Consider general Brownian bridge W(0)=0; W(T) = a. (Here "general" means: $W(T)\ne 0$).

What is the probability W(t) >= b, for all $ t \in [0, T] $ ?

Is there close simple formula in terms of a, b , T ?

EDIT Is the answer equal to $1-exp(-2b(a+b)/T) $ ? (See comment-edit below).

Can the formula be the same for any martingale type stochastic process (e.g. random walk), not only brownian motion, or it somehow depends on the details of stochastic process ? If there is such dependence how the questioned probability will change if we consider the distribition W(t) to have more and more "heavy tails" ?

(I'm sure that questions are well-known for experts, but it is somehow difficult to google the asnwer, so let me ask here).

Remark: If we consider somewhat informally related question - probability for W first hit a, before hitting (-b), (i.e. W(some T)=a and W(t < T) > b) there is simple formula P = b/(a+b), which holds true for any martingale stochastic process. The questions are somewhat different, but still resemble each other and so the simplicity of the answer P=b/(a+b) makes me hope that the answer to my question might be simple and closed form.

EDIT As MattF suggested at a comment we might look at "On the maximum of the generalized Brownian bridge" Theorem 2.1. And the answer seems to be almost there - given by the formula presented in edit above. But I am not 100% sure, since they consider more general case and we need to set u=t which causes division by zero in their formula. Hopefully it is not a big problem - the formula is continuous and just simplifies to $exp( -2\beta(\beta-\eta)/ u ) $ in a limit u=t, in their notations.

Is it correct ?

To pass from my notations - I am interested in "min greater" not "max greater", that is why I would need to change sign for their $\eta$ and chnage to "1-their answer" and $\eta -> -a$ $\beta -> b$".