Distribution related to brownian bridge Let $B(t)$ be a Brownian Bridge and $U$ is uniformly distributed on $(0,1)$. I wish to know the distribution function $B(U)$. Is it possible? As we know, $B(t)\sim N(0,t(1-t))$. But, I haven't a clue when $t$ is replaced by random variable $U$. Could anyone help me?

This may be completely wrong but its just an idea
So any distribution is characterized by its CDF $F(x)=P(X\leq x)$ well to determine this for a distribution $X\sim B(U)$, what $P(X\leq x)$ would if $U$ was a discrete uniform $S=\{\frac{1}{4},\frac{2}{4},\frac{3}{4}\}$ letting $\Phi_{t}(x)$ represent CDF for $N\left((0,t(1-t)\right)$ would be 
$$F(x)=P(X\leq x)=\sum_{t\in S}P(X\leq x |U=t)P(U=t)=\frac{1}{3}\left(\Phi_{\frac{1}{4}}(x)+\Phi_{\frac{2}{4}}(x)+\Phi_{\frac{3}{4}}(x) \right)$$
Thus my guess for if $U\sim Uni(0,1)$ then the CDF for $B(U)$ would be 
$$F(x)=\int_{0}^{1}\Phi_{t}(x)dt$$
again this is a naive guess
even if this doesn't look like it pertains to a specific known distribution I think you can still simulate from this distribution using following code in R

rB=function(size,lower.bound,upper.bound)
{
sample=NULL
for(i in 1:size)
{
U=runif(1,lower.bound,upper.bound)
BU=rnorm(1,mean=0,sd=sqrt(U*(1-U)))
sample[i]=BU
}
return(sample)
}

I also assumed that when you write $N(0,t(t-1))$ that $\sigma^{2}=t(t-1)$
