Distribution and Variance of a Brownian-Weighted Time Integral
Summary
The document derives the distribution of the time integral of a deterministic function multiplied by a Brownian motion. It uses integration by parts to rewrite the integral in terms of the Brownian value at the horizon and a stochastic integral whose integrand is the accumulated deterministic function. Because both terms are jointly Gaussian, their difference is Gaussian with mean zero.
It then computes the variance by combining the two individual variances and their covariance, with the cross term evaluated using Itô isometry. This gives a closed-form expression in terms of the accumulated function and the horizon, from which the higher moments follow from Gaussian moment formulas. The argument assumes the stated stochastic integral is well defined and uses standard Brownian motion properties. The source does not give a worked example for a particular weighting function, and the result would need adjustment if the process or its assumptions differed.
Key ideas
- Integration by parts rewrites the weighted Brownian integral using the accumulated deterministic function.
- The integral is Gaussian with zero mean under the stated Brownian motion assumptions.
- Its variance includes a covariance term between the terminal Brownian value and a stochastic integral.
- Itô isometry evaluates the covariance and supports the closed-form variance expression.
- Higher moments follow from the moments of a centered Gaussian distribution.
Tags
Full text
# Characterizing distribution of a stochastic intergal
# Characterizing distribution of a stochastic intergal
characterize the distribution of $\int_0^T f(t)Z_tdt$. In particular, verify that it is a Gaussian distribution and compute its moments.
## Answer by StackG (score 2)
https://quant.stackexchange.com/a/58630
It was shown in another question that \begin{align} \int^T_0 f(t) Z_t dt &= Z_T \int^T_0 f(u) du - \int^T_0 \Bigl( \int^s_0 f(u) du \Bigr) dZ_s\\ &= Z_T \cdot F(T) - \int^T_0 F(s) \cdot dZ_s \end{align} where I've defined the integral $F(T) = \int^T_0 f(t) dt$, and note this is just another deterministic function of time.
Both of the two terms in the solution are normally distributed with mean 0, so their difference must also be normally distributed with mean 0.
In order to calculate the variance of the combined quantity, we remember that: \begin{align} {\mathbb V}[A - B] &= {\mathbb E}[(A-B)^2] - {\mathbb E}[(A-B)]^2 \\ &= {\mathbb E}[A^2] + {\mathbb E}[B^2] - 2 \cdot {\mathbb E}[AB] \end{align} The first two terms are the variances of the two terms above, which are $F(T)^2 \cdot T$ and $\int_0^T F(t)^2 dt$ (as shown in this answer), which just leaves the remaining ${\mathbb E}[AB]$ term to calculate, for which we use the Ito Isometry: \begin{align} {\mathbb E}[AB] &= {\mathbb E}\Bigl[ F(T) \cdot Z_T \cdot \int_0^T F(t) dZ_t \Bigr] \\ &= {\mathbb E}\Bigl[ F(T) \cdot \int_0^T dZ_t \cdot \int_0^T F(t) dZ_t \Bigr] \\ &= F(T) \cdot \int_0^T F(t) dt \end{align}
so the variance is $F(T)^2 \cdot T + \int_0^T F(t)^2 dt - 2 \cdot F(T) \cdot \int_0^T F(t) dt$, and \begin{align} \int^T_0 f(t) Z_t dt \sim {\mathbb N}\Bigl( 0, F(T)^2 \cdot T + \int_0^T F(t)^2 dt - 2 \cdot F(T) \cdot \int_0^T F(t) dt \Bigr) \end{align}
Since this is gaussian in $T$, it is completely characterised by its mean and variance, and higher moments can be calculated in the usual way (odd moments are all 0, even moments as given in the table in the link)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.