Can a random variable be expressed as a sum of deterministic and random variable? Say we have a sequence of random variables $\{X_t:t\geq 0\}$ following an unknown stochastic process with distribution $X_t\sim N(\mu_X,\sigma_X^2)$. This idea came to me from the additive noise model. Say at time $t$ we have observed $x_t$. Can it, thus, be said that at time $t$ 
\begin{equation}
X_t=x_{t}+noise
\end{equation}
The math or the idea may be incorrect. If it is indeed incorrect, I would like to reiterate my question in the title, as to whether the random variable $X_t$ can be decomposed in a similar (yet correct) manner as above. Thank you!
 A: Consider linear regression: $$Y = X\beta + e$$ We decompose a random variable ($Y \sim N(X\beta, \sigma^2)$) into a deterministic part ($X\beta$) and a random part ($e \sim N(0, \sigma^2)$). So the answer is yes, we can decompose a random variable into deterministic and random components.
Though of course note that from a statistical point of view (as opposed to a probabilistic one) we won't be able to recover these components exactly - it's an estimation.
A: Yes, we can.  Consider a random variable $X$ with mean $\mu \in \mathbb{R}$.  (Note that not every random variable has a mean, so there is a loss of generality in this starting point.)  If we define the random variable $\varepsilon \equiv X-\mu$ as the deviation from the mean then we can write the original random variable as:
$$X = \mu + \varepsilon.$$
The random variables $\varepsilon$ has zero mean, by construction.  This technique is possible because whenever we have a random variable, adding a constant imposes a location-shift that yields a new random variable.  Even for a random variable that does not have a mean, it is possible to write it as a sum of a constant and a location-shifted version of that random variable.
