In the vast majority of cases, linear regression models are used in practice as opposed to the more complicated errors-in-variables models. For the sake of example, consider modelling height $Y$ vs weight $X$, or any two appropriate continuous variable of your own choosing - the following is a typical example of what can be found in textbooks/literature: $$ Y = \beta_0 + \beta_1 X + \varepsilon. $$ As far as I am aware, this corresponds to assuming we measure $X$ with no error and we measure $Y$ with error. But we practically always have error when we measure $X$. Which in this case is weight, but in most of the cases you will find in the textbooks/literature, the independent variable comes from a measurement process so it will have some measurement error.
- So when we use the above model for height and weight, is the error $\varepsilon$, which explicitly accounts for measurement error in the response variable, also implicitly accounting for measurement error in $X$? Because in reality, as I just mentioned, there is usually always measurement error in the independent variable.
- If $\varepsilon$ is not implicitly accounting for the measurement error in $X$, then how does this lack of accounting for the measurment error in $X$ manifest itself in the results obtained from linear regression? Since this linear regression model is applied practically everywhere, it seems the deliberate mistake we are making by not accounting for measurement error in $X$ is not so bad?
- Finally, I have read that when the goal is prediction, errors-in-variables provides no benefit over ordinary linear regression, why is this?