Integral of square of the Brownian Motion with respect to time

Question

Is there anything known about the distribution of: $$ \int_0^1 B_t^2 \; dt \; ?$$ I know that without squaring, the distribution of the above variable is normal with mean $0$ and variance $\frac{1}{3}$ (https://quant.stackexchange.com/questions/29504/integral-of-brownian-motion-w-r-t-time). But if underlying process is nonnegative, the distribution obviously cannot be normal and from simulation I can only conclude that it has mean $\frac{1}{2}$ (which is actually easy to prove) and variance $\frac{1}{3}$. But the exact distribution is unknown to me. Is there any source of information about such integral?

Answer 1

You have the right mean and variance for

$$ \int_0^1 B_t^2 \; dt \; ?$$

But, its distribution is not of any known functions. In fact, it is rather complex. You may find some research results in the link below.

https://projecteuclid.org/download/pdf_1/euclid.aop/1020107767

Answer 2

You can actually compute the integral with the formula $d(tY_t) = Y_t \, dt + t \, dY_t$. Fixing $Y_t = B_t^2$, you have by Itô formula that $dY_t = 2B_t dB_t + dt$ and therefore \begin{align} \int_0^1 B_t^2 d t &= \int_0^1 d(tB_t^2) - \int_0^1 td(B_t^2)\\ &= B_1^2 - \int_0^12tB_tdB_t - \int_0^1tdt\\\ &= B_1^2 - \frac12 - \int_0^12tB_tdB_t. \end{align} With this expression (and Itô isometry) you can easily compute the second moment, and have a good idea of the distribution of your object.

For the variance, let $$ Z = \int_0^1 B_t^2 d t. $$ Then, since $E(B_1^4) = 3$, applying the Itô isometry and the properties of the Itô integral (zero mean) we get \begin{align} E (Z^2) &= E\left(\left(B_1^2 - \frac12 - \int_0^12tB_tdB_t\right)^2\right) \\ &= E(B_1^4) + \frac14 + E\left(\left(\int_0^12tB_tdB_t\right)^2\right) - E (B_1^2)\\\ &\quad -2E\left(B_1^2\int_0^12tB_tdB_t\right) + E\left(\int_0^12tB_tdB_t\right) \\ &= 3 + \frac14 + \int_0^14t^2E(B_t^2)dt -1 - \frac83+ 0\\ &=2+\frac14+\int_0^14t^3 dt-\frac83\\ &= 3+\frac14-\frac83 = \frac13 + \frac14, \end{align} where we used the fact that $$ E\left(B_1^2\int_0^12tB_tdB_t\right) = \frac43 $$ Therefore, since $E(Z) = 1/2$ (easy to compute) $$ \mathrm{Var}(Z) = E(Z^2) - E(Z)^2 = \frac13. $$

Answer 3

This question has been asked before in this forum. My answer in Determining distribution of $X_t = \int_0^t W_s^2 \mathrm{d} s$ gives a couple of references to the original solution of Cameron & Martin, and a different approach of M. Kac.