# The conditional expectation of a geometric brownian motion

In this question it states that $$\mathbb{E}[e^{\sigma(W_t-W_s)}|\mathcal{F}_s] = \mathbb{E}[e^{\sigma(W_t-W_s)}],$$ and I assume that $$0 \leq s \leq t$$. The accepted answer states that this step is correct. However, how can the random variable (LHS) ever be equal to a constant (RHS) for all $$0 \leq s \leq t$$? The only situation in which it seems to be true is when $$s=0$$, since the LHS then collapses to a constant as well. Can anybody provide an explanation?

• What do you know about the increments of a brownian motion? – Andrew May 2 '19 at 8:53
• Independent and $W_t - W_s \sim N(0, t-s)$. So do you mean that $\mathbb{E}(e^{\sigma(W_t-W_s)}|\mathcal{F_s}) = \mathbb{E}(e^{\sigma(W_{t-s}-W_0)}|\mathcal{F_0}) = \mathbb{E}(e^{\sigma W_{t-s}}) = \mathbb{E}(e^{\sigma( W_t-W_s)})$? – pabk May 2 '19 at 9:06
• The independent increments imply that $W_t-W_s$ is independent of $\mathcal{F}_s$ and therefore is the conditional expectation equal to the expectation $\mathbb{E}[W_t-W_s |\mathcal{F}_s ] =\mathbb{E}[W_t-W_s]$. This also holds if you apply continuous functions on $W_t-W_s$. – Andrew May 2 '19 at 12:51
• The question you referenced says this:$$\mathbb{E}[e^{\sigma W(t)}|{\cal F}_s] = \mathbb{E}[e^{\sigma (W(t) - W(s) + W(s))}|{\cal F}_s] = \mathbb{E}[e^{\sigma (W(t) - W(s))}|{\cal F}_s]e^{W(s)}$$ Could you clarify where you get the above equation from please? – Magic is in the chain May 2 '19 at 20:32
• It's the third equality in the question – pabk May 3 '19 at 11:18