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I am using two-stage least squares to estimate a local average treatment effect, with 'treatment' being the endogenous variable and 'assignment to treatment' being the instrument. If I use my full sample, I get a LATE of about 31. But when I split my sample into two subsets and estimate a LATE for each subset, I get LATEs of about 27 and 24 for the two subsets. How can it be that the 2SLS estimate for the whole sample is larger than the 2SLS estimate for either subsample? I would have thought that it should be some sort of weighted average of the two subset estimates.

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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 complicated).

Within group regressions

Let $i$ be individual and $g$ be group. Consider the linear within group regressions on each group separately:

$$ y_{i,g} = \alpha_g + \beta_g x_{i,g} + \varepsilon_{i,g},\\ $$

Here, as usual, we assume that $\mathbb{E}(\varepsilon_{i,g}|g ) = \mathbb{E}(\varepsilon_{i,g} x_{i,g}|g) = 0$.

Notice that $\alpha_g$ and $\beta_g$ both depend on $g$, as we do the regression group by group.

To simplify the analysis, assume that we first normalized the values of $x_{i,g}$ such that $\mathbb{E}(x_{i,g}) = 0$. This could change the estimate of the intercept but not of the slope (but will simplify the maths considerably).

Pooled regression

The pooled regression then estimates a common intercept $\alpha$ and a common slope $\beta$ over the entire sample. This then takes the form: $$ y_{i,g} = \alpha + \beta x_{i,g} + \underset{\chi_{i,g}}{\underbrace{\varepsilon_{i,g} + (\alpha_g - \alpha) + (\beta_g - \beta)x_{i,g}}} $$

Computing the biases

If you run such pooled regression of $y_{i,g}$ on $x_{i,g}$, you will set the estimates $\alpha$ and $\beta$ such that the following moment conditions are satisfied: $$ \mathbb{E}(\chi_{i,g} x_{i,g}) = 0 \text{ and } \mathbb{E}(\chi_{i,g}) = 0. $$ Using the definition of $\chi_{i,g}$ above this gives: $$ \begin{align*} &\mathbb{E}(\varepsilon_{i,g}) + \mathbb{E}((\alpha_g - \alpha)) + \mathbb{E}((\beta_g - \beta) x_{i,g}) = 0,\\ &\mathbb{E}(\varepsilon_{i,g} x_{i,g}) + \mathbb{E}((\alpha_g - \alpha) x_{i,g}) + \mathbb{E}((\beta_g - \beta)x_{i,g}^2) = 0. \end{align*} $$ Simplifying and using $\mathbb{E}(\varepsilon_{i,g} x_{i,g}) = \mathbb{E}(\varepsilon_{i,g}) = \mathbb{E}(x_{i,g}) = 0$ (use the law of iterated expectations) gives: $$ \begin{align*} &\mathbb{E}(\alpha_g) - \alpha + \mathbb{E}(\beta_g x_{i,g}) - \beta \mathbb{E}(x_{i,g}) = 0,\\ \to &\mathbb{E}(\alpha_g) - \alpha + \mathbb{E}(\beta_g x_{i,g}) = 0,\\ &\mathbb{E}(\alpha_g x_{i,g}) - \alpha \mathbb{E}(x_{i,g}) + \mathbb{E}(\beta_g x_{i,g}^2) - \beta \mathbb{E}(x_{i,g}^2) = 0,\\ \to &\mathbb{E}(\alpha_g x_{i,g}) + \mathbb{E}(\beta_g x_{i,g}^2) - \beta \mathbb{E}(x_{i,g}^2) = 0 \end{align*} $$ Solving these two equations for $\alpha$ and $\beta$ gives: $$ \begin{align*} &\alpha = \mathbb{E}(\alpha_g) + \mathbb{E}(\beta_g x_{i,g}),\\ & \beta = \frac{\mathbb{E}(\alpha_g x_{i,g}) + \mathbb{E}(\beta_g x_{i,g}^2)}{\mathbb{E}(x_{i,g}^2)} \end{align*} $$ Notice that $\mathbb{E}(\beta_g x_{i,g}^2) = cov( \beta_g, x_{i,g}^2 ) + \mathbb{E}(\beta_g) \mathbb{E}(x_{i,g}^2)$, that $\mathbb{E}(\beta_g x_{i,g}) = cov(\beta_g, x_{i,g})$, that $\mathbb{E}(\alpha_g x_{i,g}) = cov(\alpha_g, x_{i,g})$ and that $\mathbb{E}(x_{i,g}^2) = var(x_{i,g})$. As such: $$ \begin{align*} &\alpha = \mathbb{E}(\alpha_g) + cov(\beta_g, x_{i,g}), \tag{1}\\ &\beta = \mathbb{E}(\beta_g) + \frac{cov(\beta_g, x_{i,g}^2) + cov(\alpha_g, x_{i,g})}{var(x_{i,g})} \tag{2} \end{align*} $$ This shows that $\alpha$ and $\beta$ are equal to the means of $\alpha_g$ and $\beta_g$ plus some terms (biases).

Illustrations

First let's focus on equation $(1)$ wich contains the bias for $\alpha$. The value of $\alpha$ will be different from $\mathbb{E}(\alpha_g)$ if $cov(\beta_g, x_{i,g}) \ne 0$, which requires higher or lower values of $x_{i,g}$ to produce systematically higher or lower values of $\beta_g$.

If $cov(\beta_{g}, x_{i,g}) > 0$, then $\alpha$ will be biased upwards. This is illustrated by the picture below. We have chosen 8 data points such that $\mathbb{E}(x_{i,g}) = 0$. For simplicity, we set $\varepsilon_{i,g} = 0$. The two subgroups are located on the two red lines. These are the within group regression lines.

One group has higher slope $(\beta_g)$ than the other group but both groups have the same value of $\alpha_g$. Notice that points on the steeper red line have (on average) a larger $x_{i,g}$ value, so $cov(\beta_g, x_{i,g}) > 0$.

As for both lines $\alpha_g = 0$, we have $\mathbb{E}(\alpha_g) = 0$. However, the regression line (in green) has a strictly positive intercept $\alpha$.

case1

Now, let's look at the biases of the $\beta$ term. First there is a bias if $cov(\beta_g, x_{i,g}^2)$ is positive. This will occur if more extreme values of $x_{i,g}$ are in groups with higher slope $\beta_g$. An example is given in the figure below. It is clear that the pooled green regression line has a slope that is steeper than the mean of the withing group red regression lines.

case2

Finally, there is a bias if $x_{i,g}$ is correlated with $\alpha_g$ which means that higher values of $x_{i,g}$ are on within group regression lines with higher intercepts. This is illustrated in the picture below: there are two lines with the same slope. We have that $cov(\beta_g, x_{i,g}^2) = 0$ as there is no variation in $\beta_g$. However points on the higher within group regression have on average higher values of $x_{i,g}$, which means that $cov(\alpha_g, x_{i,g}) > 0$. So although $\alpha = \mathbb{E}(\alpha_g)$ we have that $\beta > \mathbb{E}(\beta_g)$: the green line has a steeper slope compared to the red lines.

case3

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  • $\begingroup$ Great thanks for the really thorough answer! $\endgroup$ Jun 15 at 23:38

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