# Finding distribution of $\int_0 ^T uW_u du$

I would like to find the distribution of $\int_0 ^T uW_u du$ where $(W_u)_{u\geq0}$ is the Brownian motion.

What I have tried:

$$\int_0 ^T uW_u du = \int_0 ^T B_udu - \int_0^T \int_0^tB_sdsdt$$ by integration by parts. I know each term on the RHS is normal as any integration of a Gaussian process is again a Gaussian process.

However, I cannot conclude RHS is normals since I don't have independence of $\int_0 ^T B_udu$ and $\int_0^T \int_0^tB_sdsdt$ since $Cov(\int_0 ^T B_udu, \int_0^T \int_0^tB_sdsdt)=T^4 /8 \neq 0$

Any help is appreciated.

• In what types of settings does this sde come up? – David Addison May 8 '18 at 15:32
• @DavidAddison its just a homework type question – izimath May 8 '18 at 15:33

Using the Ito Formula

The general approach that often works for these kinds of question is to search for functions such that their Ito differential contains the terms that we are interested in. In your case, we are looking for a function $$f(t, x)$$ such that $$f_t(t, x) = t x$$. Let

$$\begin{equation} f(t, x) = \frac{1}{2} t^2 x \end{equation}$$

with

$$\begin{equation} f_t(t, x) = t x, \qquad f_x(t, x) = \frac{1}{2} t^2, \qquad f_{xx}(t, x) = 0. \end{equation}$$

Applying the Ito formula yields

$$\begin{equation} \frac{1}{2} T^2 W_T = \int_0^T u W_u \mathrm{d}u + \frac{1}{2} \int_0^T u^2 \mathrm{d}W_u \end{equation}$$

or

$$\begin{equation} \int_0^T u W_u \mathrm{d}u = \frac{1}{2} \int_0^T \left( T^2 - u^2 \right) \mathrm{d}W_u. \end{equation}$$

The Ito integral of a deterministic integrand is normally distributed with zero mean and variance

$$\begin{equation} \text{Var} \left( \int_0^T (T^2 - u^2) \mathrm{d}W_u \right) = \int_0^T \left( T^2 - u^2 \right)^2 \mathrm{d}u = \frac{8}{15} T^5 \end{equation}$$

and we conclude that

$$\begin{equation} \int_0^T u W_u \mathrm{d}u \sim \mathcal{N} \left( 0, \frac{2}{15} T^5 \right). \end{equation}$$

Discretizing the Integral

Alternatively, we could write

$$\begin{equation} \int_0^T u W_u \mathrm{d}u = \lim_{n \rightarrow \infty} X_n, \qquad X_n = \sum_{i = 1}^n \underbrace{t_i W_{t_i}}_{Y_i} \Delta_n, \end{equation}$$

where $$\Delta_n = T / n$$ and $$t_i = i \Delta_n$$. Then, each $$Y_i$$ is normal with mean zero and covariance

$$\begin{equation} \text{Cov} \left( Y_i, Y_j \right) = t_i t_j \min \left\{ t_i, t_j \right\} = i j \min \{ i, j \} \Delta_n^3 . \end{equation}$$

Letting $$\bar{Y}_n$$ be the corresponding column-vector with elements $$\left( Y_1, Y_2, \ldots, Y_n \right)'$$, we get in matrix form

$$\begin{equation} \bar{\Sigma}_n = \mathbb{E} \left[ \bar{Y}_n \bar{Y}_n' \right] = \Delta_n^3 \left[ \begin{array}{c c c c} 1 & 2 & \dots & n\\ 2 & 8 & \dots & 4 n\\ 3 & 12 & \ddots & \vdots\\ n & 4 n & \dots & n^3 \end{array} \right] \end{equation}$$

As the weighted sum of normally distributed random variables is itself normally distributed, it follows that $$X_n \sim \mathcal{N} \left( 0, \Delta_n \bar{1}_n \bar{\Sigma}_n \bar{1}_n' \Delta_n \right)$$, where $$\bar{1}_n$$ is an $$n$$-dimensional column vector of ones. We have

$$\begin{eqnarray} \text{Var} \left( X_n \right) & = & \Delta_n^5 \sum_{i = 1}^n \left( i \sum_{j = 1}^i j^2 + i^2 \sum_{j = i + 1}^n j \right)\\ & = & \Delta_n^5 \left( \sum_{i = 1}^n \left( \frac{1}{3} i^4 + \mathcal{O} \left( i^3 \right) + i^2 \left( \frac{1}{2} n^2 - \frac{1}{2} i^2 + \mathcal{O}(i) + \mathcal{O}(n) \right) \right) \right)\\ & = & \Delta_n^5 \left( \frac{1}{15} n^5 + \frac{1}{6} n^5 - \frac{1}{10} n^5 + \mathcal{O}\left( n^4 \right) \right) \\ & = & \Delta_n^5 \left( \frac{2}{15} n^5 + \mathcal{O}\left( n^4 \right) \right) \end{eqnarray}$$

Note that $$\Delta_n^5 = \mathcal{O} \left( n^{-5} \right)$$. Thus, we have

$$\begin{equation} \lim_{n \rightarrow \infty} \text{Var} \left( X_n \right) = \frac{2}{15} T^5, \end{equation}$$

just like before.

Another approach consists in using the Fubini theorem to write that \begin{align} \int_0^T u W_u du &= \int_0^T \int_0^u u\, dW_v\, du \tag{$W_u = \int_0^u dW_v$} \\ &= \int_0^T \int_v^T u\, du\, dW_v \tag{Fubini}\\ &= \frac{1}{2}\int_0^T (T^2 - v^2) dW_v \end{align} This is an Itô integral. Since the integrand is deterministic it is Gaussian distributed with zero mean and variance given by the Itô isometry $$\int_0^T u W_u du \sim N\left(0, \frac{1}{4}\int_0^T (T^2 - v^2)^2 dv \right)$$

First, you should apply the integration by parts:

$\int_0^T uW_udu = \frac{u^2}{2}W_u \mid_0^T - \int_0^T\frac{u^2}{2}dW_u$, where both parts are normally distributed. The variance of the first summand is $(\frac{T^2}{2})^2T = \frac{T^5}{4}$ by definition, the variance of the second summand can be found by Ito Isometry: $Var \left( \int_0^T\frac{u^2}{2}dW_u \right) = E \left(\int_0^T\frac{u^2}{2}dW_u \right)^2 = \int_0^TE\frac{u^4}{4}du = \frac{T^5}{20}$. These two terms are correlated so we have to calculate the co-variance (again using Ito Isometry):

$Cov(\frac{T^2}{2}W_T, \int_0^T\frac{u^2}{2}dW_u) = \frac{T^2}{2}E(W_T\int_0^T\frac{u^2}{2}dW_u) = \frac{T^2}{2}E(\int_0^TdW_u\int_0^T\frac{u^2}{2}dW_u) = \\ \frac{T^2}{2}\int_0^T\frac{u^2}{2}du = \frac{T^5}{12}$

So, as $Var(X-Y) = Var(X) + Var(Y) - 2Cov(X,Y)$, we obtain that:

$Var(\int_0^T uW_udu) = \frac{T^5}{4} + \frac{T^5}{20} - \frac{T^5}{6} = \frac{2T^5}{15}$,

so as the expectation of Ito integral is 0, $\int_0^T uW_udu$ ~ $N(0, \frac{2T^5}{15})$

• There are a few mistakes on the variance: the two terms on the right hand side are correlated and the variance of the first term is not $T$ as there is coefficient. However, the whole idea is correct, in particular, the integration by parts. Upvoted. – Gordon May 8 '18 at 14:11
• @Gordon, yes, sure. I didn't mentioned the covariance because the author of the question already knows how to calculate it. And yes, got a mistake in variance of the first term,sorry! – Anton Tsches May 8 '18 at 14:44
• You may edit your answer to correct it. – Gordon May 8 '18 at 14:45