Brownian bridge in different forms On the wikipedia page for a Brownian bridge (https://en.wikipedia.org/wiki/Brownian_bridge), it says that the Brownian bridge is given by $B(t) = W(t) - \frac{t}{T}W(T)$. It further goes on to say that another representation of this is given by $B(t) = \frac{T-t}{\sqrt{T}}W(\frac{t}{T-t}).$ Are these two exactly the same? I don't see how they are an exact match of each other.
 A: Wikipedia is saying that given a Wiener process $W(t)$, the two processes you mentioned are both Brownian bridges. But they aren't necessarily the same Brownian bridge. Analogously, if $X$ is a normally distributed random variable, then $e^X$ and $e^{-X}$ are both lognormally distributed, but they aren't the same.
A: The two processes are the same in law but not the same pathwise.
A Brownian bridge, say on $[0,1]$, is by definition a process $(X_t)$ such that

*

*$(X_t)$ has continuous sample paths almost surely.

*$(X_t)$ is Gaussian, i.e. $(X_{t_1},\cdots, X_{t_k})$ has joint Gaussian distribution for any $t_1,\cdots, t_k$ (with mean zero).

*For any $t_1 \leq t_2$, $Cov(X_{t_1}, X_{t_2}) = t_1 (1-t_2)$.

If $(W_t)$ is a standard Brownian motion, then it's clear that,
$$
B^{(1)}_t = W_t- {t}W_1, \mbox{ and  } B^{(2)}_t = (1-t) W_{\frac{t}{1-t}}
$$
are both Brownian bridges. In particular, for any $t_1,\cdots, t_k$, the random vectors
$(B^{1}_{t_1},\cdots, B^{1}_{t_k})$ and $(B^{2}_{t_1},\cdots, B^{2}_{t_k})$ have the same distribution. In other words, $B^{(1)}_t$ and  $B^{(2)}_t$ are equal in law.
It's also clear that they are not the same pathwise.
For a given sample path of $W$, which is a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$, the corresponding paths
$$
g^{1}(t) = f(t) - t f(1) \; \mbox{  and   } \;  g^{(2)}_t = (1-t) f( \frac{t}{1-t}),
$$
of $B^{1}_t$ and $B^{2}_t$ respectively, are not the same.
Both $B^{1}_t$ and $B^{2}_t$ are weak solutions to the SDE
$$
dX_t = -\frac{t}{1-t} X_t dt + dW^{0}_t,
$$
driven by different Brownian motions $W^{0}_t$'s---neither of which is $W_t$.
(A trivial analogue for random variables would be as follows: Let $X$ be a random variable, defined on some probability space, that has $N(0,1)$ distribution. Then $-X $ also has $N(0,1)$ distribution---$X$ and $-X$ are equal in law. On the other hand, $X$ and $-X$ are not equal almost surely, or "almost surely pathwise" in a trivial sense.)
