# Can we express a one-dimensional raised Bessel Bridge as a function of a single Brownian Motion?

A Bessel Bridge is a Brownian Motion, conditioned such that $B(0) = B(1) = 0$ and $B([0, 1]) \ge 0$. A raised Bessel Bridge is a generalization of this: it's a Brownian Motion conditioned such that $C(0) = a, C(1) = b, C([0, 1]) \ge 0$ for some nonnegative constants $a, b$.

My end goal is to find a density function for the integral of a random realization of a raised Bessel Bridge. I plan to approach this problem as follows: the Feynman-Kac formula gives an expression for $E(e^{-u \int_0^1 V(x(t)) dt})$, where $x(t)$ is a Brownian Motion, $u$ is a constant, and $V$ is any function. If I choose $V$ such that $V(x(t))$ is a raised Bessel Bridge, then this information is sufficient to compute the desired density function (simply take a Fourier Transform).

So, my question: what function $V$ makes the process $V(x(t))$ a raised Bessel Bridge, where $x(t)$ is a Brownian Motion?

There is no such function, simply because the raised bridge is a nonhomogeneous Markov process (its drift away from zero becomes small as time approaches $1$).
Moreover, there is no such function that also depends on $t$: both the Brownian motion and the bridge have quadratic variation $dt$, so for any process $V(x(t),t)$ the quadratic variation would be $\left( \frac{\partial V}{\partial x} \right)^2 dt$, so to get $dt$ again you would have no choice but $V(x,t) = \pm x + f(t)$, which obviously doesn't fit.
On the other hand, I guess it should be possible to find the joint distribution of the Markov process $\left( B(t), \inf_{[0,t]} B, \int_{[0,t]} B(s) ds \right)$ by solving a PDE, and then condition on $B(1)$ and $\inf \ge 0$.