The classical simple linear regressoion model is $$ Y = \beta_0 + \beta_1 X + \varepsilon. \tag{1} $$

On page 3 of these slides, the author says if there are measurement errors in the outcome then we have $$ Y = \beta_0 + \beta_1 X + \varepsilon, \quad \text{where} \quad Y^* = Y + U, \tag{2} $$ or equivalently $$ Y^* = \beta_0 + \beta_1 X + (\varepsilon + U). $$

I was under the impression that the error $\varepsilon$ in the classical linear regression model already accounted for the measurement error in $Y$, but these slides seem to say that it means something else.

So if $U$ is the measurement error in $Y$, what does $\varepsilon$ represent in both the simple linear regression model $(1)$, and and the measurement error model $(2)$?

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    $\begingroup$ Not all variation in $Y$ (conditional on $X$) can necessarily be attributed to measurement: that variation can be inherent in the underlying population (or process) regardless of how the values are measured. Consider (height, weight) data for people, for instance. $\endgroup$ – whuber Sep 18 '20 at 11:53
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    $\begingroup$ Sometimes part of the thinking here is that measurement errors can be systematic as well as random. More broadly, the same statistical methods can be applied to a range of problems: at one end, measurement error is just about the only source of variation, while at the other measurement error may hardly exist but variability associated with other uncontrolled predictors is the big deal. For example, if you survey your friends and neighbours they will usually know how many cars or cats they have but you won't be able to account for all the variability. $\endgroup$ – Nick Cox Sep 18 '20 at 12:25

$\varepsilon$ represents variation in $Y$ that is not explained by variation in $X$. However, $Y$ may or may not be the observed variable. If $Y$ is the observed variable then $\varepsilon$ will include any associated measurement error. If $Y$ is unobserved but $Y^*=Y+U$ is an observed measurement of $Y$ then, by definition of the measurement process $Y\to Y^*$, $\varepsilon$ does not include measurement error.


Reading the same slide, it says "All that has happened is that the error variance is bigger" -> so U just become part of of $\epsilon$, i.e. both models are equivalent.


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