The Brownian bridge is a stochastic process defined as a Brownian motion with the condition that it comes back to the origin at time $t=1$. A Brownian bridge $X$ can be obtained from a Brownian motion using $$X_t=B_t-tB_1.\tag1$$ This definition extends to any final time $T$ and any final point $Y$ in any dimension $d$ using the same definition $$X^{d,Y}_t=B^{d}_t-\frac tT(B^d_T-Y),$$ where $B^d$ is a $d$ dimensional Brownian motion.

I was wondering if the probability density of $X$ was known in the literature and if so, if someone could provide me with a reference.

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    $\begingroup$ @ Tom-tom: Well yes it is fully known because one property of the Brownian bridge is to be a Gaussian process, a little bit of googling shall give you the full story. Best regards $\endgroup$ – TheBridge Oct 14 '15 at 13:48
  • $\begingroup$ Thanks for your answer. Unfortunately, I already heavily googled it. The Gaussian properties are simple, but this does not mean that the probability density function should be easy to compute ! Have a look at this article by Johnson and Killeen, you will see that simple statistical properties of the Brownian bridge can be very technical to establish. $\endgroup$ – Tom-Tom Oct 14 '15 at 14:09
  • $\begingroup$ @TheBridge. You were actually right, it's simple (see my answer below). I still have ofund no reference for this result, but it doesn't matter anymore. $\endgroup$ – Tom-Tom Oct 14 '15 at 15:22

Edited to remove useless material

So silly of me, the answer is straightforward. Consider a Brownian bridge $X$ in $d$ dimensions, starting at $x=0$ at time $t=0$ and reaching its final point $R$ at time $t>0$. Then the caracteristic function is determined by the equation (1) in the Question : $$ \mathbb{E}[X_u]=\frac utR, \qquad \mathbb{E}[X_u^2]-\mathbb{E}[X_u]^2=\frac{u(t-u)}t.$$ It follows that the density probability of $X_u$ is a Gaussian distribution centered on $\frac utR$ with variance $\frac{u(t-u)}t$ or explicitely $$p_{X_u}(x)=\left(\frac{t}{2\pi u(t-u)}\right)^{d/2}\exp\left(-\frac{t\left(\frac utR-x\right)^2}{2u(t-u)}\right).$$

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    $\begingroup$ In other words, it's the Gaussian density with mean $\frac{u}{t} R$ and variance $\frac{u(t-u)}{t}$. Since you knew it was a Gaussian process, all you had to find out was the mean and variance of $B_u$, which I'm sure you must have found in the references you consulted. $\endgroup$ – Nate Eldredge Oct 14 '15 at 15:28
  • $\begingroup$ @NateEldredge. Sure. Actually, only these points are necessary, my computation is superfluous. Thanks ! $\endgroup$ – Tom-Tom Oct 14 '15 at 15:32

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