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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.

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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.

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  • $\begingroup$ Thank you very much for you quick response! $\endgroup$ – Eva Facchini Jan 15 at 16:22

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