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) $ 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.

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 $\eta -> -a$ $\beta -> b$".

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.

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    $\begingroup$ By googling I find: "the maximum of the generalized Brownian bridge", theorem 2.1. researchgate.net/publication/… $\endgroup$ – Matt F. May 5 '17 at 20:54
  • $\begingroup$ @MattF. Thank you very very much ! I hope that the paper gives the answer, however I am not completely sure - they consider more general case than I need and we need to set u=t to get my case, which however causes division by zero in their formula - it very might be that it is not a problem we just need to write $\infty$ and their formula simplifies at lot. However I am not 100% sure. $\endgroup$ – Alexander Chervov May 6 '17 at 12:23

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.

  • $\begingroup$ Thank you very much ! The answer seems to be correct - it is correct in limiting cases and also seems same (up to notations) as the paper quoted. But I am not expert in stoch.proc. so I cannot fully understand your argument with reflection principle, if you can give more details it would be very helpful. What is $P^b$ and why it equals to what you write ? $\endgroup$ – Alexander Chervov May 14 '17 at 9:06
  • $\begingroup$ A reffrence to Reflection principle which close to your discussion and written for simplyminded persons would be also very helpful . Thanks in advance! $\endgroup$ – Alexander Chervov May 14 '17 at 9:39
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    $\begingroup$ $P^b$ is the law of the process starting at $b$. The equality is just shift-invariance of increments of Brownian motion. As for reflection principle, just google it... en.wikipedia.org/wiki/Reflection_principle_(Wiener_process) $\endgroup$ – ofer zeitouni May 18 '17 at 19:59
  • $\begingroup$ Wiki does not seem to give similar reflection principle that you use $\endgroup$ – Alexander Chervov May 18 '17 at 20:54
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    $\begingroup$ Of course it does. The probability in question is to start at $-b$ and end at $c$ w/out touching 0. The reflection principle (and a union-exclusion) tells you that this is the same as hitting $c$ w/out the constraint, minus the probability of hitting $-c$. This is what I wrote. If you prefer and are more analytcally inclined, solve the heat equation with Dirichlet boundary conditions at $0$. $\endgroup$ – ofer zeitouni May 18 '17 at 23:30

Consider the standard Wiener process $W = (W_t, 0 \leq t \leq 1)$ and let $m = \inf_{0\leq t\leq 1} W_t$, $M = \sup_{0\leq t\leq 1} W_t$. If $a\leq 0 \leq b$ and $a\leq a' < b' \leq b$, then

$$ \mathbb{P}(a<m\leq M < b; a' < W_1 < b') = \sum_{k=-\infty}^\infty \left(\Phi(2k(b-a)+b')-\Phi(2k(b-a)+a')\right)\\ - \sum_{k=-\infty}^\infty \left(\Phi(2k(b-a)+2b-a')-\Phi(2k(b-a)+2b-b')\right) $$

This can be found as for example Theorem 9.10 in Billingsley's Convergence of Probability Measures. A similar result for the standard Brownian bridge can be found using the previous result, and is Theorem 9.39 in the same text. It says that:

$$ \mathbb{P}\left(a < \inf_t W_t^\circ\leq \sup_t W_t^\circ \leq b\right) = \sum_{k=-\infty}^\infty \left(e^{-2k^2(b-a)^2}-e^{-2(b+k(b-a))^2}\right) $$

Of course, these results hold considering $t\in[0,1]$ and $W^\circ(1)=0$, but these should not be too hard to transfer to the case you are thinking of.

The derivation of these distribution functions relies on analysing the simple random walk, and then using Donsker's theorem to take a limit of this case.

  • $\begingroup$ Thank you for the answer ! However it seems a little bit different from what I ask. You consider W simultaneosly greater m and smaller M, but in my question I do no impose second condition. Any way thank you very much for your answer ! $\endgroup$ – Alexander Chervov May 14 '17 at 9:08
  • $\begingroup$ I believe your desired distribution can be had simply by taking the limit as $a\to -\infty$. Similarly, if you desire that $W_1=\alpha$, the right distribution should be the limit of $P(a< m \leq M < b; \alpha < W_1 < \alpha + \epsilon)$ as $\epsilon\to 0$ and $a\to -\infty$. At least, that is how one gets distributions for the Brownian bridge from the above. (i.e. the case $\alpha = 0$.) $\endgroup$ – Vilhelm Agdur May 14 '17 at 12:39

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