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.


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.

  • 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

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