Finding the probability of an increment of a Brownian Bridge

Question

I've been given a process $X_t=(1-t)B(\frac{t}{1-t}),0\leq t\leq 1$ where $B(\frac{t}{1-t})$ is a standard Brownian Motion.

So far, I have proven this is a Brownian bridge. Now I need to find $P(X_{4/5}-X_{3/5}>1/6).$

I'm sure the solution to this is something stupidly obvious, but in finding nothing between Google and what we've learned so far in calculating the probability of increments of gaussian processes, I'd really appreciate some assistance in moving in the right direction.

Answer 1

Let $t=3/5$ and $s=4/5$. Note that $\frac t {1-t} <\frac s {1-s}$. Thus $$X_{4/5}-X_{3/5}=(1-4/5)[B(\frac s {1-s})-B(\frac t {1-t})]+(4/5-3/5)B(\frac t {1-t})\sim N(0, \sigma^{2})$$ where $\sigma^{2}=(1-s)^{2}(\frac s {1-s}-\frac t {1-t})+(t-s)^{2}\frac t {1-t}$.

Answer 2

I actually ended up finding my own solution before checking back here by representing $X_{4/5}-X_{3/5}$ as

$\begin{bmatrix}1&-1\end{bmatrix}$ $\begin{bmatrix}X_{3/5}\\X_{4/5}\end{bmatrix}$

and getting the parameters of this increment by treating it as an affine linear transformation of a joint gaussian vector. This ended up giving me the same variance of $\frac{4}{25}$ as the other answer to this question so I'm pretty confident it's the right answer.