# Integral of brownian increments

I'm stuck at a problem and I'm not sure on how to proceed. My question is how would one go about and integrate the following $$\sigma\int_{t}^{T}\mathrm{e}^{a\cdot u}\cdot (W_{u}-W_{t})du.$$ I've been stuck with this problem for quite some time. Thanks in advanced!

By the stationarity of the increments of Brownian motion, we have the following equality in law and an almost sure equality coming from a simple change of variable $$\sigma \int_t^T e^{au} \left(W_u - W_t\right)du \overset{law}{=} \sigma \int_t^T e^{au}W_{u-t}dt = \sigma \int_0^{T-t} e^{a\left(u + t\right)}W_{u}du = \sigma e^{at}\int_0^{T-t}e^{au}W_udu$$

To compute $$\int_0^{T-t}e^{au}W_udu$$,

#### Edit: The argument after this point contains key error. $$W_{T-t}$$ is not independent of $$\int_0^{T-t}e^{au}dW_u$$. It is still possible to compute $$\int_0^{T-t}e^{au}W_udu$$ by using the fact that it is a gaussian process. Thank you to LucaMac for the catch and see their answer for details.

we use integration by parts to get $$\int_0^{T-t}e^{au}W_udu = \frac{1}{a}e^{a\left(T-t\right)}W_{T-t} - \frac{1}{a}\int_0^{T-t}e^{au}dW_u$$ As a Wiener integral, $$\int_0^{T-t}e^{au}dW_u \sim \mathcal{N}\left(0, e^{2a\left(T-t\right)} - 1\right)$$ and is independent of $$W_{T-t}$$ . Furthermore $$e^{a\left(T-t\right)}W_{T-t} \sim \mathcal{N}\left(0, e^{2a\left(T-t\right)}\left(T-t\right)\right)$$, therefore we have $$\int_0^{T-t}e^{au}W_udu = \frac{1}{a}\left[e^{a\left(T-t\right)}W_{T-t} - \int_0^{T-t}e^{au}dW_u\right] \sim \mathcal{N}\left(0, \frac{e^{2a\left(T-t\right)}\left(T-t + 1\right) - 1}{a^2}\right).$$ Finally multiplying by the $$\sigma e^{at}$$ factor left off at the beginning we get that your integral has a normal distribution with mean 0 and variance $$\sigma^2e^{2at}\frac{e^{2a\left(T-t\right)}\left(T-t + 1\right) - 1}{a^2}$$:

$$\sigma \int_t^T e^{au} \left(W_u - W_t\right)du \sim \mathcal{N}\left(0, \sigma^2e^{2at}\frac{e^{2a\left(T-t\right)}\left(T-t + 1\right) - 1}{a^2}\right).$$

• Thank you a lot for the detailed explanation. I'm very grateful for your answer! Commented Dec 15, 2021 at 17:56
• Hello, very sorry to point this out, but they are not independent. Take $a=0$ for example Commented Dec 20, 2021 at 20:39
• @LucaMac Thank you for the catch! I sloppily applied independence of increments. Your computation of $\int_0^{T-t} e^{au} W_u du$ looks correct. Commented Dec 21, 2021 at 16:49

Once we get to the point of $$\int_0^{T-t} e^{au}W_udu$$ (thanks Shiva), we could conclude in the following way:

it is Gaussian since limit of Gaussians (the step-wise constant approximating functions are clearly Gaussians);

its mean is $$\mathbb E \Big[\int_0^{T-t} e^{au}W_udu\Big] = \int_0^{T-t} e^{au}\mathbb E[W_u]du = 0;$$

and its variance is $$\mathbb E\Big[\Big(\int_0^{T-t} e^{au}W_udu\Big)^2\Big] = \mathbb E\Big[\Big(\int_0^{T-t} e^{au}W_udu\Big)\Big(\int_0^{T-t} e^{av}W_vdv\Big)\Big] = \mathbb E\Big[\iint_{[0,T-t]^2}e^{a(u+v)}W_uW_vdudv\Big] = \iint_{[0,T-t]^2}e^{a(u+v)}\mathbb E[W_uW_v]dudv = \iint_{[0,T-t]^2}e^{a(u+v)}\min(u,v)dudv = 2\int_0^{T-t}\int_0^ue^{a(u+v)}vdvdu.$$

Let's stop for a second and evaluate $$\int_0^u e^{cv}vdv$$, we have $$\int_0^u e^{cv}vdv = \frac1ce^{cu}u - \frac1c\int_0^ue^{cv}dv = \frac{cue^{cu} - e^{cu} + 1}{c^2}.$$

This means that the variance is equal to $$\frac2{a^2}\int_0^{T-t}aue^{2au} -e^{2au}+ e^{au}du = \frac2a\int_0^{T-t}ue^{2au}du + \frac2{a^2}\int_0^{T-t}-e^{2au}+e^{au}du = \frac1{2a^3}(2a(T-t)e^{2a(T-t)}-e^{2a(T-t)}+1) - \frac1{a^3}(e^{2a(T-t)}-1)+\frac2{a^3}(e^{a(T-t)}-1) = \frac{2a(T-t)e^{2a(T-t)}-3e^{2a(T-t)}+4e^{a(T-t)}-1}{2a^3}$$

• I didn't check all you calculations but I agree with you in your main point. Commented Dec 21, 2021 at 10:42
• I eventually go at the same results as you by using Stochastic Fubini's Theorem. The only thing where we differ is in your final result for the variance is that you have a $-5$ in the numerator, where I have $-1$. Other than that I agree with your addition to Shiva's response. Commented Dec 21, 2021 at 18:40
• Very sorry Marc, but the $-5$ looks correct to me (take the limit $a\to 0$ to see that it needs to be $-5$) Commented Dec 22, 2021 at 14:06
• I'm pretty sure I'm right. Again, correct me if I'm wrong, but I'll leave my calculations in an answer to my main question, since I cannot write it here :) Commented Dec 23, 2021 at 22:19
• you are right, the should be a missing "-" sign in my calculations somewhere Commented Dec 24, 2021 at 0:30

This is a response to Luca's answer. \begin{align} \frac{2}{a^{2}}\int_{0}^{T-t} au\mathrm{e}^{2au} - \mathrm{e}^{2au} + \mathrm{e}^{au} du &= \frac{2}{a^{2}} \left(\int_{0}^{T-t} au\mathrm{e}^{2au}du - \int_{0}^{T-t}\mathrm{e}^{2au}du + \int_{0}^{T-t}\mathrm{e}^{au} du\right)\\ &= \frac{2}{a^{2}}\left(\left[\frac{(2au - 1)\mathrm{e}^{2au}}{4a}\right]_{0}^{T-t} - \left[\frac{\mathrm{e}^{2au}}{2a}\right]_{0}^{T-t} + \left[\frac{\mathrm{e}^{au}}{a}\right]_{0}^{T-t}\right)\\ &= \frac{2}{a^{2}}\left(\frac{(2a(T-t) - 1)\mathrm{e}^{2a(T-t)} + 1}{4a} - \frac{\mathrm{e}^{2a(T-t)}-1}{2a} + \frac{\mathrm{e}^{a(T-t)} - 1}{a}\right)\\ &= \frac{2}{a^{2}}\left(\frac{(2a(T-t) - 1)\mathrm{e}^{2a(T-t)} + 1}{4a} - \frac{2\mathrm{e}^{2a(T-t)}-2}{4a} + \frac{4\mathrm{e}^{a(T-t)} - 4}{4a}\right)\\ &= \frac{2}{a^{2}}\left(\frac{(2a(T-t) - 1)\mathrm{e}^{2a(T-t)} + 1 - 2\mathrm{e}^{2a(T-t)} + 2 + 4\mathrm{e}^{a(T-t)} - 4}{4a}\right)\\ &= \frac{2}{a^{2}}\left(\frac{2a(T-t)\mathrm{e}^{2a(T-t)} - \mathrm{e}^{2a(T-t)} - 2\mathrm{e}^{2a(T-t)} + 4\mathrm{e}^{a(T-t)} - 1}{4a}\right)\\ &= \frac{2}{a^{2}}\left(\frac{2a(T-t)\mathrm{e}^{2a(T-t)} - 3\mathrm{e}^{2a(T-t)} + 4\mathrm{e}^{a(T-t)} - 1}{4a}\right)\\ &= \frac{2a(T-t)\mathrm{e}^{2a(T-t)} - 3\mathrm{e}^{2a(T-t)} + 4\mathrm{e}^{a(T-t)} - 1}{2a^{3}}\\ \end{align}