Proving a Brownian Motion Integral Is Jointly Gaussian
Summary
The document shows how to establish that Brownian motion at a fixed time and its time integral have a joint normal distribution. The key step is integration by parts: the integral of Brownian motion can be rewritten as a stochastic integral with a deterministic weighting function. Any linear combination of the two variables is then a single stochastic integral against Brownian motion, which is normally distributed; this verifies joint Gaussianity through the linear-combination criterion.
The argument is concise and gives a direct alternative to solving for a characteristic function. It relies on standard properties of Brownian stochastic integrals and applies at a fixed time. The discussion does not calculate the joint mean, covariance matrix, or extend the result to more general processes, so those details would need separate derivation for applications.
Key ideas
- Integration by parts rewrites the time integral of Brownian motion as a weighted stochastic integral.
- Every linear combination of Brownian motion and its time integral is itself a Brownian stochastic integral.
- A stochastic integral with a deterministic integrand is normally distributed.
- Checking that every linear combination is normal establishes joint Gaussianity.
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Full text
# Show that $(W_t, \int_0^t W_s ds)$ has a normal joint distribution
# Show that $(W_t, \int_0^t W_s ds)$ has a normal joint distribution
I have to show that, if $W_t$ is a 1-d Brownian motion then $\biggl(W_t, \int_0^t W_s ds\biggr)$ has normal distribution. Hint: apply Ito formula to this bivariate process. Any idea or suggestion on how to solve it?
I tried to show that with characteristic function approach, since the marginal distributions have both normal distribution
If I want to show that the couple is bivariate gaussian I have to prove that: $$\forall \lambda_1 , \lambda_2 \in \mathbb{R}: Z_t=\lambda_1 W_t + \lambda_2 \int_0^t W_s ds \ \text{ is normal} $$
$dZ_t = \lambda_1dW_t+\lambda_2W_tdt$ and if I compute $\phi_{Z_t}(\eta)$ and $d(\exp{i\eta Z_t})$, then in the end I get: $$\phi_{Z_t}(\eta)=\int_0^t \mathbb{E}(\exp{(i\eta Z_s)} \cdot i\eta\lambda_2 W_s)ds-\int_0^t \mathbb{E}(\phi_{Z_s}) \cdot \eta^2 \cdot 1/2 \cdot ds$$ and I don't know how to solve the first integral.
## Answer by Gordon (score 7, accepted)
https://quant.stackexchange.com/a/43514
Note that \begin{align*} \int_0^t W_s ds &= tW_t -\int_0^t sdW_s \tag{1}\\ &= \int_0^t (t-s)dW_s. \end{align*} Then, for $\lambda_1, \lambda_2 \in \mathbb{R}$, \begin{align*} \lambda_1 W_t + \lambda_2 \int_0^t W_s ds &= \lambda_1\int_0^t dW_s + \lambda_2 \int_0^t (t-s)dW_s\\ &=\int_0^t \big(\lambda_1 + \lambda_2(t-s)\big)dW_s, \end{align*} which is normal.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.