Is the error term a sum of r.v.?

Question

`If in a econometric model I have:

$y = \beta x + u$

where u is the error term, we have:

$u = y - \beta x$

Supposing that $\beta=1$, $y\sim N(0,1)$, $x \sim N(0,1)$ and $x$, $y$ are independent. Is $u$ the sum of two independent standard normal r.v.s?

The answer should be that it isn't but I'm not able to figure it out why.

Answer 1

In the second equation, "Supposing that β=1, y∼N(0,1), x∼N(0,1) and x, y are independent" is defining another model and hence another distribution on $u$, which is unrelated with the $u$ introduced in the first equation.

This mistake is called the fiducial fallacy, named after Fisher's attempt at turning likelihoods into probability distributions. When inverting $y = \beta x + u$ where $x,u\stackrel{\text{iid}}{\sim}\mathbf{N}(0,1)$, into $u=y-\beta x$, $y$ and $x$ are not independent and $u$ remains, both marginally and conditionally (on $x$), a $\mathbf{N}(0,1)$ variate.

Indeed, if we set from the original model, where $y=\beta x +u$, $x,u\stackrel{\text{iid}}{\sim}\mathbf{N}(0,1)$, the joint density of $(x,u)$ is $$\exp\{-(x^2+u^2)/2\}/2\pi\tag{1}$$ hence the joint density of $(y,x)$ is $$\exp\{-(x^2+[y-\beta x]^2)/2\}/2\pi\tag{2}$$ as the Jacobian is equal to one and the joint density of $(y,u)$ is $$\exp\{-(u^2+\beta^{-2}[y-u]^2)/2\}/2\pi\beta\tag{3}$$ as the Jacobian is equal to $1/\beta$. Therefore the

• marginal distribution of $u$ when integrating $x$ out in (1) or $y$ in (3) is $\mathbf{N}(0,1)$
• conditional distribution of $u$ given $x$ in (1) is $\mathbf{N}(0,1)$
• conditional distribution of $u$ given $y$ in (3) is $\mathbf{N}((1+\beta^2)^{-1}y,(1+\beta^{-2})^{-1})$
• distribution of $u$ as the transform $u=y-\beta x$ of $(x,y)$ distributed from (2) is $\mathbf{N}(0,1)$

Answer 2

From the specified value $\beta=1$ you have $u = y - \beta x = y - x = y + (-x)$, and since $-x$ and $y$ are random variables, $u$ is clearly a sum of random variables.