My endogenous variable is very likely to be measured with error, but also confounded by unobservables. I have an instrument that rises with the value of the endogenous variable, at the same time it is exogenous to the the error in the outcome equation.
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An instrumental variable will help with both the OVB and the measurement error given that it is uncorrelated with the error. Because what does this instrument do? If you have an endogenous variable $D$, an outcome $Y$, an instrument $Z$ and other exogenous variables $X$, the problem of the regression $$Y_i = \alpha + \delta D_i + X'_i \beta + \epsilon_i$$ is that $\text{Cov}(D_i,\epsilon_i) \neq 0$ which may be due to omitted variable bias, measurment error, or both. For example, let's say that $\epsilon_i = e_i + W_i$, where $e_i$ is a stochastic error and $W_i$ is the omitted variable which is such that $\text{Cov}(D_i,W_i)\neq 0$ which is the source of the endogeneity problem. Say you then estimate your first stage with the instrument, $$D_i = a + \pi Z_i + X'_i \gamma + \eta_i$$ remember that the regression splits the variation in the dependent into an explained and an unexplained part $\eta_i = u_i + W_i$, where $u_i$ again is a stochastic error term., i.e. $$\text{total sum of squares} = \text{explained sum of squares} + \text{residual sum of squares}$$ When you predict $\widehat{D}_i = a + \pi Z_i + X'_i \gamma$, this separates the "good" variation of $D_i$ from the "bad" variation in $\eta_i$. In the second stage, $$Y_i = \alpha + \delta \widehat{D}_i + X'_i \beta + \epsilon_i$$ you still have $W_i$ in the error term but this time it's not a problem because $\text{Cov}(\widehat{D}_i,W_i) = 0$ and this solves your endogeneity problem, allowing for consistent estimation of $\delta$. This example used the case of omitted variable bias but the same reasoning applies to measurement error and even to both problems together.
Andy
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