# How does average treatment effect relate to averages of individual treatment effects? [closed]

I have a balanced panel with two groups, A and B, and run a standard fixed effects regression. I want to derive how the $\beta$ coefficients of these two regressions are related: \begin{eqnarray} y_{it} &=& \beta x_{it} & + \alpha_i + \epsilon_{it} \\ y_{it} &=& \mathbb{I}(i=A) \beta_A x_{it} + \mathbb{I}(i=B) \beta_B x_{it} & +\alpha_i + \epsilon_{it} \end{eqnarray}

In my application, the case arises that the estimate of the average treatment effect is very far from the unweighted averages of the interacted treatment effect estimates, i.e. $\hat{\beta} \neq \frac{\hat{\beta}_A + \hat{\beta}_B}{2}$.

## So I want to understand how OLS does the "weighting" of the two groups when computing $\hat{\beta}$

What I would ideally like to find is a formula that relates $\beta$, $\beta_i$ and the covariances of my $x_{it}$s.

## closed as unclear what you're asking by kjetil b halvorsen, mdewey, Michael Chernick, Carl, Robert LongJun 11 '18 at 10:43

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Well, for a start, there is no $\beta$, only $\beta_A$ and $\beta_B$. So what is this $\hat{\beta}$ supposed to be estimating? – Reinstate Monica May 20 '18 at 1:27
• @Ben, thanks for the comment & edit. I clarified my point: in my original specification, $\hat{beta}$ was not supposed to depend on the group but I have reason to suspect this may not be true in the data I have. – user3096626 May 20 '18 at 10:25
• Possible duplicate of Intuition for recursive least squares – kjetil b halvorsen May 20 '18 at 10:50
• – kjetil b halvorsen May 20 '18 at 10:51
• @kjetilbhalvorsen thanks for the pointer, I would say my question is a special case of RLS. What would be my aim for this question is to answer this special case using just the elementary methods I am familiar with (i.e. basic variance/covariance operations) to get the insight what is going on – user3096626 May 20 '18 at 11:40

Let $\tilde{y}_{it}$ and $\tilde{x}_{it}$ denote the group-demeaned variables. Now we run three bivariate regressions: \begin{eqnarray} \tilde{y}_{it} &=& \beta \tilde{x}_{it} & + \epsilon_{it} \\ \tilde{y}_{it} &=& {\beta}_A \tilde{x}_{it} & + \epsilon_{it} & \text{ if } i=A \\ \tilde{y}_{it} &=& {\beta}_B \tilde{x}_{it} & + \epsilon_{it} & \text{ if } i=B \\ \end{eqnarray}

Now write the estimate of each coefficient as a function of expected values: \begin{eqnarray} \hat{\beta}_{A} = \frac{Cov(\tilde{x}_{At},\tilde{y}_{At})}{Var(\tilde{x}_{At})} = \frac{E(\tilde{x}_{At} \times \tilde{y}_{At}) - E(\tilde{x}_{At})E(\tilde{y}_{At})}{ E(\tilde{x}_{At}^2) - E(\tilde{x}_{At})^2} = \frac{E(\tilde{x}_{At} \times \tilde{y}_{At})}{ E(\tilde{x}_{At}^2)} \end{eqnarray}

where the last equality follows by definition because the tilde variables have been de-meaned by group. Proceeding analogously for group and then $\beta$, I can write my overall coefficient as:

\begin{eqnarray} \hat{\beta} =& \left(\frac{E(\tilde{x}_{At}^2)}{E(\tilde{x}_{At}^2) + E(\tilde{x}_{Bt}^2)}\right)\hat{\beta}_A + \left(\frac{E(\tilde{x}_{Bt}^2)}{E(\tilde{x}_{At}^2) + E(\tilde{x}_{Bt}^2)}\right)\hat{\beta}_B \\ = & \left(\frac{Var(\tilde{x}_{At})}{Var(\tilde{x}_{At}) + Var(\tilde{x}_{Bt})}\right)\hat{\beta}_A + \left(\frac{Var(\tilde{x}_{Bt})}{Var(\tilde{x}_{At}) + Var(\tilde{x}_{Bt})}\right)\hat{\beta}_B \end{eqnarray}

Thus, the "average treatment effect" is a weighted average of the individual treatment effects, where the weight is proportional to the variance of the explanatory variable within the group.

# Numerical Example

require(dplyr); n = 10^6; set.seed(1)
sim.l <- data.frame(t = 1:n) %>% mutate(
e.common = rnorm(n),
x.A = 5 + e.common + rnorm(n),
x.B = (e.common*4 + rnorm(n))/15,
y.A = x.A + rnorm(n),
y.B = x.A + rnorm(n)
)
# most of the variation in x.B is driven by common shock, but var x.B << var x.A
round(cov(sim.l[,-c(1:2)]), digits = 2)

sim.p <- sim.l %>% melt(id.var = 't') %>% mutate(
var = str_extract(variable,'^(.)'),
group = str_extract(variable,'(.)\$')
) %>% dcast(t + group ~  var, value.var = 'value');
lm(y ~ x:group + group, data=sim.p)
lm(y ~ x + group, data=sim.p)


This reproduces the case where the Average treatment effect (shown in column 2) is almost exactly equal to the treatment effect of group A shown in column 1:

===========================================
Model 1         Model 2
-------------------------------------------
(Intercept)       -0.00           -0.46 ***
(0.00)          (0.00)
groupB             5.00 ***        5.46 ***
(0.00)          (0.00)
x:groupA           1.00 ***
(0.00)
x:groupB           3.53 ***
(0.00)
x                                  1.09 ***
(0.00)
-------------------------------------------
R^2                0.49            0.41
Num. obs.    2000000         2000000
RMSE               1.24            1.33
===========================================
*** p < 0.001, ** p < 0.01, * p < 0.05


Based on the formula derived above, we can exactly reproduce these coefficients:

      x.A    x.B    y.A    y.B
x.A 2.0032 0.2669 2.0034 2.0028
x.B 0.2669 0.0756 0.2670 0.2667
y.A 2.0034 0.2670 3.0050 2.0038
y.B 2.0028 0.2667 2.0038 3.0045

beta_A = 2.0028/2.0032
beta_B = 0.2667/0.0756
beta   = (0.0756*beta_b   + 2.0028*beta_a)/(2.0028 + 0.0756)