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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?

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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 $-m$.

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