Given a Levy process $X$ at different points in time $s$ and $t$, and if I have an expression like this:

$$\mathbb{E}[X_t \cdot \mathbb{E}[X_s]]$$

I want to know if I can use partial averaging to say that this is equal to

$$\mathbb{E}[X_t\cdot X_s].$$

Is that correct?

  • 4
    $\begingroup$ The first expression is equal to $\mathbb E[X_t]\mathbb E[X_s]$. Can you see why this is not equal to $\mathbb E[X_tX_s]$, when $X_t$ is (say) Brownian motion? $\endgroup$ – Mike Earnest Mar 12 at 22:59
  • $\begingroup$ okay, yes I see that! But can you please tell me why $\mathbb{E}[X_t\mathbb{E}[X_s]]=\mathbb{E}[X_t]\mathbb{E}[X_s]$? I'm sorry if I'm not seeing obviously. $X_t$ is not necessarily independent of $X_s$. The increments are independent but not the specific points. $\endgroup$ – jaja Mar 13 at 8:36
  • 1
    $\begingroup$ For any constant, $k$, $\mathbb E[kX_t]=k\mathbb E[X_t]$. Let $k$ be the constant $E[X_s]$. $\endgroup$ – Mike Earnest Mar 13 at 18:16

let $X$ be a Levy process. Note that $\Bbb E[X_s]$ is then a real number. Therefore $$ \Bbb E[\Bbb E[X_s] \cdot X_t] = \Bbb E[X_s] \cdot \Bbb E[X_t]. $$

As mentioned in the comments, this is (in general) not equal to $\Bbb E[X_s \cdot X_t]$.

  • $\begingroup$ Why does it matterthat it is a real number? $\endgroup$ – jaja Mar 13 at 9:59
  • 2
    $\begingroup$ One of the most basic properties for expectations states that for any real number $c$ we have that $\Bbb E[c\cdot X] = c \cdot \Bbb E[X]$. $\endgroup$ – Cettt Mar 13 at 10:09

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.