# Does partial averaging allow moving increments in and out of an expectation?

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?

• 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? – Mike Earnest Mar 12 at 22:59
• 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. – jaja Mar 13 at 8:36
• For any constant, $k$, $\mathbb E[kX_t]=k\mathbb E[X_t]$. Let $k$ be the constant $E[X_s]$. – 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 $$\Bbb E[X_s \cdot X_t]$$.
• One of the most basic properties for expectations states that for any real number $c$ we have that $\Bbb E[c\cdot X] = c \cdot \Bbb E[X]$. – Cettt Mar 13 at 10:09