# Error of linear regression as a result of the coefficients

I'm trying to resolve a paradox. Suppose that I have a bunch of data points $\{x_i,y_i\}$ and calculate a slope and intercept, $m$ and $b$ such that $y=mx+b$. Also, I can calculate the error in $m$, $\sigma_m$ and the error in $b$, $\sigma_b$. Suppose further that the true model is a line $y=ax+c +\epsilon$ where $\epsilon$ is an error term that is normally distributed and has a standard deviation of $\sigma$. If I were to calculate the error in my calculated value as a result of the errors in my coefficients I would get $error^2 = \sigma_m^2 x^2 +\sigma_b^2$. This error increases with increasing $x$, a very strange property given that the standard deviation of y at any given point x is constant, i.e. $\sigma$. What am I missing? What is the resolution of this paradox?

Update: The standard error of the slope $\sigma_m$ is given by $\frac{\sigma^2}{\sum_i (x_i -\bar{x})^2}$ so $\sigma_m$ goes down inversely with the standard deviation of the X values. I thought that perhaps this resolved the paradox. However, x,and $\bar{x}$ are not constrained by the standard deviations. All of the x values can be within two units of each other and yet x may be a thousand, million, billion or more.

• Can you write your definition and derivation of $error^2$? – Emre Mar 15 '16 at 8:17

You have some inherent error in $m$, say that you predict $m=m_0+a$, where $m_0$ is the true value and where you can control the size of $a$ in terms of your error bounds on $m$. Then you get:
$$y=(m_0+a)x+b=m_0x+ax+b.$$
The $ax$ term is what will cause your error to grow as you increase/decrease $x$ toward $\pm\infty$. In other words, your predicted line will ultimately be infinitely far from the true line (as can be visualized by drawing two lines with slightly different slopes).