Brownian bridge $Z_t$ is a diffusion process distributed as Brownian motion $B_t$ conditioned on the event $B_1 = 0$. It is rather well-studied, and allows for a Markov-like SDE representation. I wonder whether such representations are known for Brownian motion conditioned on $B_1\in \{x_1,\dots,x_n\}$. Even if this is the case, are the probabilities $P(Z_1 = x_i)$ fixed?

A similar question from a different perspective: if I would like to construct a continuous Markovian diffusion $Z_t$ with a property $Z_1 \in \{x_1,\dots,x_n\}$, are there known examples of such processes in the literature? I am particularly interested in the case $n = 2$.


1 Answer 1


One representation of $B_t$ knowing that $B_1$ can take value $x_0$ with probability $p$ and $x_1$ with probaiblity $1-p$ is the following.

Take 2 Brownian Bridges $B_1(t)=B_t^{0\to x_0}$ and $B_2(t)=B_t^{0\to x_1}$ that take respectively the value $x_0$ and $x_1$ at time 1 and a Bernoulli random variable $\beta$ with parameter $p$, then the process you are looking for is the following :

$$B^{x_0,x_1}(t)= \beta.B_1(t)+(1-\beta).B_2(t)$$

with $\beta$ being $\mathcal{F}_0$-measurable (i.e. measurable for the initial $\sigma$- algebra).

In a way you can view this process as a $\mathcal{F}_0$-measurable mixture of Brownian Bridges.

You can add refinements for example the time where the final value is decided could be dependent on a stopping time, this would only add a step for the Bernoulli random variable and the two Bridges at that time.

Best regards.

  • 1
    $\begingroup$ I thought of this kind of mixture, but it does not seem to be Markov given its natural filtration. $\endgroup$
    – SBF
    Mar 5, 2015 at 8:13
  • $\begingroup$ @ Ilya : I am not sure that a Markovian process with these features actually exists. My intuition tells me that for Markovian property to be included in such a process, this would entail that even at time $1-\epsilon$ you cannot tell for one path for sure which point the process will choose which I don't think is possible. I believe that at one point in time your process must know its way to the final point (even if it's a stopping time) so it can not be Markovian because one instant after that if Markovian and if path of both option coexist then you have forgotten what was decided about that $\endgroup$
    – TheBridge
    Mar 5, 2015 at 23:05

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