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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $\Bbb E[X_s \cdot X_t]$.

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  • $\begingroup$ Why does it matterthat it is a real number? $\endgroup$ – jaja Mar 13 at 9:59
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    $\begingroup$ 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

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