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?
 A: Let $p_t$ denote the PDF of $B(t)$ and assume that $U$ is independent of $B$ with PDF $f_U$, then the distribution of $B(U)$ has PDF 
$$
q(\ )=\int p_t(\ )f_U(t)\mathrm dt.
$$
In the present case, $U$ is uniform on $(0,1)$ and, for every $t$ in $(0,1)$,
$$
p_t(x)=\frac1{\sqrt{2\pi t(1-t)}}\mathrm e^{-x^2/(2t(1-t))},
$$
hence
$$
q(x)=\int_0^1\frac1{\sqrt{2\pi t(1-t)}}\mathrm e^{-x^2/(2t(1-t))}\mathrm dt.
$$
The change of variable $4t(1-t)=1/u^2$ yields
$$
q(x)=\frac2{\sqrt{2\pi}}\int_0^\infty\mathrm e^{-2x^2(1+u^2)}\frac{\mathrm du}{1+u^2}.
$$
Differentiating this and identifying $q'(x)$ yields finally

$$
q(x)=2\int_{|x|}^\infty\mathrm e^{-2u^2}\mathrm du.
$$

Note that, to compute some characteristics of the distribution of $B(U)$, one may find more convenient to bypass the PDF $q$ and to go back to the definition of $B(U)$, for example,
$$
E(B(U))=\int_0^1E(B(t))\mathrm dt=0,
$$
and
$$
E(B(U)^2)=\int_0^1E(B(t)^2)\mathrm dt=\int_0^1t(1-t)\mathrm dt=\frac16.
$$
Likewise, for every suitable measurable function $A$,
$$
E(A(B(U)))=\int_0^1E(A(B(t)))\mathrm dt=\int_0^1\int_\mathbb RA(x)p_t(x)\mathrm dx\mathrm dt.
$$
A: 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 $\sigma^{2}=t(t-1)$
