# Measurement error as additive or subtractive?

We have population $Y_i = a + b X_i + e_i$ with $e_i \sim N(0,\sigma^2_e)$, but only have access to $X_i^*$ which is the real $X_i$ combined with some measurement error $u_i \sim N(0,\sigma^2_u)$.

I want to explore the residual variance-covariance matrix $\Sigma$ of the specification with the measurement error under the condition that $Cov(e_i,u_i)=\mathbb{E}[e_iu_i] > 0$. However, I am not sure whether to consider the measurement error as additive or subtractive around $X_i$. This could leave us with either $\Sigma =\mathbb{E}[(\mathbf{e}+b\mathbf{u})(\mathbf{e}+b\mathbf{u})^T]$ or $\Sigma =\mathbb{E}[(\mathbf{e}-b\mathbf{u})(\mathbf{e}-b\mathbf{u})^T]$

Consider (1) $X_i^* = X_i + u_i$, then after rearranging the regression equation the error term will be $(e_i + bu_i)$ and the positive covariance will actually increase the elements of the diagonal of $\Sigma$ and also our Type II errors. However if we choose (2) $X_i^* = X_i - u_i$ then the positive covariance will subtract from each element of the diagonal.

As an example of what I'm talking about, take scenario (1). Then $\textrm{diag}(\Sigma)_1=\mathbb{E}[(e_1 + b u_1)(e_1 + bu_1)] = \sigma^2_e + b^2\sigma^2_u + 2b\mathbb{E}[u_1e_1].$ However if we take (2) then we'd actually subtract off $2b\mathbb{E}[u_1e_1]$.

My problem lies in the fact that I don't see a reason to choose $(1)$ over $(2)$ or vice versa. If the error wasn't $N(0,...)$ then it'd be an easy choice one way or the other.

Or perhaps I'm not respecting that the $cov()$ changes sign dependent on whether I take $(1)$ or $(2)$.

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I guess I don't get the point of the question. $E[e_iu_i]\gt0$ means different things depending on whether the error is additive or subtractive. In the former case, you are stipulating that a positive error in $X$ tends to be associated with a positive error in $Y$. In the latter case, you are stipulating that a positive error in $X$ tends to be associated with a negative error in $Y$. Another way to put it is that you might as well assume a "positive" measurement error; having done that, you need to decide whether to assume $E[e_iu_i]$ is positive or negative. – whuber Sep 8 '12 at 22:38