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I need an help with the following: let $(B_t)_t$ a Brownian motion. Compute the distribution of $X:=\int_0^1 B_t dt$. Integrating by parts we have that: $$\int_0^1 B_t dt=B_1-\int_0^1 t dB_t.$$ Now, I know that $B_1\sim N(0,1)$, $\int_0^1 t dB_t\sim N(0,1/3)$, if they were independent, I could say that $X\sim N(0,4/3)$.

But is it true that $B_1, \int_0^1 t dB_t$ are independent? Why? Is this the right way to solve the exercise?

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1 Answer 1

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Since $B_1 = \int_{0}^{1}\mathrm dB_t$, we have $$ \int_{0}^{1}B_t\ \mathrm dt = \int_{0}^{1}(1-t)\ \mathrm dB_t, $$ the limit of sums of independent gaussian random variables, with distribution $\mathcal N(0, \int_{0}^{1}(1-t)^2\ \mathrm dt)$.

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