the integral represents a type of weighted average of the stock price over $[0,T]$. But you cannot calculate this average merely based on the stock price over $[0,1]$.
The probability that $\int_{1}^{T} B_{s}(\omega) d s$ takes any particular fixed value is zero (this integral is normally distributed). Hence the LHS and RHS are almost surely not equal. However, as @Kevin pointed out the LHS and RHS are equal in distribution that is these two random variables have the same moments.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.