Brownian Motion
Integrated Brownian Motion
Let \(W_t\) be a standard Wiener process and define
\[
X_t=\int_0^t W_\tau\,d\tau.
\]
We want the distribution of \(X_t\) and whether \((X_t)_{t\ge0}\) is a martingale.
Distribution
\(X_t\) is Gaussian because it is a linear functional of a Gaussian process.
\[
X_t\sim \mathcal N\!\left(0,\frac{t^3}{3}\right)
\]
Interview shortcut: linear combination/integral of Gaussian random variables ⇒ Gaussian. Then compute only mean and variance.
Step 1
Mean
\[
\begin{aligned}
\mathbb E[X_t]
&=\mathbb E\!\left[\int_0^t W_\tau\,d\tau\right] \\
&=\int_0^t \mathbb E[W_\tau]\,d\tau \\
&=0.
\end{aligned}
\]
Step 2
Variance
Because the mean is zero,
\[
\begin{aligned}
\operatorname{Var}(X_t)
&=\mathbb E[X_t^2] \\
&=\int_0^t\int_0^t
\mathbb E[W_sW_u]\,ds\,du.
\end{aligned}
\]
For Brownian motion,
\[
\mathbb E[W_sW_u]=\operatorname{Cov}(W_s,W_u)=\min(s,u).
\]
\(\min(s,u)=s\)
\(\min(s,u)=u\)
\(s\)
\(u\)
Split the square along \(s=u\). By symmetry,
\[
\operatorname{Var}(X_t)
=2\int_0^t\int_0^u s\,ds\,du.
\]
Inner integral:
\[
\int_0^u s\,ds=\frac{u^2}{2}.
\]
Therefore,
\[
2\int_0^t\frac{u^2}{2}\,du
=\int_0^t u^2\,du
=\frac{t^3}{3}.
\]
Hence
\[
\boxed{X_t\sim\mathcal N\!\left(0,\frac{t^3}{3}\right)}.
\]
Step 3
Is \(X_t\) a martingale?
For \(s
\[
X_t
=\int_0^s W_\tau\,d\tau+\int_s^t W_\tau\,d\tau
=X_s+\int_s^t W_\tau\,d\tau.
\]
For \(\tau>s\), write
\[
W_\tau=W_s+(W_\tau-W_s).
\]
The increment \(W_\tau-W_s\) is independent of \(\mathcal F_s\) and has mean zero, so
\[
\mathbb E[W_\tau\mid\mathcal F_s]=W_s.
\]
Therefore,
\[
\begin{aligned}
\mathbb E[X_t\mid\mathcal F_s]
&=X_s+\int_s^t
\mathbb E[W_\tau\mid\mathcal F_s]\,d\tau \\
&=X_s+\int_s^t W_s\,d\tau \\
&=X_s+(t-s)W_s.
\end{aligned}
\]
Since
\[
X_s+(t-s)W_s\neq X_s
\]
in general, \(X_t\) does not satisfy the martingale property.
Final result
\[
\boxed{X_t=\int_0^t W_\tau\,d\tau
\sim \mathcal N\!\left(0,\frac{t^3}{3}\right)}
\]
\[
\boxed{(X_t)_{t\ge0}\text{ is not a martingale.}}
\]
The key reason is that future Brownian levels have conditional mean \(W_s\), not \(0\).