Deriving the Brownian Integral Over a Subinterval
Summary
The document derives the time integral of Brownian motion from s to t using the product rule for uBᵤ. The resulting expression contains both a term proportional to the Brownian value at the starting time and a stochastic integral over the interval. This distinction corrects a commonly stated formula that omits the contribution from the path before s.
The answer checks the result by expressing the integral from zero to T and subtracting the corresponding expression up to t. It also notes that the two terms in the subinterval representation are independent, and gives the resulting variance. The discussion is a mathematical derivation rather than a trading method; its conclusion assumes standard Brownian motion and the stated integration setup.
Key ideas
- Applying the product rule to uBᵤ yields an expression for the integrated process.
- The integral from s to t includes a starting-value term, (t−s)Bₛ.
- The remaining contribution is a stochastic integral weighted by t−u over the interval.
- The Brownian integral from t to T alone does not represent the time integral of Wₛ over that interval.
- Independence of the two terms gives the stated variance.
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# Integrated Brownian motion
# Integrated Brownian motion
I occasionally see a post here: Integral of brownian motion wrt. time over [t;T].
This post has the conclusion that $\int_t^T W_s ds = \int_t^T (T-s)dB_s$.
However, here is my derivation which is different from him and I don't know where I am wrong.
First, we have:
$$\mathrm{d}(uB_u) = u \mathrm{d} B_u + B_u \mathrm{d} u.$$
Then, integral from $s$ to $t$, we have:
$$tB_t - sB_s = \int_{s}^{t} u \mathrm{~d}B_u + \int_{s}^{t} B_u \mathrm{~d} u$$
\begin{equation} \begin{aligned} \int_{s}^{t} B_u \mathrm{~d}u &= tB_t - sB_s - \int_{s}^{t} u \mathrm{~d}B_u\\ &= \int_{0}^{t} t \mathrm{~d}B_u - \int_{0}^{s} s \mathrm{~d} B_u - \int_{s}^{t} u \mathrm{~d} B_u\\ &= \int_{0}^{s} t \mathrm{~d} B_u + \int_{s}^{t} t \mathrm{~d}B_u - \int_{0}^{s} s \mathrm{~d}B_u - \int_{s}^{t} u \mathrm{~d}B_u\\ &= \int_{0}^{s}(t - s)\mathrm{~d}B_u + \int_{s}^{t} (t - u) \mathrm{~d} B_u\\ &= (t - s)B_s + \int_{s}^{t} (t - u) \mathrm{~d}B_u. \end{aligned} \end{equation}
## Answer by Kurt G. (score 3)
https://quant.stackexchange.com/a/75919
Your derivation is correct. Even if we fix your obvious typo the formula $$ \textstyle\int_t^TW_s\,ds=\int_t^T(T-s)\,dW_s $$ is wrong.
There is no doubt that \begin{align} TW_T&=\textstyle\int_0^Ts\,dW_s+\int_0^TW_s\,ds\,,\\[2mm] TW_T&=\textstyle\int_0^TT\,dW_s\, \end{align} hold. Therefore, $$\boxed{\phantom{\Bigg|}\quad \textstyle\int_0^TW_s\,ds=\int_0^T(T-s)\,dW_s\,.\quad}\tag{1} $$ Subtracting two such expressions yields \begin{align} \textstyle\int_t^TW_s\,ds&=\textstyle\int_0^T(T-s)\,dW_s-\int_0^t(\color{red}{t}-s)\,dW_s\\[2mm] &=\textstyle\int_t^T(T-s)\,dW_s+\int_0^t(\color{red}{T}-s)\,dW_s-\int_0^t(\color{red}{t}-s)\,dW_s\tag{2}\\[2mm] &=\textstyle\int_t^T(T-s)\,dW_s+\int_0^t(\color{red}{T}-\color{red}t)\,dW_s\\ &=\textstyle\int_t^T(T-s)\,dW_s+(\color{red}{T}-\color{red}t)\,W_t\,.\tag{3} \end{align} Which is what you have shown.
Someone in that other post must have thought that the last two terms in (2) cancel.
The two terms in (3) are independent due to the independence of the increments of BM. Therefore, the variance of $\int_t^TW_s\,ds$ should be $$ \frac{(T-t)^3}{3}+(T-t)^2t=(T-t)^2\frac{T+2t}{3}\,. $$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.