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$$\def\indep{\perp\!\!\!\perp} \def\Er{\mathrm{E}} \def\R{\mathbb{R}} \def\En{{\mathbb{E}_n}} \def\Pr{\mathrm{P}} \newcommand{\norm}[1]{\left\Vert {#1} \right\Vert} \newcommand{\abs}[1]{\left\vert {#1} \right\vert} \DeclareMathOperator*{\argmax}{arg\,max} \DeclareMathOperator*{\argmin}{arg\,min}$$

Using machine learning to estimate causal effects¶

Double debiased machine learning¶

• @chernozhukov2018, @chernozhukov2017

• Parameter of interest $\theta \in \R^{d_\theta}$

• Nuisance parameter $\eta \in T$

• Moment conditions $$\Er[\psi(W;\theta_0,\eta_0) ] = 0 \in \R^{d_\theta}$$ with $\psi$ known

• Estimate $\hat{\eta}$ using some machine learning method

• Estimate $\hat{\theta}$ using cross-fitting

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Cross-fitting¶

• Randomly partition into $K$ subsets $(I_k)_{k=1}^K$
• $I^c_k = {1, …, n} \setminus I_k$
• $\hat{\eta}_k =$ estimate of $\eta$ using $I^c_k$
• Estimator: $$\begin{align*} 0 = & \frac{1}{K} \sum_{k=1}^K \frac{K}{n} \sum_{i \in I_k} \psi(w_i;\hat{\theta},\hat{\eta}_k) \\ 0 = & \frac{1}{K} \sum_{k=1}^K \En_k[ \psi(w_i;\hat{\theta},\hat{\eta}_k)] \end{align*}$$

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Assumptions¶

• Linear score $$\psi(w;\theta,\eta) = \psi^a(w;\eta) \theta + \psi^b(w;\eta)$$
• Near Neyman orthogonality: $$\lambda_n := \sup_{\eta \in \mathcal{T}_n} \norm{\partial \eta \Er\left[\psi(W;\theta_0,\eta_0)[\eta-\eta_0] \right] } \leq \delta_n n^{-1/2}$$

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Assumptions [assumptions-1]¶

• Rate conditions: for $\delta_n \to 0$ and $\Delta_n \to 0$, we have $\Pr(\hat{\eta}_k \in \mathcal{T}_n) \geq 1-\Delta_n$ and $$\begin{align*} r_n := & \sup_{\eta \in \mathcal{T}_n} \norm{ \Er[\psi^a(W;\eta)] - \Er[\psi^a(W;\eta_0)]} \leq \delta_n \\ r_n' := & \sup_{\eta \in \mathcal{T}_n} \Er\left[ \norm{ \psi(W;\theta_0,\eta) - \psi(W;\theta_0,\eta_0)}^2 \right]^{1/2} \leq \delta_n \\ \lambda_n' := & \sup_{r \in (0,1), \eta \in \mathcal{T}_n} \norm{ \partial_r^2 \Er\left[\psi(W;\theta_0, \eta_0 + r(\eta - \eta_0)) \right]} \leq \delta_n/\sqrt{n} \end{align*}$$
• Moments exist and other regularity conditions

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Proof outline:¶

• Let $$\begin{align*} \hat{J} = & \frac{1}{K} \sum_{k=1}^K \En_k [\psi^a(w_i;\hat{\eta}_k)], \\ J_0 = & \Er[\psi^a(w_i;\eta_0)], \\ R_{n,1} = & \hat{J}-J_0 \end{align*}$$

• Show: $$\small \begin{align*} \sqrt{n}(\hat{\theta} - \theta_0) = & -\sqrt{n} J_0^{-1} \En[\psi(w_i;\theta_0,\eta_0)] + \\ & + (J_0^{-1} - \hat{J}^{-1}) \left(\sqrt{n} \En[\psi(w_i;\theta_0,\eta_0)] + \sqrt{n}R_{n,2}\right) + \\ & + \sqrt{n}J_0^{-1}\underbrace{\left(\frac{1}{K} \sum_{k=1}^K \En_k[ \psi(w_i;\theta_0,\hat{\eta}_k)] - \En[\psi(w_i;\theta_0,\eta_0)]\right)}_{R_{n,2}} \end{align*}$$

• Show $\norm{R_{n,1}} = O_p(n^{-1/2} + r_n)$

• Show $\norm{R_{n,2}}= O_p(n^{-1/2} r_n’ + \lambda_n + \lambda_n’)$

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Proof outline: Lemma 6.1¶

Lemma 6.1

1. If $\Pr(\norm{X_m} > \epsilon_m | Y_m) \to_p 0$, then $\Pr(\norm{X_m}>\epsilon_m) \to 0$.

2. If $\Er[\norm{X_m}^q/\epsilon_m^q | Y_m] \to_p 0$ for $q\geq 1$, then $\Pr(\norm{X_m}>\epsilon_m) \to 0$.

3. If $\norm{X_m} = O_p(A_m)$ conditional on $Y_m$ (i.e. for any $\ell_m \to \infty$, $\Pr(\norm{X_m} > \ell_m A_m | Y_m) \to_p 0$), then $\norm{X_m} = O_p(A_m)$ unconditionally

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Proof outline: $R_{n,1}$¶

$$R_{n,1} = \hat{J}-J_0 = \frac{1}{K} \sum_k \left( \En_k[\psi^a(w_i;\hat{\eta}_k)] - \Er[\psi^a(W;\eta_0)] \right)$$

• $$\norm{\En_k[\psi^a(w_i;\hat{\eta}_k)] - \Er[\psi^a(W;\eta_0)]} \leq U_{1,k} + U_{2,k}$$ where $$\begin{align*} U_{1,k} = & \norm{\En_k[\psi^a(w_i;\hat{\eta}_k)] - \Er[\psi^a(W;\hat{\eta}_k)| I^c_k]} \\ U_{2,k} = & \norm{ \Er[\psi^a(W;\hat{\eta}_k)| I^c_k] - \Er[\psi^a(W;\eta_0)]} \end{align*}$$

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Proof outline: $R_{n,2}$¶

• $R_{n,2} = \frac{1}{K} \sum_{k=1}^K \En_k\left[ \psi(w_i;\theta_0,\hat{\eta}_k) - \psi(w_i;\theta_0,\eta_0) \right]$
• $$\sqrt{n} \norm{\En_k\left[ \psi(w_i;\theta_0,\hat{\eta}_k) - psi(w_i;\theta_0,\eta_0) \right]} \leq U_{3,k} + U_{4,k}$$ where

$$\small \begin{align*} U_{3,k} = & \norm{ \frac{1}{\sqrt{n}} \sum_{i \in I_k} \left( \psi(w_i;\theta_0, \hat{\eta}_k) - \psi(w_i;\theta_0,\eta_0) - \Er[ \psi(w_i;\theta_0, \hat{\eta}_k) - \psi(w_i;\theta_0,\eta_0)] \right) } \\ U_{4,k} = & \sqrt{n} \norm{ \Er[ \psi(w_i;\theta_0, \hat{\eta}_k) | I_k^c] - \Er[\psi(w_i;\theta_0,\eta_0)]} \end{align*}$$

• $U_{4,k} = \sqrt{n} \norm{f_k(1)}$ where

$$f_k(r) = \Er[\psi(W;\theta_0,\eta_0 + r(\hat{\eta}_k - \eta_0)) | I^c_k] - \Er[\psi(W;\theta_0,\eta_0)]$$

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Asymptotic normality¶

$$\sqrt{n} \sigma^{-1} (\hat{\theta} - \theta_0) = \frac{1}{\sqrt{n}} \sum_{i=1}^n \bar{\psi}(w_i) + O_p(\rho_n) \leadsto N(0,I)$$

• $\rho_n := n^{-1/2} + r_n + r_n’ + n^{1/2} (\lambda_n +\lambda_n’) \lesssim \delta_n$

• Influence function $$\bar{\psi}(w) = -\sigma^{-1} J_0^{-1} \psi(w;\theta_0,\eta_0)$$ with $$\sigma^2 = J_0^{-1} \Er\left[ \psi(w;\theta_0,\eta_0) \psi(w;\theta_0,\eta_0)'\right] (J_0^{-1})'$$

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Creating orthogonal moments¶

• Need $$\partial \eta\Er\left[\psi(W;\theta_0,\eta_0)[\eta-\eta_0] \right] \approx 0$$

• Given an some model, how do we find a suitable $\psi$?

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Orthogonal scores via concentrating-out¶

• Original model: $$(\theta_0, \beta_0) = \argmax_{\theta, \beta} \Er[\ell(W;\theta,\beta)]$$
• Define $$\eta(\theta) = \beta(\theta) = \argmax_\beta \Er[\ell(W;\theta,\beta)]$$
• First order condition from $\max_\theta \Er[\ell(W;\theta,\beta(\theta))]$ is $$0 = \Er\left[ \underbrace{\frac{\partial \ell}{\partial \theta} + \frac{\partial \ell}{\partial \beta} \frac{d \beta}{d \theta}}_{\psi(W;\theta,\beta(\theta))} \right]$$

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Orthogonal scores via projection¶

• Original model: $m: \mathcal{W} \times \R^{d_\theta} \times \R^{d_h} \to \R^{d_m}$ $$\Er[m(W;\theta_0,h_0(Z))|R] = 0$$
• Let $A(R)$ be $d_\theta \times d_m$ moment selection matrix, $\Omega(R)$ $d_m \times d_m$ weighting matrix, and $$\begin{align*} \Gamma(R) = & \partial_{v'} \Er[m(W;\theta_0,v)|R]|_{v=h_0(Z)} \\ G(Z) = & \Er[A(R)'\Omega(R)^{-1} \Gamma(R)|Z] \Er[\Gamma(R)'\Omega(R)^{-1} \Gamma(R) |Z]^{-1} \\ \mu_0(R) = & A(R)'\Omega(R)^{-1} - G(Z) \Gamma(R)'\Omega(R)^{-1} \end{align*}$$
• $\eta = (\mu, h)$ and $$\psi(W;\theta, \eta) = \mu(R) m(W;\theta, h(Z))$$

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Example: average derivative¶

• $x,y \in \R^1$, $\Er[y|x] = f_0(x)$, $p(x) =$ density of $x$

• $\theta_0 = \Er[f_0’(x)]$

• Joint objective $$\min_{\theta,f} \Er\left[ (y - f(x))^2 + (\theta - f'(x)^2) \right]$$

• Solve for minimizing $f$ given $\theta$ $$f_\theta(x) = \Er[y|x] - \theta \partial_x \log p(x) + f''(x) + f'(x) \partial_x \log p(x)$$

• Concentrated objective: $$\min_\theta \Er\left[ (y - f_\theta(x))^2 + (\theta - f_\theta'(x)^2) \right]$$

• First order condition at $f_\theta = f_0$ gives $$0 = \Er\left[ (y - f_0(x))\partial_x \log p(x) + (\theta - f_0'(x)) \right]$$

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Example : average derivative with endogeneity¶

• $x,y \in \R^1$, $p(x) =$ density of $x$

• Model : $\Er[y - f(x) | z] = 0$ $\theta_0 = \Er[f_0’(x)]$

• Joint objective: $$\min_{\theta,f} \Er\left[ \Er[y - f(x)|z]^2 + (\theta - f'(x))^2 \right]$$

• then $$f_\theta(x) = (T^\ast T)^{-1}\left((T^\ast \Er[y|z])(x) - \theta \partial_x \log p(x)\right)$$

  • where $T:\mathcal{L}^2_{p} \to \mathcal{L}^2_{\mu_z}$ with $(T f)(z) = \Er[f(x) |z]$
  • and $T^\ast :\mathcal{L}^2_{\mu_z} \to \mathcal{L}^2_{p}$ with $(T^\ast g)(z) = \Er[g(z) |x]$

• Orthogonal moment condition : $$0 = \Er\left[ \Er[y - f(x) | z] (T (T^\ast T)^{-1} \partial_x \log p)(z) + (\theta - f'(x)) \right]$$

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Example: average elasticity¶

• Demand $D(p)$, quantities $q$, instruments $z$ $$\Er[q-D(p) |z] = 0$$

• Average elasticity $\theta \Er[D’(p)/D(p) | z ]$

• Joint objective : $$\min_{\theta,D} \Er\left[ \Er[q - D(p)|z]^2 + (\theta - D'(p)/D(p))^2 \right]$$

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Example: control function¶

$$\begin{align*} 0 = & \Er[d - p(x,z) | x,z] \\ 0 = & \Er[y - x\beta - g(p(x,z)) | x,z] \end{align*}$$

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Treatment heterogeneity¶

• Potential outcomes model
  • Treatment $d \in {0,1}$
  • Potential outcomes $y(1), y(0)$
  • Covariates $x$
  • Unconfoundedness or instruments
• Objects of interest:
  • Conditional average treatment effect $s_0(x) = \Er[y(1) - y(0) | x]$
  • Range and other measures of spread of conditional average treatment effect
  • Most and least affected groups

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Fixed, finite groups¶

• $G_1, …, G_K$ finite partition of support $(x)$

• Estimate $\Er[y(1) - y(0) | x \in G_k]$ as above

• pros: easy inference, reveals some heterogeneity

• cons: poorly chosen partition hides some heterogeneity, searching partitions violates inference

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Generic Machine Learning Inference on Heterogenous Treatment Effects in Randomized Experiments¶

• @cddf2018

• Use machine learning to find partition with sample splitting to allow easy inference

• Randomly partition sample into auxillary and main samples

• Use any method on auxillary sample to estimate $$S(x) = \widehat{\Er[y(1) - y(0) | x]}$$ and $$B(x) = \widehat{\Er[y(0)|x]}$$

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Generic Machine Learning Inference on Heterogenous Treatment Effects in Randomized Experiments [generic-machine-learning-inference-on-heterogenous-treatment-effects-in-randomized-experiments-1]¶

• Define $G_k = 1{\ell_{k-1} \leq S(x) \leq \ell_k}$

• Use main sample to regress with weights $(P(x)(1-P(X)))^{-1}$ $$y = \alpha_0 + \alpha_1 B(x) + \sum_k \gamma_k (d-P(X)) 1(G_k) + \epsilon$$

• $\hat{\gamma}_k \to_p \Er[y(1) - y(0) | G_k]$

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Best linear projection of CATE¶

• Randomly partition sample into auxillary and main samples

• Use any method on auxillary sample to estimate $$S(x) = \widehat{\Er[y(1) - y(0) | x]}$$ and $$B(x) = \widehat{\Er[y(0)|x]}$$

• Use main sample to regress with weights $(P(x)(1-P(X)))^{-1}$ $$y = \alpha_0 + \alpha_1 B(x) + \beta_0 (d-P(x)) + \beta_1 (d-P(x))(S(x) - \Er[S(x)]) + \epsilon$$

• $\hat{\beta}0, \hat{\beta}_1 \to_p \argmin \Er[(s_0(x) - b_0 - b_1 (S(x)-E[S(x)]))^2]$

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Inference on CATE¶

• Inference on $\Er[y(1) - y(0) | x] = s_0(x)$ challenging when $x$ high dimensional and/or few restrictions on $s_0$

• Pointwise results for random forests : @wager2018, @athey2016

• Recent review of high dimensional inference : @bcchk2018

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Random forest asymptotic normality¶

• @wager2018

• $\mu(x) = \Er[y|x]$

• $\hat{\mu}(x)$ estimate from honest random forest

  • honest $=$ trees independent of outcomes being averaged

  • sample-splitting or trees formed using another outcome

• Then $$\frac{\hat{\mu}(x) - \mu(x)}{\hat{\sigma}_n(x)} \leadsto N(0,1)$$

  • $\hat{\sigma}_n(x) \to 0$ slower than $n^{-1/2}$

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Random forest asymptotic normality¶

• Results are pointwise, but what about?
  • $H_0: \mu(x_1) = \mu(x_2)$
  • ${x: \mu(x) \geq 0 }$
  • $\Pr(\mu(x) \leq 0)$

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Uniform inference¶

• @bcchk2018
• @bccw2018

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Bibliography¶

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