RD Estimator Bias

Author

Dor Leventer

Published

June 1, 2023

Setup

Let $Y,R$ denote outcome and running variable. Set cutoff $c=0$, and assume RD sharp treatment rule $T=\boldsymbol{1}\left(R\geq0\right)$. $q=2$ (second order polynomial for the bias estimation) $p=1$ (first order polynomial for the outcome regression estimation), $\nu=0$ (0th order derivative), sample size $n$ of pairs $\left(Y_{i},R_{i}\right)_{i=1}^{n}$ observed data points.

Estimator

Say we are interested in estimating the intercept from above the cutoff.

Denote by $I_{+,i}=\boldsymbol{1}\left(R_{i}\geq c\right)$ the indicator whether the observation is above the cutoff, and by $\mathcal{I}_+=\left\{ i:I_{+,i}=1\right\}$ the set of units above the cutoff, and its size by $n_{+}=\left|\mathcal{I}_{+}\right|$. Also denote the following matrices

\[\begin{equation} Y=\left[\begin{array}{c} Y_{1}\\ \vdots\\ Y_{n} \end{array}\right]_{n_{+}\times1},\quad X\left(h\right)=\left[\begin{array}{cc} 1 & R_{1}\\ \vdots\\ 1 & R_{n} \end{array}\right]_{n_{+}\times2} \end{equation}\]

\[\begin{equation} W_{+}=\left[\begin{array}{ccc} K_{h}\left(R_{1}\right) & & 0\\ & \ddots\\ 0 & & K_{h}\left(R_{n}\right) \end{array}\right]_{n_{+}\times n_{+}} \end{equation}\] where $K_{h}\left(r\right)=K\left(r/h\right)/h$ for some kernel function $K$ and bandwidth $h$.

We can write the estimator of the weighted least squares using a first order polynomial by \[\begin{equation} \widehat{\beta}_{+,1}=\arg\min_{a,b}\left\{ \sum_{i:i\in\mathcal{I}_{+}}K_{h}\left(R_{i}\right)\left[Y-a-bR_{i}\right]^{2}\right\} \end{equation}\]

and in matrix form

\[\begin{equation} \widehat{\beta}_{+,1}=\left(X^{\prime}W_{+}X\right)^{-1}X^{\prime}W_{+}Y \end{equation}\]

Estimator CCT 2014 format

For ease of presentation, assume the first $i=1$ until $i=n_{+}$ units are above the cutoff, i.e. $I_{+,i}=1$. Denote the following vectors and matrices

\[\begin{align*} r_{p}\left(r\right) & =\left[1,r,...,r^{p}\right]^{\prime}\\ X_{+,p}\left(h\right) & =\left[r_{p}\left(R_{1}/h\right),...,r_{p}\left(R_{n_{+}}/h\right)\right]\\ S_{+,p} & =\left[\begin{array}{c} \left(X_{1}/h\right)^{p}\\ \vdots\\ \left(X_{n_{+}}/h\right)^{p} \end{array}\right]_{n_{+}\times1} \end{align*}\]

which together make up the matrices

\[\begin{align*} \Gamma_{+,p}\left(h\right) & =X_{+,p}\left(h\right)^{\prime}W_{+}\left(h\right)X_{+,p}\left(h\right)/n\\ \vartheta_{+,p,q}\left(h\right) & =X_{+,p}\left(h\right)^{\prime}W_{+}\left(h\right)S_{+,q}\left(h\right)/n \end{align*}\]

where we set $p=1$, $q=2$ unless stated otherwise.

Using this notation, we can re-write the estimator as

\[\begin{equation} \widehat{\beta}_{+,1}=H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}X_{+,1}\left(h\right)^{\prime}W_{+}\left(h\right)Y/n \end{equation}\]

where \[\begin{equation} H_{1}=\left[\begin{array}{cc} 1 & 0\\ 0 & h^{-1} \end{array}\right] \end{equation}\]

(its a nice exercise to check whether the two estimators are equivalent)

Taylor Expansion

Denote the conditional outcome mean by $\mu\left(r\right)=\mathbb{E}\left[Y\mid R=r\right]$ and its derivative via $\mu^{\left(v\right)}\left(r\right)=d^{v}\mu\left(r\right)/dr^{v}$.

The (3rd order) taylor expansion to the right of the cutoff at $r=c=0$ is given by \[\begin{equation} \mu_{+}\left(r\right)=\mu_{+}^{\left(0\right)}\left(0\right)\frac{r^{0}}{0!}+\mu_{+}^{\left(1\right)}\left(0\right)\frac{r^{1}}{1!}+\mu_{+}^{\left(2\right)}\left(0\right)\frac{r^{2}}{2!}+\mu_{+}^{\left(3\right)}\left(0\right)\frac{r^{3}}{3!}+TR \end{equation}\] where $TR$ is the remainder from the taylor expansion, which is needed to achive equiality and not approximation of the above equation.

Now, denote for each unit $i$ the taylor expansion \[\begin{equation} \mu\left(R_{i}\right)=\mu\left(0\right)+\mu^{\left(1\right)}\left(0\right)\frac{R_{i}^{1}}{1!}+\mu^{\left(2\right)}\left(0\right)\frac{R_{i}^{2}}{2!}+\mu^{\left(3\right)}\left(0\right)\frac{R_{i}^{3}}{3!}+TR_{i} \end{equation}\]

Denote the vector of taylor expansions at 0 via \[\begin{equation} M=\mu\left(0\right)\left[\begin{array}{c} 1\\ \vdots\\ 1 \end{array}\right]+\mu^{\left(1\right)}\left(0\right)\left[\begin{array}{c} R_{1}\\ \vdots\\ R_{n_{+}} \end{array}\right]+\frac{\mu^{\left(2\right)}\left(0\right)}{2}\left[\begin{array}{c} R_{1}^{2}\\ \vdots\\ R_{n_{+}}^{2} \end{array}\right]+\frac{\mu^{\left(3\right)}\left(0\right)}{6}\left[\begin{array}{c} R_{1}^{3}\\ \vdots\\ R_{n_{+}}^{3} \end{array}\right]+\left[\begin{array}{c} TR_{1}\\ \vdots\\ TR_{n_{+}} \end{array}\right] \end{equation}\]

Now manipulate this a bit to get \[\begin{align*} M & =\mu\left(0\right)\left[\begin{array}{c} 1\\ \vdots\\ 1 \end{array}\right]+h\mu^{\left(1\right)}\left(0\right)\left[\begin{array}{c} R_{1}h^{-1}\\ \vdots\\ R_{n_{+}}h^{-1} \end{array}\right]+\frac{h^{2}\mu^{\left(2\right)}\left(0\right)}{2}\left[\begin{array}{c} R_{1}^{2}h^{-2}\\ \vdots\\ R_{n_{+}}^{2}h^{-2} \end{array}\right]+\frac{h^{3}\mu^{\left(3\right)}\left(0\right)}{6}\left[\begin{array}{c} R_{1}^{3}h^{-3}\\ \vdots\\ R_{n_{+}}^{3}h^{-3} \end{array}\right]+o_{p}\left(h^{3}\right)\\ & =X_{1}\left(h\right)\left[\begin{array}{c} \mu\left(0\right)\\ h\mu^{\left(1\right)}\left(0\right) \end{array}\right]+\frac{h^{2}\mu^{\left(2\right)}\left(0\right)}{2}S_{+,2}+\frac{h^{3}\mu^{\left(3\right)}\left(0\right)}{6}S_{+,3}+o_{p}\left(h^{3}\right) \end{align*}\]

Bias Statement

For $s=v=0$, $l=p=1$, and $\mathcal{R}_{n}=\left[R_{1},...,R_{n}\right]^{\prime}$, we can characterize the conditional expectation of the WLS estimator by \[\begin{align*} \mathbb{E}\left[\widehat{\beta}_{+,1}\left(h\right)\mid\mathcal{R}_{n}\right] & =H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}X_{+,1}\left(h\right)^{\prime}W_{+}\left(h\right)M/n\\ & =H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}X_{+,1}\left(h\right)^{\prime}W_{+}\left(h\right)\\ & \left(X_{1}\left(h\right)\left[\begin{array}{c} \mu\left(0\right)\\ h\mu^{\left(1\right)}\left(0\right) \end{array}\right]+\frac{h^{2}\mu^{\left(2\right)}\left(0\right)}{2}S_{+,2}+\frac{h^{3}\mu^{\left(3\right)}\left(0\right)}{6}S_{+,3}+o_{p}\left(h^{3}\right)\right)/n\\ & =H_{1}\left[\begin{array}{c} \mu\left(0\right)\\ h\mu^{\left(1\right)}\left(0\right) \end{array}\right]\\ & +H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}X_{+,1}\left(h\right)^{\prime}W_{+}\left(h\right)S_{+,2}/n\frac{h^{2}\mu^{\left(2\right)}\left(0\right)}{2}\\ & +H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}X_{+,1}\left(h\right)^{\prime}W_{+}\left(h\right)S_{+,3}/n\frac{h^{3}\mu^{\left(3\right)}\left(0\right)}{6}+o_{p}\left(h^{3}\right)\\ & =\left[\begin{array}{c} \mu\left(0\right)\\ \mu^{\left(1\right)}\left(0\right) \end{array}\right]\\ & +H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}\vartheta_{+,1,2}\left(h\right)\frac{h^{2}\mu^{\left(2\right)}\left(0\right)}{2}\\ & +H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}\vartheta_{+,1,3}\left(h\right)\frac{h^{3}\mu^{\left(3\right)}\left(0\right)}{6}+o_{p}\left(h^{3}\right) \end{align*}\] where the small o is not divided by $n$ since it is cancelled out with the $n^{-1}$ in $\Gamma$.

Denote $B_{+v,p,q}=e_{v}H_{p}\left(\Gamma_{+,p}\left(h\right)\right)^{-1}\vartheta_{+,p,q}\left(h\right)\frac{h^{q}\mu^{\left(q\right)}\left(0\right)}{q!}$. Then for the intercept estimator $\widehat{\mu}_{+,1}^{\left(0\right)}\left(0\right)=e_{0}\widehat{\beta}_{1,+}$ we get \[\begin{equation} \mathbb{\mathbb{E}}\left[\widehat{\mu}_{+,1}^{\left(0\right)}\left(0\right)\mid\mathcal{R}_{n}\right]=\mu_{+}\left(0\right)+e_{0}B_{+,0,1,2}+e_{0}B_{+,0,1,3}+o_{p}\left(h^{3}\right) \end{equation}\]

--- title: "RD Estimator Bias" author: "Dor Leventer" date: "June 2023" format: html: theme: light: cosmo dark: darkly toc: true toc-location: left page-layout: article embed-resources: true code-tools: true editor: visual --- # Setup Let $Y,R$ denote outcome and running variable. Set cutoff $c=0$, and assume RD sharp treatment rule $T=\boldsymbol{1}\left(R\geq0\right)$. $q=2$ (second order polynomial for the bias estimation) $p=1$ (first order polynomial for the outcome regression estimation), $\nu=0$ (0th order derivative), sample size $n$ of pairs $\left(Y_{i},R_{i}\right)_{i=1}^{n}$ observed data points. # Estimator Say we are interested in estimating the intercept from above the cutoff. Denote by $I_{+,i}=\boldsymbol{1}\left(R_{i}\geq c\right)$ the indicator whether the observation is above the cutoff, and by $\mathcal{I}_+=\left\{ i:I_{+,i}=1\right\}$ the set of units above the cutoff, and its size by $n_{+}=\left|\mathcal{I}_{+}\right|$. Also denote the following matrices \begin{equation} Y=\left[\begin{array}{c} Y_{1}\\ \vdots\\ Y_{n} \end{array}\right]_{n_{+}\times1},\quad X\left(h\right)=\left[\begin{array}{cc} 1 & R_{1}\\ \vdots\\ 1 & R_{n} \end{array}\right]_{n_{+}\times2} \end{equation} \begin{equation} W_{+}=\left[\begin{array}{ccc} K_{h}\left(R_{1}\right) & & 0\\ & \ddots\\ 0 & & K_{h}\left(R_{n}\right) \end{array}\right]_{n_{+}\times n_{+}} \end{equation} where $K_{h}\left(r\right)=K\left(r/h\right)/h$ for some kernel function $K$ and bandwidth $h$. We can write the estimator of the weighted least squares using a first order polynomial by \begin{equation} \widehat{\beta}_{+,1}=\arg\min_{a,b}\left\{ \sum_{i:i\in\mathcal{I}_{+}}K_{h}\left(R_{i}\right)\left[Y-a-bR_{i}\right]^{2}\right\} \end{equation} and in matrix form \begin{equation} \widehat{\beta}_{+,1}=\left(X^{\prime}W_{+}X\right)^{-1}X^{\prime}W_{+}Y \end{equation} # Estimator CCT 2014 format For ease of presentation, assume the first $i=1$ until $i=n_{+}$ units are above the cutoff, i.e. $I_{+,i}=1$. Denote the following vectors and matrices \begin{align*} r_{p}\left(r\right) & =\left[1,r,...,r^{p}\right]^{\prime}\\ X_{+,p}\left(h\right) & =\left[r_{p}\left(R_{1}/h\right),...,r_{p}\left(R_{n_{+}}/h\right)\right]\\ S_{+,p} & =\left[\begin{array}{c} \left(X_{1}/h\right)^{p}\\ \vdots\\ \left(X_{n_{+}}/h\right)^{p} \end{array}\right]_{n_{+}\times1} \end{align*} which together make up the matrices \begin{align*} \Gamma_{+,p}\left(h\right) & =X_{+,p}\left(h\right)^{\prime}W_{+}\left(h\right)X_{+,p}\left(h\right)/n\\ \vartheta_{+,p,q}\left(h\right) & =X_{+,p}\left(h\right)^{\prime}W_{+}\left(h\right)S_{+,q}\left(h\right)/n \end{align*} where we set $p=1$, $q=2$ unless stated otherwise. Using this notation, we can re-write the estimator as \begin{equation} \widehat{\beta}_{+,1}=H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}X_{+,1}\left(h\right)^{\prime}W_{+}\left(h\right)Y/n \end{equation} where \begin{equation} H_{1}=\left[\begin{array}{cc} 1 & 0\\ 0 & h^{-1} \end{array}\right] \end{equation} (its a nice exercise to check whether the two estimators are equivalent) # Taylor Expansion Denote the conditional outcome mean by $\mu\left(r\right)=\mathbb{E}\left[Y\mid R=r\right]$ and its derivative via $\mu^{\left(v\right)}\left(r\right)=d^{v}\mu\left(r\right)/dr^{v}$. The (3rd order) taylor expansion to the right of the cutoff at $r=c=0$ is given by \begin{equation} \mu_{+}\left(r\right)=\mu_{+}^{\left(0\right)}\left(0\right)\frac{r^{0}}{0!}+\mu_{+}^{\left(1\right)}\left(0\right)\frac{r^{1}}{1!}+\mu_{+}^{\left(2\right)}\left(0\right)\frac{r^{2}}{2!}+\mu_{+}^{\left(3\right)}\left(0\right)\frac{r^{3}}{3!}+TR \end{equation} where $TR$ is the remainder from the taylor expansion, which is needed to achive equiality and not approximation of the above equation. Now, denote for each unit $i$ the taylor expansion \begin{equation} \mu\left(R_{i}\right)=\mu\left(0\right)+\mu^{\left(1\right)}\left(0\right)\frac{R_{i}^{1}}{1!}+\mu^{\left(2\right)}\left(0\right)\frac{R_{i}^{2}}{2!}+\mu^{\left(3\right)}\left(0\right)\frac{R_{i}^{3}}{3!}+TR_{i} \end{equation} Denote the vector of taylor expansions at 0 via \begin{equation} M=\mu\left(0\right)\left[\begin{array}{c} 1\\ \vdots\\ 1 \end{array}\right]+\mu^{\left(1\right)}\left(0\right)\left[\begin{array}{c} R_{1}\\ \vdots\\ R_{n_{+}} \end{array}\right]+\frac{\mu^{\left(2\right)}\left(0\right)}{2}\left[\begin{array}{c} R_{1}^{2}\\ \vdots\\ R_{n_{+}}^{2} \end{array}\right]+\frac{\mu^{\left(3\right)}\left(0\right)}{6}\left[\begin{array}{c} R_{1}^{3}\\ \vdots\\ R_{n_{+}}^{3} \end{array}\right]+\left[\begin{array}{c} TR_{1}\\ \vdots\\ TR_{n_{+}} \end{array}\right] \end{equation} Now manipulate this a bit to get \begin{align*} M & =\mu\left(0\right)\left[\begin{array}{c} 1\\ \vdots\\ 1 \end{array}\right]+h\mu^{\left(1\right)}\left(0\right)\left[\begin{array}{c} R_{1}h^{-1}\\ \vdots\\ R_{n_{+}}h^{-1} \end{array}\right]+\frac{h^{2}\mu^{\left(2\right)}\left(0\right)}{2}\left[\begin{array}{c} R_{1}^{2}h^{-2}\\ \vdots\\ R_{n_{+}}^{2}h^{-2} \end{array}\right]+\frac{h^{3}\mu^{\left(3\right)}\left(0\right)}{6}\left[\begin{array}{c} R_{1}^{3}h^{-3}\\ \vdots\\ R_{n_{+}}^{3}h^{-3} \end{array}\right]+o_{p}\left(h^{3}\right)\\ & =X_{1}\left(h\right)\left[\begin{array}{c} \mu\left(0\right)\\ h\mu^{\left(1\right)}\left(0\right) \end{array}\right]+\frac{h^{2}\mu^{\left(2\right)}\left(0\right)}{2}S_{+,2}+\frac{h^{3}\mu^{\left(3\right)}\left(0\right)}{6}S_{+,3}+o_{p}\left(h^{3}\right) \end{align*} # Bias Statement For $s=v=0$, $l=p=1$, and $\mathcal{R}_{n}=\left[R_{1},...,R_{n}\right]^{\prime}$, we can characterize the conditional expectation of the WLS estimator by \begin{align*} \mathbb{E}\left[\widehat{\beta}_{+,1}\left(h\right)\mid\mathcal{R}_{n}\right] & =H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}X_{+,1}\left(h\right)^{\prime}W_{+}\left(h\right)M/n\\ & =H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}X_{+,1}\left(h\right)^{\prime}W_{+}\left(h\right)\\ & \left(X_{1}\left(h\right)\left[\begin{array}{c} \mu\left(0\right)\\ h\mu^{\left(1\right)}\left(0\right) \end{array}\right]+\frac{h^{2}\mu^{\left(2\right)}\left(0\right)}{2}S_{+,2}+\frac{h^{3}\mu^{\left(3\right)}\left(0\right)}{6}S_{+,3}+o_{p}\left(h^{3}\right)\right)/n\\ & =H_{1}\left[\begin{array}{c} \mu\left(0\right)\\ h\mu^{\left(1\right)}\left(0\right) \end{array}\right]\\ & +H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}X_{+,1}\left(h\right)^{\prime}W_{+}\left(h\right)S_{+,2}/n\frac{h^{2}\mu^{\left(2\right)}\left(0\right)}{2}\\ & +H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}X_{+,1}\left(h\right)^{\prime}W_{+}\left(h\right)S_{+,3}/n\frac{h^{3}\mu^{\left(3\right)}\left(0\right)}{6}+o_{p}\left(h^{3}\right)\\ & =\left[\begin{array}{c} \mu\left(0\right)\\ \mu^{\left(1\right)}\left(0\right) \end{array}\right]\\ & +H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}\vartheta_{+,1,2}\left(h\right)\frac{h^{2}\mu^{\left(2\right)}\left(0\right)}{2}\\ & +H_{1}\left(\Gamma_{+,1}\left(h\right)\right)^{-1}\vartheta_{+,1,3}\left(h\right)\frac{h^{3}\mu^{\left(3\right)}\left(0\right)}{6}+o_{p}\left(h^{3}\right) \end{align*} where the small o is not divided by $n$ since it is cancelled out with the $n^{-1}$ in $\Gamma$. Denote $B_{+v,p,q}=e_{v}H_{p}\left(\Gamma_{+,p}\left(h\right)\right)^{-1}\vartheta_{+,p,q}\left(h\right)\frac{h^{q}\mu^{\left(q\right)}\left(0\right)}{q!}$. Then for the intercept estimator $\widehat{\mu}_{+,1}^{\left(0\right)}\left(0\right)=e_{0}\widehat{\beta}_{1,+}$ we get \begin{equation} \mathbb{\mathbb{E}}\left[\widehat{\mu}_{+,1}^{\left(0\right)}\left(0\right)\mid\mathcal{R}_{n}\right]=\mu_{+}\left(0\right)+e_{0}B_{+,0,1,2}+e_{0}B_{+,0,1,3}+o_{p}\left(h^{3}\right) \end{equation}