Conditional probability of a geometric Brownian motion

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Given a GBM $S_t$ with parameters $\mu$ and $\sigma$, I would like to understand if it is possible to calculate the following quantity: $$ P(S_{\tau+1} \geq \alpha \mid S_\tau)\ . $$ That is, at some moment in time $\tau$, if I observe a value $S_\tau$, I would like to be able to compute the probability of observing, in the next time step, a value larger than some threshold $\alpha$.

I tried applying Baye's theorem but didn't get very far. If I write $$ P(S_{\tau+1} \geq \alpha \mid S_\tau) = \frac{P(S_\tau \mid S_{\tau+1} \geq \alpha) P(S_{\tau+1} \geq \alpha)}{P(S_\tau)}\ , $$ I know how to compute (at least numerically), $P(S_{\tau+1} \geq \alpha)$ and $P(S_\tau)$, but the conditional probability once again stumps me. Any thoughts would be appreciated.

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Way too long for a comment.

Hints:

  • The approach with Bayes' theorem does not work because $S_\tau$ is not an event of which you could calculate a probability.

  • The quantity that you want to calculate is the special case of a conditional expectation: $$ P(S_{\tau+1}\ge\alpha|S_{\tau})=E(1_{\{S_{\tau+1}\ge\alpha\}}|S_{\tau}) $$

  • The key observation is now that $$ S_{\tau+1}=S_\tau\exp\Big(\sigma (W_{\tau+1}-W_\tau)-\frac{\sigma^2}{2}+\mu\Big). $$ In other words: $S_{\tau+1}$ is a function of $S_\tau$ and the Brownian increment $W_{\tau+1}-W_\tau$ that is independent of $S_\tau$.

  • Therefore (please study the properties of those conditional expectations), the conditional expectation you seek is equal to the probability that $S_1\ge 0$ in which you replace the deterministic $S_0$ by $S_\tau$ after having done the calculations.