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?

  • $\begingroup$ What do you know about the increments of a brownian motion? $\endgroup$ – Andrew May 2 '19 at 8:53
  • $\begingroup$ Independent and $W_t - W_s \sim N(0, t-s)$. So do you mean that $\mathbb{E}(e^{\sigma(W_t-W_s)}|\mathcal{F_s}) = \mathbb{E}(e^{\sigma(W_{t-s}-W_0)}|\mathcal{F_0}) = \mathbb{E}(e^{\sigma W_{t-s}}) = \mathbb{E}(e^{\sigma( W_t-W_s)})$? $\endgroup$ – pabk May 2 '19 at 9:06
  • $\begingroup$ 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 $W_t-W_s$. $\endgroup$ – Andrew May 2 '19 at 12:51
  • $\begingroup$ 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? $\endgroup$ – Magic is in the chain May 2 '19 at 20:32
  • $\begingroup$ It's the third equality in the question $\endgroup$ – pabk May 3 '19 at 11:18

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