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I am trying to find the variance of $\int_0^t B_s^2 ds$ where $B_s$ is a standard Brownian motion random variable. My approach is to represent the integral as a sum. However, I am not sure how this works in finding a variance since I need a second moment. Is there another approach in doing this? thanks.

user321627
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    Tonelli is your friend here: $$X=\int_0^tB_s^2ds\implies E(X)=\int_0^tE(B_s^2),ds,\ \ E(X^2)=2\int_0^t\int_0^sE(B_s^2B_u^2),du,ds$$ ...And computing $E(B_s^2)$ and $E(B_s^2B_u^2)$ should not be a problem. – Did Mar 27 '16 at 23:23
  • Thanks! For the first implication, where you took the expectation inside the integral, is that by Tonelli's theorem? – user321627 Mar 28 '16 at 06:38
  • Yes, both expectations are computed using Tonelli. – Did Mar 28 '16 at 07:14
  • @Did I saw this post and really like your concise answer. Could you explain though how you got $E(X^2)$ above with the double integrals? I believe you broke the integral into parts with $\int_{s}^{t}B_u^2 du$ to take advantage of independent increments but don't know where the $2$ in front comes from. Thanks! – user136503 Apr 27 '16 at 17:00
  • @user136503 Decomposing the integral on $(0,t)\times(0,t)$ into the sum of two integrals on the triangles $0<u<s<t$ and $0<s<u<t$ and using the symmetry of the integrand $B_s^2B_u^2$, one gets $$X^2=\int_0^t\int_0^tB_s^2B_u^2duds=2\int_0^t\int_0^sB_s^2B_u^2duds.$$ – Did Apr 27 '16 at 17:08

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