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What is expectation of $$\int_0^t B(s)^2ds$$ where $B(s) is standard Brownian motion. Is the integral a well known random variable?

Sida
  • 31

1 Answers1

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The expectations is easy to calculate, using Fubini's theorem, which applies since the integrand is positive: $$ \begin{align} E\left[\int_0^t B(s)^2 ds\right] = \int_0^tE[B(s)^2]ds = \int_0^ts\,ds = \frac{t^2}{2} \end{align} $$

Milind Hegde
  • 3,914