Distribution of a random variable when noise is a multiple of a constant Let $y = ax + \eta$ represent a random variable, where $\eta$ takes values which are multiples of $\alpha$ or $\beta$, where $\alpha$ and $\beta$ are some known constants. For e.g: when $\alpha = -10$ and $\beta = 15$, then $y$ can have the following values:
\begin{equation}
y = ax - 10\\
y = ax - 20\\
y = ax \pm 30 \\
y = ax + 45 \\
\cdots
\end{equation}
What is the distribution of $y$ under such conditions? Assume that $a$ and $x$ are deterministic and $a$ is known. Can I use least squares method to estimate $x$ from a set of values of $y$?
 A: 
Can I use least squares method to estimate $x$ from a set of values of $y$?

Here are some fairly weak sufficient conditions ...
We have $$y_i = ax + \eta_i,\quad i \in\{ 1,...,n\},$$
where $a$ is known. The least-squares estimator for the unknown $x$ is easily found to be $\hat{x}=\overline{y}/a$ (assuming $a\ne 0$), where
$\overline{y}=\frac{1}{n}\sum_{i=1}^ny_i.$ 
Thus: 
$$0= \frac{\partial}{\partial x} \sum_{i=1}^n{(y_i-ax)^2}=-2a\sum_{i=1}^n(y_i-ax)\implies  \hat{x}=\frac{1}{a}\,\overline{y}=\frac{1}{a}\frac{1}{n}\sum_{i=1}^ny_i.$$
The following theorem and corollary are easy to prove (proofs omitted):
Theorem. If the $\eta_i$ are pairwise independent with means $E[\eta_i]<\infty$ and variances $V[\eta_i]<\infty$, then
$$\begin{align}
E[\overline{y}]&=ax + \overline{\mu},\quad\text{where } \overline{\mu}=\frac{1}{n}\sum_{i=1}^n E[\eta_i]\\ \\
V[\overline{y}]&=\frac{1}{n}\,\overline{\sigma^2},\quad\text{where } \overline{\sigma^2}=\frac{1}{n}\sum_{i=1}^n V[\eta_i].
\end{align}$$
Corollary. If the $\eta_i$ are pairwise independent with $\frac{1}{n}\sum_{i=1}^n E[\eta_i]=0$ and $V[\eta_i]<\infty$, then
$$\hat{x}=\frac{1}{a}\,\overline{y}=\frac{1}{a}\frac{1}{n}\sum_{i=1}^ny_i$$
is an unbiased consistent estimator for $x$.
