3
$\begingroup$

If $\{X_{t}:t\geq 0\}$ is a real-valued stochastic process with independent increments then $\{X_{t}:t\geq0\}$ is a Markov process?

Let $\{ \mathcal{F}_{t} \}_{t\geq0} $ be a natural filtration of $\{X_{t}:t\geq 0\}$. I want to prove tha following assertion:

For bounded measurable function $f$, $E\left[f(X_t)|\mathcal{F}_{s} \right]=E\left[f(X_t)|\sigma({X}_{s}) \right]$ ($\forall t\geq \forall s\geq 0$)


Because independent increments means that $\forall t\geq \forall s\geq0$, $X_{t}-X_{s}$ is independent of $\mathcal{F}_{s}$ and $X_{s}$ is $\mathcal{F}_{s}$-measurable,

$E\left[f(X_t)|\mathcal{F}_{s} \right]=E\left[f(X_{t}-X_{s}+X_{s})|\mathcal{F}_{s} \right]=E\left[f(X_{t}-X_{s}+X_{s})|\sigma(X_{s}) \right]$

How do I justify last equality? I tried to use tower property, but it didn't do well. Please teach me.

$\endgroup$
2
$\begingroup$

Consider a sigma-algebra $\mathcal G$, a random variable $Y$ independent of $\mathcal G$, a random variable $Z$ measurable for $\mathcal G$, and a measurable function $u$ such that $u(Y,Z)$ is integrable. Then, $$ E[u(Y,Z)\mid\mathcal G]=v(Z), $$ where the function $v$ is defined as $$ v(z)=E[u(Y,z)]. $$ In particular, $$ E[u(Y,Z)\mid\mathcal G]=E[u(Y,Z)\mid Z]. $$ Source: your textbook.

$\endgroup$
  • 1
    $\begingroup$ Thanks! I found this statement. I wish I had studied this. $\endgroup$ – ko4 Feb 22 '14 at 11:11
  • $\begingroup$ Is that using the tower property? $\endgroup$ – guimption Nov 11 '18 at 0:45
  • 1
    $\begingroup$ @guimption Yes, based on the inclusion $\sigma(Z)\subseteq\mathcal G$. $\endgroup$ – Did Nov 11 '18 at 7:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.