Distribution of an Exponentially Weighted Brownian Integral

# Integral of brownian increments


# 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!

## Answer by Shiva (score 5, accepted)

https://quant.stackexchange.com/a/69094

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 $\sigma^2e^{2at}\frac{e^{2a\left(T-t\right)}\left(T-t + 1\right) - 1}{a^2}$:

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

## Answer by LucaMac (score 5)

https://quant.stackexchange.com/a/69177

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 $$\frac2{a^2}\int_0^{T-t}aue^{2au} -e^{2au}+ e^{au}du = \frac2a\int_0^{T-t}ue^{2au}du + \frac2{a^2}\int_0^{T-t}-e^{2au}+e^{au}du = \frac1{2a^3}(2a(T-t)e^{2a(T-t)}-e^{2a(T-t)}+1) - \frac1{a^3}(e^{2a(T-t)}-1)+\frac2{a^3}(e^{a(T-t)}-1) = \frac{2a(T-t)e^{2a(T-t)}-3e^{2a(T-t)}+4e^{a(T-t)}-1}{2a^3}$$

## Answer by Marc Allan (score 0)

https://quant.stackexchange.com/a/69228

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