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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

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I have been trying to figure this out. i was convinced that there should be no difference between 1 and 2. But then the covariance restriction seemed to make them different. I think you hit on the answer yourself. if you subtract u from the X the distribution for -u is by symmetry the same as the distribution for u. But to keep everything the same in one instance the covariance must be positive but in the other it must simply switch sign.

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