# Variance of OLS estimator with binary treatment

I know that in general, given a (stacked) regression of the form $$y = X \beta + \epsilon$$, where $$\mathbb{V}(\epsilon_i) = \sigma^2 \forall i$$, then letting $$\hat{\beta}$$ denote the OLS estimate of $$\beta$$, $$$$\mathbb{V}(\hat{\beta}) = \sigma^2 (X'X)^{-1}$$$$

This I understand. In my econometrics class, we are studying randomized trials with binary treatment, so $$X \in \{0,1\}^n$$, where $$n$$ is the number of observations. I am told that in this simple setup,

$$$$\mathbb{V}(\hat{\beta}) = \frac{\sigma^2}{p(1-n)p}$$$$

where $$p$$ is the proportion of observations $$i$$ such that $$X_i = 1$$.

I have no idea how to derive the second equation from the first. Any hints?

## 1 Answer

In the case of simple linear regression

$$\underbrace{\begin{bmatrix} y_1 \\ y_2 \\ \vdots\\ y_n \end{bmatrix}}_{Y} = \underbrace{\begin{bmatrix} 1 & x_1 \\ 1 & x_2 \\ \vdots\\ 1 & x_n \end{bmatrix} \begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix}}_{X\beta}.$$

Taking $$X'X$$ gives $$\begin{bmatrix} n & \sum_{i=1}^{n}x_i \\ \sum_{i=1}^{n}x_i & \sum_{i=1}^{n}x_i^2 \end{bmatrix}.$$

After we take the inverse (i.e., $$(X'X)^{-1}$$) , we would find that the element at $$(2,2)$$ is

$$\frac{1}{ \sum_{i=1}^{n}(x_i-\bar{x})^2 }.$$

So $$\mathbb{V}(\beta_1) = \frac{\sigma^2}{ \sum_{i=1}^{n}(x_i-\bar{x})^2 }$$.

Next, we need to make $$\sum_{i=1}^{n}(x_i-\bar{x})^2$$ look like $$np(1-p)$$.

Since $$x$$ is binary, we know $$\bar{x} = \frac{\sum_{i=1}^{n}x_i}{n} = p$$. Therefore, we can rewrite the denominator of $$\mathbb{V}(\beta_1)$$ as $$\sum_{i=1}^{n}(x_i-p)^2$$.

Expanding gives

\begin{align} \sum_{i=1}^{n}(x_i-p)^2 &= \sum_{i=1}^{n}(x_i^2 - 2px_i + p^2)\\ &= np - 2np^2 + np^2\\ &= np - np^2\\ &= np(1-p).\\ \end{align}

Therefore, $$\mathbb{V}(\beta_1) = \frac{\sigma^2}{np(1-p)}$$, which corresponds to $$\mathbb{V}(\beta)$$ in your question.

• Thank you, this was super clear! Commented Apr 17, 2022 at 18:38