1
$\begingroup$

Let $(B_t)$ a Brownian motion and $(\mathcal F_t)$ its natural filtration. In my lecture, to prove that $(B_t)$ has the markov property they do as following : les $f$ measurable and bounded.

$$\mathbb E[f(B_t)\mid \mathcal F_s]=\mathbb E[f(B_t-B_s+B_s)\mid \mathcal F_s]=\mathbb E[f(B_t-B_s+x)]|_{x=B_s}=\mathbb E[f(B_t)\mid B_s].$$

And I do not agree with the last equality. For me $$\mathbb E[f(B_t-B_s+x)]|_{x=B_s}=\mathbb E[f(B_{t-s}+x)]|_{x=B_s}=\mathbb E[f(B_{t-s}+B_s)\mid B_s],$$ so how did they found that $\mathbb E[f(B_t)\mid B_s ]$ ?

$\endgroup$

1 Answer 1

1
$\begingroup$

Notice that $$\mathbb E[f(B_t)\mid \mathcal F_s]=\mathbb E[f(B_t)\mid B_s],$$ if and only if $$\mathbb E[f(B_t)\mid \mathcal F_s]=\varphi (B_s),$$for some measurable function $\varphi $. Therefore, having $$\mathbb E[f(B_t)\mid \mathcal F_s]=\left.\mathbb E[f(B_t-B_s+x)] \right|_{x=B_s},$$ implies automatically that $$\mathbb E[f(B_t)\mid \mathcal F_s]=\mathbb E[f(B_t)\mid B_s].$$


Anyway. Instead of writing $$\left.\mathbb E[f(B_t-B_s+x)]\right|_{x=B_s}=\left.\mathbb E[f(B_{t-s}+x)]\right|_{x=B_s},$$ remark that $$\left.\mathbb E[f(B_t-B_s+x)]\right|_{x=B_s}=\mathbb E[f(B_t-B_s+B_s)\mid B_s]=\mathbb E[f(B_t)\mid B_s],$$ as wished.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .