Distribution of $\int_0^t s^2 B(s) \, ds$ where $B$ is a standard Brownian motion

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I guess $\int_0^t s^2 B(s) \, ds$ is a normal distribution as $\int_0^t B(s) \, ds$ but I don't know how to argue that.

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From $$ \frac{t^3}{3}B_t=\int_0^ts^2\,B_s\,ds+\int_0^t\frac{s^3}{3}\,dB_s $$ it follows that $$ \int_0^ts^2\,B_s\,ds=\int_0^t\frac{t^3-s^3}{3}\,dB_s $$ which is Gaussian with mean zero and variance $$ \int_0^t\frac{(t^3-s^3)^2}{9}\,ds\,. $$ Can you proceed?