Proof of $ x = \mu + \epsilon $ for a random $x$ from population $N(\mu, \sigma)$

Question

Let's say $x$ is a random variable drawn from a normally distributed population with mean $\mu$ and variance $\sigma^2$. Then we can write $x$ in terms of $\mu$ and a random error component $\epsilon$ as: $$ x = \mu + \epsilon $$

Can someone explain (or point me to some resources) which proves this.

More specifically, what does the noise term $\epsilon$ consist of, and is it related somehow to the variance $\sigma^2$ of the distribution?

Answer 1

You assume that noise is normally distributed with zero mean $\varepsilon \sim \mathcal{N}(0, \sigma^2)$, then you take $X = \varepsilon + \mu$ and since $E(Y+c) = E(Y) + c$ (where $Y$ is a random variable and $c$ is a constant) the new mean is $\mu$ rather then $0$.