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I am doing some work with measuring the distance between distributions, and someone pointed out to me that I should look into calculating the integrated squared difference of two brownian bridge processes. Rather than re-invent the wheel, what I would like to ask is if there are any known results for the integrated squared difference of two brownian bridge processes (does it have a known distribution)? In particular,

$$\int_0^1(B_{1t} - B_{2t})^2dt$$

where $B_{1t}$ and $B_{2t}$ are independent brownian bridge processes.

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  • $\begingroup$ Are the two Brownian bridges independent? If yes, there difference is $\sqrt{2}$ times a Brownian bridge, so the distribution is the same as $2\int_0^1 B_t^2 dt$, where $B$ is a Brownian bridge. $\endgroup$
    – Christophe Leuridan
    Commented Nov 21, 2022 at 17:25
  • $\begingroup$ @ChristopheLeuridan the two bridge processes can be assumed to be independent. Is it straightforward to show what you stated, or is there literature on this? The natural follow-up question would be does the integral $\int_0^1B_t^2dt$ have a closed form (or known distribution)? $\endgroup$
    – John Smith
    Commented Nov 21, 2022 at 17:36
  • $\begingroup$ See projecteuclid.org/journals/annals-of-probability/volume-30/… $\endgroup$
    – Christophe Leuridan
    Commented May 19, 2023 at 12:10

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For every Gaussian centered process $X$, if one takes an independent copy $X'$, then the process $X+X'$ has the same distribution as $\sqrt{2}X$ (to see that, compare their finite dimensional marginals).

I found this reference https://projecteuclid.org/journals/annals-of-probability/volume-30/issue-1/On-the-Distribution-of-the-Square-Integral-of-the-Brownian/10.1214/aop/1020107767.full?tab=ArticleFirstPage

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  • $\begingroup$ Would you be able to elaborate on your solution, i.e., showing that it is indeed $\sqrt{2}B_t$. I know you give a reason above, but it still eludes me. Or rather, is a Gaussian centered process a brownian bridge? $\endgroup$
    – John Smith
    Commented May 19, 2023 at 4:21
  • $\begingroup$ If $X$ and $X'$ are independent Gaussian random variable with values in $\mathbb{R}^d$ and distribution $\mathcal{N}(0,C)$, the distribution of $X+X'$ (and also $X-X'$) is $\mathcal{N}(0,2C)$, so $X+X'$ (and also $X-X'$) has the same distribution as $\sqrt{2}X$. The same holds for centered Gaussian processes since their distribution is determined by finite-dimensional distributions. In particular, this holds for Brownian bridge, Brownian motion, Ornstein Uhlenbeck processes. $\endgroup$
    – Christophe Leuridan
    Commented May 19, 2023 at 12:12

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