Conditional probability of geometric brownian motion

Question

I created paths using GBM to implement The stochastic mesh method. But the method requires the conditional distribution, given some S(t) the probability of S(t+1).

I've searched and can't find this formula, Does someone know it? Is there any other way to calculate the conditional probability?

UPDATE

Let me explain a little better, I'm working on the stochastic mesh method The method require I create N paths over M steps of time. I choose GBM to do this, so in a single path simulation I have

$$ S = S_{0}, S_{1}, ... ,S_{M} $$

Next I need to calculate the continuation value for wich I need Weights

$$ w_{ij} = p_{ij} / \sum_k p_{kj} $$ $$ p_{ij} = P (S_{t + \bigtriangleup t} \in A \mid S{t} = x) $$ The last is the formula I'm looking for the GBM

Answer 1

Applying Itô's lemma to the Black-Scholes SDE and integrating from $t$ to $t+\Delta t$ gives: $$ S_{t+\Delta t} = S_t e^{(r-\frac{1}{2}\sigma^2)\Delta t + \sigma \sqrt{\Delta t}Z} $$ with $Z \sim N(0,1)$, showing that $S_{t+\Delta t}$ given $S_t$ is log-normally distributed.

It is then straightforward to write, for any compact $\mathcal{A} = [a_1,a_2]$ with $0 < a_1 \leq a_2$ and under the risk-netral measure $\mathbb{Q}$

\begin{align} \mathbb{Q}\left(S_{t+\Delta t} \in \mathcal{A} \vert S_t = x\right) &= \mathbb{Q}\left( x e^{ (r-\frac{1}{2}\sigma^2)\Delta t + \sigma \sqrt{\Delta t}Z} \in [a_1, a_2] \right) \\ &= \mathbb{Q} \left( \ln(x) + (r-\frac{1}{2}\sigma^2)\Delta t + \sigma \sqrt{\Delta t}Z \in [\ln(a_1), \ln(a_2)] \right) \\ &= \mathbb{Q} \left( Z \in [a_1^*, a_2^*] \right) \\ &= \Phi(a_2^*) - \Phi(a_1^*) \end{align} where we have used the fact that the natural logarithm is a monotone increasing bijective function and defined $$\Phi(x) = \text{Pr}(X \leq x), X \sim N(0,1)$$ to be the normal cumulative distribution function along with $$ a_i^* = \frac{ \ln\left(\frac{x}{a_i}\right) + \left(r - \frac{1}{2}\sigma^2\right)\Delta t }{\sigma\sqrt{\Delta t}} $$

Answer 2

1. Note that $${{f}_{W(t)\left| W(s) \right.}}\left(x\left| y \right. \right)=\frac{{{f}_{ W(s),W(t)}}\left( x,y \right)}{{{f}_{ W(s)}}\left( y \right)}=\frac{1}{\sqrt{2\pi(t-s)}}\exp \left[-\frac{{{(x-y)}^{2}}}{2(t-s)} \right]$$

2. By application of Ito's lemma we have $$ln\,S_{t+\Delta t}=ln\,S_t+\left((\mu-\frac{1}{2}\sigma^2)\Delta t+\sigma(W_{t+\Delta t}-W_t)\Delta t\right)$$ Indeed $ln\,S_{t}$ has a Normal distribution. $$$$

3. $P(S_{t+\Delta t}<x|S_{t}<y)$=$P(ln\,S_{t+\Delta t}<ln\,x|ln\,S_{t}<ln\,y)$