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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!

Carl
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    Imagine that the deterministic part of the process is the mean $\mu$, which can be time-varying ($\mu_t$), and the stochastic part is $e_t \sim N(0, \sigma^2)$. Then you've decomposed $X_t$ into the sum of a deterministic part ($\mu_t$) and a stochastic part ($e_t$). – jbowman Apr 08 '20 at 16:41
  • The answer trivially is yes: define $x_t$ to be whatever you want and subtract it from $X_t.$ Evidently you have some restrictive concepts concerning how $x_t$ might be chosen and what constitutes "noise" (for instance, usually noise terms are centered around zero). Would you like to refine your question to narrow down the possibilities? – whuber Feb 03 '23 at 15:05
  • Yes, see the reparamaterization trick used in autoencoders for an example https://stats.stackexchange.com/questions/199605/how-does-the-reparameterization-trick-for-vaes-work-and-why-is-it-important – Estimate the estimators Feb 03 '23 at 15:17

2 Answers2

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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.

Ben
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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.

RyanFrost
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