Consider general Brownian bridge W(0)=0; W(T) = a. (Here ["general"](https://en.wikipedia.org/wiki/Brownian_bridge#General_case) 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) $ or $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. 




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**EDIT**
As MattF suggested at a comment we might look at ["On the maximum of the generalized Brownian bridge"](https://www.researchgate.net/publication/236984395_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 $\eta -> -a$ $\beta -> b$".


However the answer in my notations $1-exp(-2b(b+a)/T)$ does not seems to be reasonable, because if a=b, the answer should be equal to "1", which seems to contradict the formula.


It seems $exp(-2b(a-b)/T)$, for a>=b would be more reasonable since at least we get "1" for a=b, but it seems not to correspond to the formula from the paper or I made some stupid mistake passing from one notations to the other.



Probability of general Brownian (or non) bridge to be higher than given parameter?
For Brownian bridge, this is more or less a standard exercise. It is simpler to do it first for Brownian motion and then move to Brownian bridge. There are two steps. I assume $b<0$, otherwise the answer is $0$. Let $c=a-b>0$.


1) By the reflection principle,
$$P(W\_t\geq b \; \forall t\in [0,T], W\_T\in [a,a+dx])=P^{-b}(W\_t\geq 0 \;\forall t\in [0,T], W\_T\in [c,c+dx])=
\frac{1}{\sqrt{2\pi T}} \left(e^{-a^2/2T}-e^{-(a-2b)^2/2T}\right)dx:=F(a,b,T)dx.$$


2) We also have
$$P(W\_T\in [a,a+dx])=\frac{1}{\sqrt{2\pi T}}e^{-a^2/2T}dx:= G(a,T)dx.$$


So the answer to your question is $F(a,b,T)/G(a,T)$.


For more general random walks, it is hopeless to get explicit formulae (since you do not have a reflection principle, unless the random walk is e.g. simple random walk or a variant of that), and even the analogue of $G$ is not explicit. But asymptotics are available, see 
F. Caravenna, A local limit theorem for random walks conditioned to stay positive, Probab. Theory Related Fields, 133:508--530, 2005.


