I was reading on conditional probability with respect to a partition of a sample space, and I came across the following example:

Let $(N_t:t\geq0)$ be the Poisson process. Given fixed times $0\leq s\leq t$, let $A=[N_s=0]$ and for all $n\in\mathbb N$, let $B_n=[N_t=n]$. Letting $\mathscr B$ be the $\sigma$-algebra generated by the $B_n$'s, we get that $$P(A|\mathscr B)=\left(1-\frac st\right)^{N_t}.$$

The two crucial ingredients of this is that increments in the Poisson process are independent and Poisson distributed, and so we get that \begin{align*} P(A|B_n)&=\frac{P(N_s=0\cap N_t=n)}{P(N_t=n)}\\ &=\frac{P(N_s=0\cap N_t-N_s=n)}{P(N_t=n)}\\ &=\frac{P(N_s=0)P(N_t-N_s=n)}{P(N_t=n)}, \end{align*} and then we easily compute from here.

To test my understanding, I tried to see if I could come up with a similar example with Brownian motion $(W_t:t\geq 0)$.

Of course, the events of the form $[W_t=a]$ for $a\in\mathbb R$ have probability zero, so I decided to define $$A=[W_s\leq 0]$$ and $$B_n=\big[W_t\in[n,n+1)\big],\qquad n\in\mathbb Z.$$ (Of course $P(A)=1/2$ by symmetry, but it seems an interesting to compute the probability that the Brownian motion is negative at time $s$ knowing in what interval $[n,n+1)$ the motion lies at some future time).

However, I'm having problems replicating the computations for my example. Here, we have that \begin{align*} P(A|B_n)&=\frac{P(W_s\leq0\cap W_t\in[n,n+1))}{P(W_t\in[n,n+1))}.\\ \end{align*} The problem here is that, since $W_s$ does not have a fixed value, I cannot write $$P(W_s\leq0\cap W_t\in[n,n+1))$$ as $$P(W_s\leq0\cap W_t-W_s\in E)$$ for some nice set $E$ (I would get $W_t-W_s\in[n-W_s,n-W_s+1)$, which I can't deal with).

Any help/hint on how to deal with this would be greatly appreciated!

  • $\begingroup$ Shouldn't that be $(1-{s\over t})$, not $(1+{s\over t})$? $\endgroup$ – user940 Jul 1 '15 at 15:51
  • $\begingroup$ You are correct. Fixed. $\endgroup$ – user75206 Jul 1 '15 at 15:53

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