## Introduction

We introduce sketching, which is an R package that provides a variety of sketching methods. Researchers may perform regressions using a sketch of data of size $$m$$ instead of the full sample of size $$n$$ for a variety of reasons. Lee and Ng (2022) considers the case when the regression errors do not have constant variance and heteroskedasticity robust standard errors would normally be needed for test statistics to provide accurate inference. It is shown in Lee and Ng (2022) that estimates using data sketched by random projections will behave as if the errors were homoskedastic. Estimation by random sampling would not have this property. Before demonstrating the R package, we first provide theoretical backgrounds.

## Least Squares Estimation with Heteroskedastic Errors

Given $$n$$ observations $$\{ (y_i, X_i,Z_i): i=1,\ldots,n \}$$, we consider a linear regression model: \begin{align*} y_i = X_i^T \beta_0 + e_i, \; \end{align*} where $$y_i$$ is the scalar dependent variable, $$X_i$$ is a $$p \times 1$$ vector of regressors, $$\beta_0$$ is a $$p \times 1$$ vector of unknown parameters. The innovation $$e_i$$ is said to be (conditionally) homoskedastic if $$E[e_i^2|X_i]=E[e_i^2]$$. Otherwise, $$e_i$$ is said to be heteroskedastic. In matrix form, the model given above can be written as \begin{align*} y = X \beta_0 + e, \end{align*} where $$y$$ and $$e$$ are $$n \times 1$$ vectors whose $$i$$-th rows are $$y_i$$ and $$e_i$$, respectively, and $$X$$ is the $$n \times p$$ matrix of regressors whose $$i$$-th row is $$X_i^T$$.

We focus on the exogenous regressor case when $$\mathbb E(e_iX_i)=0$$. The least squares estimator $$\hat\beta_{OLS}:=(X^TX)^{-1} X^T y$$ is $$\sqrt{n}$$ consistent and asymptotically normal, i.e., $\sqrt{n}(\hat\beta_{OLS}-\beta_0)\rightarrow_d N(0,V_1)$ as $$n\rightarrow\infty$$, where $V_1 := [\mathbb E(X_iX_i^T) ]^{-1} \mathbb E(e_i^2X_iX_i^T) [\mathbb E(X_iX_i^T)]^{-1}$ is the heteroskedasticity-robust asymptotic variance matrix. Under homoskedasticity, $$V_1$$ becomes $V_0:= \mathbb E(e_i^2) [\mathbb E(X_iX_i^T)]^{-1}.$

Consider testing $$H_0:\beta_2= \bar\beta_2$$, where the subscript 2 refers to an element of $$\beta$$. The point estimates $$\hat\beta$$ can be used to test the null hypothesis employing the $$t$$ test
$\frac{\sqrt{n}(\hat\beta_2-\bar \beta_2)}{\sqrt{[\hat V]_{22}}},$ where $$\hat V$$ is an estimate of either $$V_1$$ or $$V_0$$. The distribution of this test under the null hypothesis crucially depends on the correct standard error $$\sqrt{[\hat V]_{22}/n}$$ being used. Using $$\hat V_0$$ when the robust estimator $$\hat V_1$$ should have been used would lead to inaccurate inference, in the sense of rejecting the null hypothesis too often or not enough.

## Sketched Least Squares Estimation

A sketch of the data $$(y, X)$$ is $$(\tilde{y}, \tilde{X})$$, where $$\tilde{{y}} = \Pi {y}$$, $$\tilde{{X}} = \Pi {X}$$, and $$\Pi$$ is usually an $$m \times n$$ random matrix. The sketched least squares estimator is $$\tilde\beta_{OLS}:=(\tilde X^T\tilde X)^{-1}\tilde X^T \tilde y$$. Even though the sketched regression is based on a sample of size $$m$$, $$\tilde X^T \tilde X$$ and $$\tilde X^T \tilde y$$ can be written as weighted moments in a sample of size $$n$$ as $$\tilde X^T\tilde X = X^T \Pi^T \Pi X$$ and $$\tilde X^T \tilde y = X^T \Pi^T \Pi y$$.

Note that for a general random $$\Pi$$ whose $$(k,i)$$ element is $$\Pi_{ki}$$, the difference between the full and the sketched moments is of the form \begin{align*} %\label{u-stat-form-general} \begin{split} &n^{-1} \left( U^T \Pi^T \Pi V - U^T V \right) \\ %&= n^{-1} \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^{m} U_i \Pi_{ki} \Pi_{kj} V_j %- n^{-1} \sum_{i=1}^n U_i V_i \\ &= n^{-1} \sum_{i=1}^n \psi_i U_i V_i + n^{-1} \sum_{i=1}^n \sum_{j=1, j \neq i}^n U_i \varphi_{ij} V_j \\ &=: T_{n1} + T_{n2}, \end{split} \end{align*} where $$U \in \mathbb{R}^n$$ and $$V \in \mathbb{R}^n$$ are vectors of certain i.i.d. random variables $$(U_i, V_i) \in \mathbb{R}^2$$ that are independent of $$\Pi$$, \begin{align*} \psi_i := \sum_{k=1}^{\textrm{r.dim}(\Pi)} \Pi_{ki}^2 - 1, \; \varphi_{ij} := \sum_{k=1}^{\textrm{r.dim}(\Pi)} \Pi_{ki} \Pi_{kj}, \end{align*} and $$\textrm{r.dim}(\Pi) \in \{m, n\}$$ denotes the row dimension of $$\Pi$$.

There are two classes of sketching schemes to consider. Random sampling schemes have $$\varphi_{ij} = 0$$ for all $$i \neq j$$ because there is only one non-zero entry in each row of $$\Pi$$. In such cases, $$T_{2n}$$ is negligible and $$T_{1n}$$ is the leading term. The second class is random projection schemes with which $$T_{1n}$$ is asymptotically negligible and $$T_{2n}$$ is the leading term.

As examples, we consider Bernoulli sampling (BS) from the first type and countsketch (CS) from the second type. Under BS, the sampling probability is determined by i.i.d. Bernoulli random variables with success probability $$m/n$$. Thus, $$\Pi = \sqrt{\frac{n}{m}} B$$ is an $$n\times n$$ matrix (not $$m\times n$$), where $$B$$ is a diagonal sampling matrix. Under CS, each column of $$\Pi$$ has only one non-zero entry taking on value $$\{ +1, -1\}$$ randomly drawn with equal probability and located uniformly at random.

We now state formal results from Lee and Ng (2022), which makes the following assumptions.

1. The data $$\mathcal{D}_n := \{(y_i, X_i) \in \mathbb{R}^{1+p}: i=1,\ldots,n \}$$ are independent and identically distributed (i.i.d.).

2. $$\mathbb E(e_iX_i)=0$$, $$\mathbb{E}(y_i^4)<\infty$$, $$\mathbb{E}(\| X_i \|^4)<\infty$$, and $$\mathbb{E} ( X_i X_i^T )$$ has full rank $$p$$.

3. The random matrix $$\Pi$$ is independent of $$\mathcal{D}_n$$.

4. $$m = m_n \rightarrow \infty$$ but $$m/n \rightarrow 0$$ as $$n \rightarrow \infty$$, while $$p$$ is fixed.

Then,

1. Under BS, $$m^{1/2} ( \tilde{\beta}_{OLS} - \hat{\beta}_{OLS} \,) \rightarrow_d N (0, V_1 )$$.

2. Under CS, $$m^{1/2} ( \tilde{\beta}_{OLS} - \hat{\beta}_{OLS} \,) \rightarrow_d N (0, V_0 )$$.

Though this theoretical result indicates that both sampling schemes yield asymptotically normal estimates, their asymptotic variances are different, and asymptotic normality holds for different reasons.

## Numerical Illustration

We begin by calling the sketching package and fix the seed for reproducibility.

library(sketching)
seed <- 220526
set.seed(seed)  

To illustrate the usefulness of the package, we use the well-known Angrist and Krueger (1991) dataset. A particular extract of their dataset is included in the package. Specifically, we look at the ordinary least squares (OLS) and two stage least squares (2SLS) estimates of the return to education in columns (1) and (2) of Table IV in their paper. The dependent variable $$y$$ is the log weekly wages, the covariates $$X$$ include years of education, the intercept term and 9 year-of-birth dummies $$(p=11)$$. Following Angrist and Krueger (1991), the instruments $$Z$$ are a full set of quarter-of-birth times year-of-birth interactions $$(q=30)$$. Their idea was that season of birth is unlikely to be correlated with workers’ ability but can affect educational attainment because of compulsory schooling laws. The full sample size is $$n = 247,199$$.

We now define the variables accordingly.

Y <- AK$LWKLYWGE intercept <- AK$CNST
X_end <- AK\$EDUC
X_exg <- AK[,3:11]
X <- cbind(X_exg, X_end)
Z_inst <- AK[,12:(ncol(AK)-1)]
Z <- cbind(X_exg, Z_inst)
fullsample <- cbind(Y,intercept,X)
n <- nrow(fullsample)
d <- ncol(X)

### How to Choose $$m$$

We start with how to choose $$m$$ in this application. Lee and Ng (2020) highlights the tension between a large $$m$$ required for accurate inference, and a small $$m$$ for computation efficiency. From the algorithmic perspective, $$m$$ needs to be chosen as small as possible to achieve computational efficiency. For example, we may set \begin{align*}%\label{smallest-m} m_1 = C_m p \log p \; \text{ or } \; m_1 = C_m p^2, \end{align*} where $$C_m$$ is a constant that needs to be chosen by a researcher. However, statistical analysis often cares about the variability of the estimates in repeated sampling and a larger $$m$$ may be desirable from the perspective of statistical efficiency. An inference-conscious guide $$m_2$$ can be obtained as in Lee and Ng (2020) by targeting the power at $$\bar\gamma$$ of a one-sided $$t$$-test for given nominal size $$\bar\alpha$$. For pre-specified effect size $$c^T(\beta^0-\beta_0)$$, $m_2(m_1)=m_1 S^2(\bar\alpha,\bar\gamma)\left[\frac{\text{se}(c^T \tilde\beta_{OLS})} {c^T(\beta^0-\beta_0)]}\right]^2,$ where $$S(\alpha,\gamma):=\Phi^{-1}(\gamma) +\Phi^{-1}(1-\alpha)$$.

Alternatively, a data-oblivious sketch size for a pre-specified $$\tau_2(\infty)$$ is defined as $\begin{equation*} m_3=n\frac{S^2(\bar\alpha,\bar\gamma)}{\tau_2^2(\infty)}. \end{equation*}$ Note that $$m_3$$ only requires the choice of $$\bar \alpha, \bar\gamma,$$ and $$\tau_2(\infty)$$ which, unlike $$m_2$$, can be computed without a preliminary sketch. The condition $$m / n \rightarrow 0$$ can be viewed as $$\tau_2 (\infty) \rightarrow \infty$$ as $$n \rightarrow \infty$$.

We focus on the data-oblivious sketch size $$m_3$$, as it is simpler to use. We set the target size $$\alpha= 0.05$$ and the target power $$\gamma = 0.8$$. Then, $$S^2(\bar\alpha,\bar\gamma) = 6.18$$. It remains to specify $$\tau_2(\infty)$$, which can be interpreted as the value of $$t$$-statistic when the sample size is really large.

### OLS Estimation Results

For OLS, we take $$\tau_2(\infty) = 10$$, resulting in $$m = 15,283$$ (about 6% of $$n$$).

# choice of m (data-oblivious sketch size)
target_size <- 0.05
target_power <- 0.8
S_constant <- (stats::qnorm(1-target_size) + stats::qnorm(target_power))^2
tau_limit <- 10
m_ols <- floor(n*S_constant/tau_limit^2)
print(m_ols)
#>  15283

As a benchmark, we first obtain the OLS estimate using the full sample.

ys <- fullsample[,1]
reg <- as.matrix(fullsample[,-1])
fullmodel <- lm(ys ~ reg - 1)
# use homoskedasticity-only asymptotic variance
ztest <- lmtest::coeftest(fullmodel, df = Inf)
est <- ztest[(d+1),1]
se <- ztest[(d+1),2]
print(c(est,se))
#>  0.0801594610 0.0003552066
# use heteroskedasticity-robust asymptotic variance
ztest_hc <- lmtest::coeftest(fullmodel, df = Inf,
vcov = sandwich::vcovHC, type = "HC0")
est_hc <- ztest_hc[(d+1),1]
se_hc <- ztest_hc[(d+1),2]
print(c(est_hc,se_hc))
#>  0.0801594610 0.0003946747

We now obtain the OLS estimates using a Bernoulli subsampling.

subsample <- sketch(fullsample, m_ols, method = "bernoulli")
ys <- subsample[,1]
reg <- subsample[,-1]
submodel <- lm(ys ~ reg - 1)
# use homoskedasticity-only asymptotic variance
ztest <- lmtest::coeftest(submodel, df = Inf)
est <- ztest[(d+1),1]
se <- ztest[(d+1),2]
print(c(est,se))
#>  0.080631991 0.001443075
# use heteroskedasticity-robust asymptotic variance
ztest_hc <- lmtest::coeftest(submodel, df = Inf,
vcov = sandwich::vcovHC, type = "HC0")
est_hc <- ztest_hc[(d+1),1]
se_hc <- ztest_hc[(d+1),2]
print(c(est_hc,se_hc))
#>  0.080631991 0.001603487

As another example of random sampling, we now consider uniform sampling.

subsample <- sketch(fullsample, m_ols, method = "unif")
ys <- subsample[,1]
reg <- subsample[,-1]
submodel <- lm(ys ~ reg - 1)
# use homoskedasticity-only asymptotic variance
ztest <- lmtest::coeftest(submodel, df = Inf)
est <- ztest[(d+1),1]
se <- ztest[(d+1),2]
print(c(est,se))
#>  0.078158594 0.001467223
# use heteroskedasticity-robust asymptotic variance
ztest_hc <- lmtest::coeftest(submodel, df = Inf,
vcov = sandwich::vcovHC, type = "HC0")
est_hc <- ztest_hc[(d+1),1]
se_hc <- ztest_hc[(d+1),2]
print(c(est_hc,se_hc))
#>  0.078158594 0.001630221

We now move to random projection schemes. First, we consider countsketch.

subsample <- sketch(fullsample, m_ols, method = "countsketch")
ys <- subsample[,1]
reg <- subsample[,-1]
submodel <- lm(ys ~ reg - 1)
# use homoskedasticity-only asymptotic variance
ztest <- lmtest::coeftest(submodel, df = Inf)
est <- ztest[(d+1),1]
se <- ztest[(d+1),2]
print(c(est,se))
#>  0.07818373 0.00142449
# use heteroskedasticity-robust asymptotic variance
ztest_hc <- lmtest::coeftest(submodel, df = Inf,
vcov = sandwich::vcovHC, type = "HC0")
est_hc <- ztest_hc[(d+1),1]
se_hc <- ztest_hc[(d+1),2]
print(c(est_hc,se_hc))
#>  0.07818373 0.00146307

Next, we consider Subsampled Randomized Hadamard Transform (SRHT). That is, $$\Pi = \sqrt{\frac{n}{m} } S H D$$, $$S \in \mathbb{R}^{m \times n}$$ is a uniform sampling matrix with replacement, $$H \in \mathbb{R}^{n \times n}$$ is a normalized Walsh-Hadamard transform matrix, and $$D \in \mathbb{R}^{n \times n}$$ is a diagonal Rademacher matrix with i.i.d. entries of $$\pm 1$$.

subsample <- sketch(fullsample, m_ols, method = "srht")
ys <- subsample[,1]
reg <- subsample[,-1]
submodel <- lm(ys ~ reg - 1)
# use homoskedasticity-only asymptotic variance
ztest <- lmtest::coeftest(submodel, df = Inf)
est <- ztest[(d+1),1]
se <- ztest[(d+1),2]
print(c(est,se))
#>  0.077694934 0.001417144
# use heteroskedasticity-robust asymptotic variance
ztest_hc <- lmtest::coeftest(submodel, df = Inf,
vcov = sandwich::vcovHC, type = "HC0")
est_hc <- ztest_hc[(d+1),1]
se_hc <- ztest_hc[(d+1),2]
print(c(est_hc,se_hc))
#>  0.077694934 0.001420585

For each sketching scheme, only one random sketch is drawn; hence, the results can change if we redraw sketches. Note that the intercept term is included in the full sample data before applying sketching methods. This is important for random sketching schemes as the observations across different rows are randomly combined.

Remarkably, all sketched estimates are 0.08, reproducing the full sample estimate up to the second digit. The sketched homoskedasticity-only standard errors are also very much the same across different methods. The Eicker-Huber-White standard error (i.e., heteroskedasticity-robust standard error) is a bit larger than the homoskedastic standard error with the full sample. As expected, the same pattern is observed for Bernoulli and uniform sampling, as these sampling schemes preserve conditional heteroskedasticity.

### 2SLS Estimation Results

We now move to 2SLS estimation. For 2SLS, as it is more demanding to achieve good precision, we take $$\tau_2(\infty) = 5$$, resulting in $$m = 61,132$$ (about 25% of $$n$$).

fullsample <- cbind(Y,intercept,X,intercept,Z)
n <- nrow(fullsample)
p <- ncol(X)
q <- ncol(Z)
# choice of m (data-oblivious sketch size)
target_size <- 0.05
target_power <- 0.8
S_constant <- (qnorm(1-target_size) + qnorm(target_power))^2
tau_limit <- 5
m_2sls <- floor(n*S_constant/tau_limit^2)
print(m_2sls)
#>  61132

As before, we first obtain the 2SLS estimate using the full sample.

ys <- fullsample[,1]
reg <- as.matrix(fullsample[,2:(p+2)])
inst <- as.matrix(fullsample[,(p+3):ncol(fullsample)])
fullmodel <- ivreg::ivreg(ys ~ reg - 1 | inst - 1)
# use homoskedasticity-only asymptotic variance
ztest <- lmtest::coeftest(fullmodel, df = Inf)
est <- ztest[(d+1),1]
se <- ztest[(d+1),2]
print(c(est,se))
#>  0.07685568 0.01504165
# use heteroskedasticity-robust asymptotic variance
ztest_hc <- lmtest::coeftest(fullmodel, df = Inf,
vcov = sandwich::vcovHC, type = "HC0")
est_hc <- ztest_hc[(d+1),1]
se_hc <- ztest_hc[(d+1),2]
print(c(est_hc,se_hc))
#>  0.07685568 0.01512252

The 2SLS estimate of the return to education is 0.769; both types of standard errors are almost the same and less precisely estimated than the OLS estimates. Both types of standard errors are almost identical across all sketches for 2SLS.

We now consider a variety of sketching schemes.

# sketching methods for 2SLS
methods <- c("bernoulli","unif","countsketch","srht")
results_2sls <- array(NA, dim = c(length(methods),3))
for (met in 1:length(methods)){
method <- methods[met]
# generate a sketch
subsample <- sketch(fullsample, m_2sls, method = method)
ys <- subsample[,1]
reg <- as.matrix(subsample[,2:(p+2)])
inst <- as.matrix(subsample[,(p+3):ncol(subsample)])
submodel <- ivreg::ivreg(ys ~ reg - 1 | inst - 1)
# use homoskedasticity-only asymptotic variance
ztest <- lmtest::coeftest(submodel, df = Inf)
est <- ztest[(d+1),1]
se <- ztest[(d+1),2]
# use heteroskedasticity-robust asymptotic variance
ztest_hc <- lmtest::coeftest(submodel, df = Inf,
vcov = sandwich::vcovHC, type = "HC0")
est_hc <- ztest_hc[(d+1),1]
se_hc <- ztest_hc[(d+1),2]
results_2sls[met,] <- c(est, se, se_hc)
}
rownames(results_2sls) <- methods
colnames(results_2sls) <- c("est", "non-robust se","robust se")
print(results_2sls)
#>                    est non-robust se  robust se
#> bernoulli   0.08090466    0.02358229 0.02362023
#> unif        0.05960437    0.02412623 0.02467241
#> countsketch 0.10289271    0.02297033 0.02615567
#> srht        0.08491405    0.02403900 0.02412490

The sketched 2SLS estimates vary more than the sketched OLS estimates, reflecting that the 2SLS estimates are less precisely estimated than the OLS estimates. As in the full sample case, both types of standard errors are similar across all sketches for 2SLS.