Random Variable Defined in terms of it's conditional expectation Suppose $X_t$ is defined so that $X_{t+1} = a_t + \varepsilon_{t+1}$ where $a_t = \mathbb{E}[X_{t+1}|X_t]$ and $\mathbb{E}[\varepsilon_{t+1}|X_t] = 0$.
How does one even define a random variable this way? This seems to be a circular definition unless I'm simply missing something?
I'm then asked to calculate the variance of this random variable, requiring me to show that $\operatorname{Cov}(X_t,\varepsilon_t) = 0$, but I don't know how to begin with that until I've correctly interpreted the definition.

 A: The relation $X_{t+1}=m_t+\varepsilon_{t+1}$ (with $\mathbb E_t\varepsilon_{t+1}=0$)     is not supposed to defined in a unique way $X_t$ (just consider $X_{t+1}=f_t\left(X_t\right)$ where $f_t\colon\mathbb R\to\mathbb R    $  is a Borel function. 
We have $$\operatorname{Var}\left(X_{t+1}\right)=\operatorname{Var}\left(m_t\right)+2\operatorname{Cov}\left(m_t,\varepsilon_{t+1}\right) +\operatorname{Var}\left(\varepsilon_{t}\right),$$
so it suffices to prove that 


*

*$\operatorname{Cov}\left(m_t,\varepsilon_{t+1}\right) =0$ and 

*$\operatorname{Var}\left(\varepsilon_{t}\right)=\mathbb E\left[ \operatorname{Var}_t\left(\varepsilon_{t+1}\right)\right].        $


The first bullet follows from the fact that  $\varepsilon_{t+1}$ has zero mean      and since $m_t$ is $\sigma\left(X_t\right)$-measurable, we have     $$\mathbb E\left[ m_t \varepsilon_{t+1} \right]= \mathbb E\left[ \mathbb E_t\left[ m_t \varepsilon_{t+1}\right]  \right]= \mathbb E\left[ m_t \mathbb E_t\left[ \varepsilon_{t+1}\right]  \right].$$
The second bullet is a consequence of $\mathbb E_t\left[  \varepsilon_{t+1}\right]=0$. Indeed, we have $\operatorname{Var}_t\left(\varepsilon_{t+1}\right)=\mathbb E_t \left[ \varepsilon_{t+1}^2\right] $.
