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Let $$dr_t=adt+\eta dW_t$$ so $$r_t=at+\eta W_t$$ so $$r_t \sim \mathcal{N} (at, \eta).$$ $$\int_0^t r_s= \int_0^t(as+\eta W_s)dt=a\frac{t^2}{2}+\eta \int_0^t W_s ds$$

I can write $$\int_0^t W_s ds=W_tt-\int_0^ts dW_s=\int_0^t(t-s)dW_s$$

This is a brownian motion with mean $0$ and variance $\frac{t^3}{3}$. Could i say that $X_t=\int_0^t r_s ds \sim \mathcal{N}(a\frac{t^2}{2},\eta \frac{t^3}{3}) ?$ If this is true what about $-X_t$? It is $-X_t \sim \mathcal{N}(-a\frac{t^2}{2},\eta \frac{t^3}{3})?$

Thanks to all.

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