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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.

namasikanam
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  • $X=\int_{0}^{t} s^2 B(s) , \mathrm{d}s$ is a "linear combination" of jointly normal variables, hence is again normal. In fact, since each $B(s)$ is centered, $X$ is also centered. So you only need to figure out the variance of $X$ in order to completely specify the distribution of $X$. – Sangchul Lee Dec 27 '22 at 15:02
  • @SangchulLee I understand a linear combination of jointly normal variables is again normal. But an integration is defined as a limitation. Why this limitation of these normal random variables is still normal? – namasikanam Dec 27 '22 at 15:11
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    It is a well-known fact that the limit in distribution of normal distributions is again normal. I believe the proof can be found in many textbooks, or even in this community (for instance, see this). – Sangchul Lee Dec 28 '22 at 01:40

1 Answers1

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Too long for a comment.

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?

Kurt G.
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