how to get brownian bridges maximum and minimum's joint distribution I'm wondering how to get brownian bridge's maximum and minimum's joint distribution.Brownian bridge's definition is as below.

$B_t = W_t - tW_1$ for $t \in [0,1]$, W is wiener process

How to get it? In fact, I didn't get maximum's distribution of brownian bridge. I cannot catch any idea.. I tried to get this distribution via conditional probability, that fix $W_1$, but it is still hard one. Can you help me?
edit) I think that it is concerned with joint distribution of brownian motion's maximum and minimum, and W(1). Then how to attain this one?

Let $M$ and $m$ be the Max and min of $B_t$. Let us find the joint cdf of $M$ and $-m$, defined by
$$
P(\{M< a\}\cap \{-m< b\}),
$$
In fact, the argument you can use is exactly the continuous version of the discrete problem here: A generalization of reflection. Namely, Brownian motion is the limit of a discrete random walk, and there is an exact formula for the number of random walks of a given length who stay between two horizontal lines, using the reflection principle.
Applying that logic, we get that
$$
P(M<a,-m<b)
=\frac1{P(W_1=0)}\left(
\sum_{k\in \mathbb Z} P\big(W_1=2k(a+b)\big) - 
\sum_{k\in \mathbb Z} P\big(W_1=2k(a+b)+2a\big) 
\right)
$$
Note that this does not quite make sense, as all the probabilities are zero, but you have to understand this as a limit. Anyways, what you get is
$$
P(M<a,m<-b)=\sum_k\exp(-2k^2(a+b)^2)-\sum_k\exp\left(-2\big(k(a+b)+a\big)^2\right)
$$
This gives the joint cdf of $M$ and $-m$.
