In section 4.1 of this paper, the authors talk about the PLR model:

\begin{aligned} &Y=D \theta_{0}+g_{0}(X)+U, \quad E_{P}[U \mid X, D]=0 \\ &D=m_{0}(X)+V, \quad E_{P}[V \mid X]=0 \end{aligned}

where the parameter of interest is the regression coefficient $\theta_0$. In equation (4.4), the authors provide the score function

$$\psi(W ; \theta, \eta):=\{Y-\ell(X)-\theta(D-m(X))\}(D-m(X)), \quad \eta=(\ell, m)$$ where $W = (Y, D, X)$ and stated that "it's easy to see that $\theta_0$ satisfies the orthogonality condition $\partial_{\eta} E_{P} \psi\left(W ; \theta_{0}, \eta_{0}\right)[\eta-$ $\left.\eta_{0}\right]=0$, for $\eta_{0}=\left(\ell_{0}, m_{0}\right)$, where $\ell_{0}(X)=E_{P}[Y \mid X]$."

I have two questions:

  1. How did the authors arrive at the score function $\psi(W;\theta,\eta)$?
  2. How can one show that the orthogonality condition is satisfied?

1 Answer 1


The first question is somewhat difficult to answer, because it requires speculation about the author's psychology, but let me at least try to give some intuition that it is a sensible score. Recall that what it means for $\psi(W;\theta,\eta)$ to be a score is that at the true values of $\eta_0 = (\ell_0,m_0)$, we have the moment condition

$$E[\psi(W;\theta,\eta_0)] = 0$$

Let us now write out what that entails given the $\psi$ defined above. We end up with the equation

$$E[(Y - \ell_0(X) - \theta_0(D-m_0(X)))\cdot(D-m_0(X))] = 0$$ Using the linearity of expectations and rearranging, we can get a closed form expression for $\theta_0$ given the above equation: $$\theta_0 = \frac{E[(Y-\ell_0(X))(D-m_0(X))]}{E[(D-m_0(X))^2]} = \frac{E[\mathrm{Cov}(Y,D|X)]}{E[\mathrm{Var}(D|X)]}$$

I like to think about this expression for $\theta_0$ in terms of the Frisch-Waugh-Lovell (FWL) theorem. Recall that this theorem states that in the linear model $Y = \beta D + \gamma X + \varepsilon$, $\beta$ is numerically equivalent to the outcome of the model $r_Y = \beta r_D + \delta$ where $r_Y = Y - \mathrm{L}(Y|X_2)$ and $r_D = Y - \mathrm{L}(D|X_2)$ are respectively the residuals from predicting $Y$ and $D$ using the $X$'s (here $\mathrm{L}(A|B)$ is defined to be the best linear predictor of $A$ given $B$). Recall additionally that when $D$ is scalar, the OLS is just the covariance of the outcome and the predictor divided by the variance of the predictor, i.e. $\beta = \frac{\mathrm{Cov}(r_Y,r_D)}{\mathrm{Var}(r_D)}$. Note that the expression for $\theta$ can thus be thought of as the "nonparametric" analogue of the FWL theorem: rather than taking the residual from the best linear predictor, we take the residual from the best predictor, i.e. the conditional expectation function.

Now, addressing your second point, let $\delta_\ell(X),\delta_m(X)$ be two test functions respectively perturbing $\ell$ and $m$. Then Neyman orthogonality states that for choices of $\delta_\ell$ and $\delta_m$, we have

$$\frac{\mathrm d E[\psi(W;\theta,\eta_0 + r(\delta_\ell,\delta_m))]}{\mathrm d r} = 0$$ where the derivative is taken around the point where $r=0$. To prove this, we can simply expand out the definition of $\psi$ to obtain

$$\begin{aligned}E[\psi(W;\theta,\eta_0 + r(\delta_\ell,\delta_m))] &= E[(Y - \ell_0(X)-r \delta_\ell(X))(D - m_0(X) - r\delta_m(X))]\\ &- \theta E[(D-m_0(X)-r\delta_m(X))^2] \end{aligned}$$ Let us first check that the derivative of the first term is mean 0. To do so, we note that by differentiating under the expectation sign around $r=0$, we have $$\begin{aligned} \frac{\mathrm d}{\mathrm dr} E[(Y - \ell_0(X)-r \delta_\ell(X))(D - m_0(X) - r\delta_m(X))] &= E\left[\frac{\mathrm d}{\mathrm dr} (Y - \ell_0(X)-r \delta_\ell(X))(D - m_0(X) - r\delta_m(X))\right]\\ &= -E[(Y-\ell_0(X))\delta_m(X) + \delta_\ell(X)(D- m_0(X))]\\ &= -E\left[\underbrace{E[Y-\ell_0(X)|X]}_{=0} \delta_m(X)\right] - E\left[E[\delta_\ell(X)\underbrace{E[D-m_0(X)|X]}_{=0}\right] = 0 \end{aligned}$$ Note that the two terms above are mean 0 as a result of the fact that $\ell_0$ and $m_0$ are by definition conditional expectation functions.

In light of the above, all that remains to be checked is that the limit of the third term goes to 0 as $r\to 0$. Specifically, we must show $$\frac{\theta\mathrm dE[(D-m_0(X)-r\delta_m(X))^2]}{\mathrm dr} = 0$$ Now, differentiating under the integral sign again, we have $$\begin{aligned}\frac{\mathrm d \theta E[(D-m_0(X)-r\delta_m(X))^2]}{\mathrm dr} &= \theta E\left[\frac{\mathrm d}{\mathrm dr}D-m_0(X)-r\delta_m(X))^2\right] \\&= -2 \theta E[(D-m_0(X))\delta_m(X)]\\ &= \theta E[E[(D-m_0(X))\delta_m(X) | X]]\\ &= \theta E[\underbrace{E[(D-m_0(X))|X]}_{=0}\delta_m(X)]\\ &= \theta E[0|X] = 0\end{aligned}$$ So once again, after some manipulation, this third term equalling 0 is due to the definition of $m_0$ as the conditional expectation function of $D$.

  • $\begingroup$ Thanks. For the second question, do you mean $\delta_m$ instead of $\delta_D$? Also, what exactly is $r(\delta_{\ell}, \delta_m)$ equal to? $\endgroup$
    – Adrian
    Mar 14, 2022 at 3:34
  • $\begingroup$ Is $E[\psi(W, \theta_0, \eta_0 +r(\delta_{\ell}, \delta_m))]$ equal to $E[(Y- \ell_0(X) - r(\ell(X) - \ell_0(X)) - \theta_0(D-m_0(X)-r(m(X)-m_0(X)))(D - m_0(X) - r(m(X) - m_0(X))]=E[(Y- \ell_0(X) - r\delta_{\ell}(X) - \theta_0(D-m_0(X)-r\delta_m(X))(D - m_0(X) - r\delta_m(X))]?$ $\endgroup$
    – Adrian
    Mar 14, 2022 at 3:44
  • $\begingroup$ Yeah exactly, the $\delta$’s are meant to represent arbitrary values of the deviations $\ell-\ell_0$ or $m-m_0$. And yeah, that’s my bad for switching notation in the middle of the derivation. Now edited. $\endgroup$ Mar 14, 2022 at 13:34
  • $\begingroup$ Thanks so much. Could you also elaborate on the expansion you wrote for "To prove this, we can simply expand out the definition of $\psi$ to obtain.."? I tried expanding it myself, but I can't seem to arrive at the same result with the 3 terms. $\endgroup$
    – Adrian
    Mar 14, 2022 at 14:46
  • $\begingroup$ Oh I seem to have made a mistake in my calculations. Let me rewrite that part $\endgroup$ Mar 14, 2022 at 15:13

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