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}


1 Answer 1


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 $$ \frac{(T-t)^3}{3}+(T-t)^2t=(T-t)^2\frac{T+2t}{3}\,. $$


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