TITLE: Law of the $L^2$ norm of a Brownian motion and related
QUESTION [6 upvotes]: Let $B_t$ be a Brownian motion with variance 1. We know that $\int_0^1 B(t) \mathrm{d} t \sim \mathcal{N}(0,1/3)$. I am interested to know what we can say about the law of the two random variables 
$X = \int_{0}^1 B(t)^2 \mathrm{d}t = \langle B,B\rangle_{L^2([0,1])}$ and $Y = \int_{0}^{1} \left( B(t) - \int_0^1 B(s)\mathrm{d}s \right)^2 \mathrm{d}t = X - \left( \int_0^1 B(s)\mathrm{d}s \right)^2$. 
Is there a well-known probability law hidden behind them? 
Thank you for attention.

REPLY [4 votes]: The distribution of Y has been studied in great details by Donati-Martin and Yor in their work: Fubini's theorem for double Wiener integrals and the variance of the Brownian motion path
http://archive.numdam.org/ARCHIVE/AIHPB/AIHPB_1991__27_2/AIHPB_1991__27_2_181_0/AIHPB_1991__27_2_181_0.pdf
In particular it is proved that in distribution
$Y=^{law}\int_0^1 | W_s |^2 ds$
where W is a standard complex Brownian bridge.