16

The meaning of the words first Some people use the word "IV estimator" to refer to any estimator that uses instrumental variables. To them, IV estimators contain 2SLS, LIML, k-class estimators, and others, so 2SLS is a special case of IV. For example, the title of Bekker's (1994, Econometrica) paper is "Alternative approximations to the distribution of ...


7

Actual availability of regressors may be an issue here, but if all four mentioned variables are available, the situation is as @Michael mentioned in a comment: Since $X_2$ is correlated with $Y$, it should be included in the regression specification as a "control". $$Y = \beta_0 + \beta_1X_1 + \beta_2X_2 + u$$ This is intuitive, but it also takes care ...


6

IV estimators are 2SLS estimators. An IV estimator is the sample analog of the form: $\beta = \frac{Cov(Y, Z)}{Cov(X, Z)}$, where $Y$ is the outcome variable, $X$ is the endogenous variable, and $Z$ is the instrumental variable. It can be shown that the 2SLS is of the above form. The advantage of 2SLS estimators over other IV estimators is that 2SLS can ...


6

Let me take an example based on the estimation of returns to education, which has been a well-studied problem. The usual result is that researchers find the 2SLS estimate to be larger than the OLS estimate by approximately 25%-50%, e.g. Card (1999, 2001). Three reasons : An omitted variable that could be negatively correlated with the amount of education. ...


5

Endogenous variables are correlated with the error terms and z is correlated with endogenous variable. Doesn't this imply that z is correlated with error terms? No it doesn't. For mean-centered variables for simplicity, we have for linear models, ENDOGENEITY: $E(xu) \neq 0$ RELEVANCE : $E(xz) \neq 0$ The above conditions do not necessarily imply ...


5

In standard linear regression model $$y = x^\top \beta + \epsilon$$ with exogeneity $\mathbb E[x\epsilon] = \mathbf 0$ you have $K$ parameters because $\beta$ is $K \times 1$ and you have $K$ equations $$\mathbb E[x\epsilon] = \mathbb E[x(y - x^\top \beta)] = \mathbf 0 \Leftrightarrow \mathbb E[xy] = \mathbb E[xx^\top]\beta,$$ where $\mathbb E[xy] = \mathbb ...


5

What you are describing is the so called "forbidden regression", which (in general) does not give consistent estimates. This is a summary of the notes of Ben Williams Consider a (nonlinear) first stage regression of $X$ on the instruments $Z$ giving fitted values (e.g. using a log-log specification): $$ \hat X = \hat \mu(Z) $$ Consider the ...


4

In general regression coefficients obtained from a regression on the entire sample need not be equal to the mean of the coefficients obtained from regressions on the subgroups of the sample. I will discuss the case of a benchmark linear regression but I guess the same should hold more general for IV estimators (although the maths may become a lot more ...


4

Yes, you are correct about the conclusion that $J_{n1}=0$ and hence $C_{n}=J_{n}-J_{n1}=J_{n}$. However, despite the test statistics are the same, the asymptotical distributions and critical values are the same, they are derived from different procedure under different null hypothesis, I prefer to keep using the name C-test instead of Sargan-Hansen test.


3

Generally 2SLS is referred to as IV estimation for models with more than one instrument and with only one endogenous explanatory variable. You can also use two stage least squares estimation for a model with one instrumental variable. It can be shown that IV estimation equals 2SLS estimation when there is one endogenous and one instrumental variable. ...


3

Check out Estimating the Return to Schooling: Progress on Some Persistent Econometric Problems by David Card (2001) ECA I know distance from school has been used in the past as an IV..although it has been criticized.


3

There's no magic. What you have to realize is that the result is conditional on the validity of the assumptions: A) Under the assumption that there is measurement error, then yes, the average of two measurements will be on average closer to the truth than a single opinion. This is very believable. We all do this kind of thing all the time. For example, when ...


3

If your are interested in the statistic relationship between income inequality and economic growth, you should use GDP per capita growth since the others indicate level rather than growth. More generally, the choice of the variable in your (and any other) regression depends on your economics model, otherwise you are just doing some linear projections. A good ...


2

There is no contradiction in two approaches. Baum-Snow (2007) looks at the effect of the Interstate Highway construction on those MSAs which happened to lie on its way. The randomness comes from the fact that when the plan for the Interstate Highway project was considered, the aim was to connect distant areas. Through which MSAs the highway would go was ...


2

Instruments are used as a replacement for an independent variable if we think that independent variable is endogenous. That means, we think it may be correlated with our error term. So in the case of estimating money made by a twin, we have a model: $$\text{salary} = \beta_0 + \beta_1 \cdot \text{guess} + u$$ Where $u$ has standard properties mean zero and ...


2

If you have endogeneity between a dependent variable and error term the use of Instrument variables are the way to go. as long as $\mathbf{COV}(x_1,z_1)\ne0,\mathbf{COV}(x_2,z_2)\ne0,\mathbf{COV}(z_2,e)=0$ and $\mathbf{COV}(z_1,e)=0$ you can do so.


2

Formulas can be checked against infinite online resources, not here. The most widespread use of $x$ and $X$ to distinguish something, is when it is needed to emphasize what is treated in theoretical derivations as a realized value (a fixed number, $x$), and what as a random variable, $X$. When using matrix algebra though bold uppercase denotes a matrix (...


2

I hope this meets your idea for intuition, but equation 1 comes from using the Law of Total Expectation with the independence condition (Condition 1). There are four possible values of $D_i(z)-D_i(w)$. $$D_i(z)=D_i(w)=1$$ $$D_i(z)=D_i(w)=0$$ $$D_i(z)>D_i(w)$$ $$D_i(z)<D_i(w)$$. Consider the LHS of (1). $$E[(D_i(z)-D_i(w))*(Y_i(1)-Y_i(0))]$$ From the ...


2

Essentially, $Cov(u,W) = 0$ is implied by $E(u|W) = E(u)$ by the Law of iterated expectations


2

The question is not totally clear, but I will attempt to give you some guidance. To answer your first questions, confounding variables are not a type of endogenous variable. We do not observe nor are we interested in the confounding variables, which means they are not endogenous variables in our model. You later give the correct definition of an ...


2

...the controls in my IV model are correlated with my instrument? The controls should be in your model precisely because they are correlated with your instrument. In the exogeneity condition $cov(z, \epsilon) = 0$ for the instrument, $\epsilon$ is the error term after controlling for other variables. This condition may not hold without controls. Consider ...


2

This would not make instrument necessarily invalid. For some 2SLS instrument model of form: $$y_i = \beta_0 + \beta_1 \hat{x_i} + \beta_2 k_i +\epsilon_i$$ $$x_i = \pi_0 + \pi_1 z_i + \pi_2 k_i +e_i $$ where $y$ is dependent variable, $x_i$ is the endogenous regressor, $k$ some controls and $z$ instrument the main conditions for instrument validity are: $z$...


2

If yes, then how does this square with the general point that causality/exclusion cannot generally be established with statistical tests... It seems to me that "[exogeneity of IV] cannot generally be established with statistical tests" does not imply that it cannot be tested in specific cases. In this (very specific) context, the exogeneity claim ...


2

Short answer: No. Your model is $Y=\alpha + \beta X + \varepsilon$. Even when $X$ is exogenous, if you regress $Y$ on $X$, $W_1$ and $W_2$, then the OLS estimator is inconsistent (for $\beta$) unless $W_1$ and $W_2$ do not affect $Y$ on average (after controlling for $X$) or $X$ is uncorrelated with $W_1$ and $W_2$. When $X$ is endogenous, there is no reason ...


2

The simple answer is no. Randomness does not imply exogeneity. In my opinion, the following example from Deaton: "Randomization in the tropics, and the search for the elusive keys to economic development" is quite illuminating. Assume that you want to estimate the effect of having a railway-station on poverty. We have data on the poverty rate $P_c$ ...


1

You should include all controls in the second stage, even if they are not instruments. The entire effect of log(mortality) should be mediated through Exprop. This does not mean there cannot be other things causing GDP or Exprop, but log mortality cannot have a effect on GDP except through Exprop.


1

Is this legit and what is the rational behind this? Yes you will find this even as a recommendation in many textbooks (e.g. see Romer Advanced Macroeconomics pp 376) so it is legit although with a caveat. A good instrument should be correlated with the endogenous variable and be able to through it exert an effect on dependent variable. Well lags are more ...


1

I'm going to assume that by "conditions for instrumental variables to work" you mean "instrumental variables is consistent." However, there are other properties to consider, like performance in small samples, etc. $ \newcommand{\Cov}{\text{Cov}} $ In this simple case, the IV estimator in sample of size $n$ is $$ \hat \gamma_2 = \frac{\...


1

Yes it is a problem. The first stage itself has to satisfy the same assumptions that standard OLS would and $cov(Z,\epsilon_1)\neq 0$ would violate them (see A Guide to Modern Econometrics by Verbeek). Furthermore, actually the two conditions you mention are not enough. The instrument should also not be 'weak', that is the first stage should have $F$-...


1

Adding to the excellent answer by @Alecos Papadopoulos, here are two simple numerical examples with $z=x^2+v$, $z^*$ is mean-centred $z$ and $v$ is independent from $u$, in which $E(xu)\neq 0$, $E(xz^*)\neq 0$, but $E(z^*u)=0$. Example 1 (with $x$ having a symmetric distribution) $E(xu)=1.5$, $E(xz^*)=-0.5$, $E(z^*u)=0$ Example 2 (showing that the ...


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