Estimating Treatment Effects with Transform Zeros

We observe covariates $X$, treatment $T$, and outcome $Y$, and want the treatment coefficient $\theta_0$.

Robinson (1988) residualizes treatment and outcome: subtract their conditional means, then estimate the coefficient from the remaining variation.

• Let $g_0(X)=E_P[T\mid X]$, where $P$ is the observed-data law; treatment noise is $\eta=T-g_0(X)$.
• Supplied regression estimates leave covariate-dependent errors in the residuals.
• Mackey et al. (2018) use non-Gaussian treatment noise to support higher-order corrections.

For correction order $r\ge2$, Jin et al. (2025) give an absolute-error quantile upper guarantee proportional to

$$\varepsilon_{1,n}^r\varepsilon_{2,n} + C_{\theta}\varepsilon_{1,n}^{r+1} + (\gamma n)^{-1/2},$$

Here $\varepsilon_{1,n},\varepsilon_{2,n}$ bound treatment- and outcome-regression errors in $L^r$; $C_{\theta}$ bounds the coefficient; $n$ is sample size; and $\gamma\in(1/2,1)$ determines the lower $(1-\gamma)$-quantile. Their eligibility conditions apply.

Finite-order correction leaves nuisance terms that can dominate sampling error. Can treatment-noise information remove them exactly?

A transform zero produces a weight whose expectation vanishes after every real shift of treatment noise.

• Treatment-regression error is a shift depending on $X$.
• Independence of $\eta$ and $X$ turns shift cancellation into conditional cancellation.
• A contour integral combines unknown zeros without locating them individually.

The transform is $M(z)=E_P[e^{z\eta}]$, with $z$ allowed to be complex. A contour integral follows a closed circle in that complex plane.

With the conditional-mean model, independent treatment noise, fixed nonzero higher cumulant, fixed range and tail bounds, i.i.d. sampling, and sufficiently small mean absolute treatment-regression error, our estimator attains minimax mean-squared error of order $n^{-1}$.

The treatment-error condition protects the zeros needed for exact cancellation.

Let $\eta$ have law $\tfrac12 N(-1,1)+\tfrac12 N(1,1)$, where $N(m,v)$ has mean $m$ and variance $v$.

Its transform vanishes at $i\pi/2$, where $i$ is the imaginary unit. The corresponding real weight is $\sin(\pi Z_n/2)$, with $Z_n=T-\bar g_n(X)$ and supplied treatment regression $\bar g_n$.

$$\widehat\theta_{\sin,n} = \begin{cases} \displaystyle \Pi_{[-C_{\theta},C_{\theta}]} \left\{ \frac{\mathbb P_n[Y\sin(\pi Z_n/2)]} {\mathbb P_n[Z_n\sin(\pi Z_n/2)]} \right\}, &\displaystyle \mathbb P_n[Z_n\sin(\pi Z_n/2)]\ge e^{-\pi^2/8}/4,\\[3mm] 0, &\displaystyle \mathbb P_n[Z_n\sin(\pi Z_n/2)]< e^{-\pi^2/8}/4. \end{cases}$$

Here $\mathbb P_n$ is the sample average. The numerator weights outcomes; the denominator weights treatment residuals.

Projection $\Pi$ clips the ratio to the coefficient range. A small denominator triggers output zero.

Why does this weight remove covariate contamination?

The population treatment and outcome regressions are

$$g_0(X)=E_P[T\mid X],\qquad q_0(X)=E_P[Y\mid X]$$

The innovations separate the coefficient from these covariate functions:

$$\eta=T-g_0(X),\qquad \xi=Y-q_0(X)-\theta_0\eta,$$

Thus $\theta_0$ links treatment innovation $\eta$ to the outcome after removing $q_0(X)$; $\xi$ is the remaining outcome innovation.

Require $E_P[\xi\mid X,T]=0$: the conditional outcome mean is $q_0(X)+\theta_0\{T-g_0(X)\}$.

Treat supplied estimates $(\bar g_n,\bar q_n)$ as fixed before drawing the estimation sample, or condition on external training.

The supplied treatment regression is clipped to its known range. Residualizing treatment means subtracting this supplied function.

Independence: $\eta\perp X$. Every covariate value shares the same treatment-noise distribution.

Fixed non-Gaussian separation:

$$|\kappa_k(\eta)|\ge\delta.$$

Here $k=r+1\ge3$, and the cumulant $\kappa_k$ is the $k$th derivative of $\log M$ at zero. The positive separation $\delta$ stays fixed as $n$ grows.

• Range bounds: $|\theta_0|\le C_{\theta}$, $\|g_0\|_{\infty}\le C_g$, and $\|q_0\|_{\infty}\le C_q$. These bound the target and covariate contamination.
• Tail bounds: $E_P[\exp(\eta^2/\psi_{\eta}^2)]\le2$ and $E_P[\exp(\xi^2/\psi_{\xi}^2)\mid X]\le2$. The fixed scales $\psi_{\eta},\psi_{\xi}$ control the exponential weights.
• Sampling: observations are i.i.d.; all range, tail, order, and separation constants are fixed.

In the mixture example, $k=4$ and $\kappa_4(\eta)=-2$.

Compare laws satisfying the model and assumptions with the same supplied regression inputs. Denote this class by $\mathcal P_{\mathrm{NG},n}(p;\mathrm{base})$, where $p$ collects the parameters and “base” fixes those inputs.

Require a nonempty class, $n\ge2$, and $E_P|g_0(X)-\bar g_n(X)|\le\varepsilon_{1,n}$, with

$$\varepsilon_{1,n}\le \{4R_1\exp(2C_gR_1)\}^{-1},$$

Here $R_1$ is a fixed search radius determined by cumulant separation and the treatment-tail bound.

$$\frac{c}{n} \le \mathfrak R_n^{\mathrm{NG}} \le \sup_{P\in\mathcal P_{\mathrm{NG},n}(p;\mathrm{base})} E_P\!\left[(\widehat\theta_{\mathrm{spec},n}-\theta_0(P))^2\right] \le \frac{C}{n},$$

The statistic is $\widehat\theta_{\mathrm{spec},n}$. The minimax risk $\mathfrak R_n^{\mathrm{NG}}$ is the smallest worst-case MSE over estimators using those same inputs. Positive $c,C$ depend only on the fixed constants.

For the mixture example, the sine estimator has MSE at most $C/n$ under the model, independence, sampling, range and outcome-tail bounds, with $\varepsilon_{1,n}\le1/\pi$.

Suppose $M$ has a zero $z_0$ of multiplicity $\ell$: its derivatives through order $\ell-1$ vanish there.

Construct the weight

$$J_{z_0,\ell}(w)=w^{\ell-1}e^{z_0w},\qquad w\in\mathbb C.$$

Here $w$ is the residual at which we evaluate the weight. For every real shift $d$,

$$E_P\!\left[J_{z_0,\ell}(\eta+d)\right]=0.$$

Expanding the shifted polynomial uses only derivatives of $M$ through order $\ell-1$; all vanish at $z_0$.

Independence therefore gives $E_P[J_{z_0,\ell}(Z_n)\mid X]=0$. This removes any outcome component depending only on $X$.

When the denominator is nonzero,

$$\theta_0 = \frac{E_P\!\left[YJ_{z_0,\ell}(Z_n)\right]} {E_P\!\left[Z_nJ_{z_0,\ell}(Z_n)\right]}.$$

This is the population weighted-outcome ratio. The mixture’s sine weight is the imaginary part of its exponential weight.

But individual zeros and their multiplicities are unknown.

Write treatment-regression error as $D_n(X)=g_0(X)-\bar g_n(X)$. Then $Z_n=\eta+D_n(X)$.

Use observable residual and outcome transforms:

$$F_n(z)=E_P[e^{zZ_n}], \qquad G_n(z)=E_P[Ye^{zZ_n}], \qquad H_n(z)=E_P[e^{z\{g_0(X)-\bar g_n(X)\}}].$$

$F_n$ averages exponential residual weights; $G_n$ weights outcomes with them. The unobserved $H_n$ is the transform of regression error.

Independence gives

$$F_n(z)=M(z)H_n(z)$$

Thus residualization multiplies the innovation transform by a nuisance factor.

Let $b_n(X)=q_0(X)-\theta_0D_n(X)$ and $B_n(z)=E_P[b_n(X)e^{zD_n(X)}]$. The conditional-mean model gives

$$G_n(z)=\theta_0(P)F_n'(z)+M(z)B_n(z).$$

The prime means differentiation in $z$. Dividing by $F_n$ separates a coefficient term $\theta_0(P)F_n'/F_n$ from covariate contamination $B_n/H_n$.

Choose a counterclockwise circle $C_j$ with:

• No zero of $F_n$ on its boundary.
• No zero of $H_n$ on or inside it.
• A positive enclosed zero count $N_{C_j}(F_n)$, including multiplicities.

$$(\theta_0(P):\mathbb C) = \Theta_{C_j}(F_n,G_n) = \{N_{C_j}(F_n)\,2\pi i\}^{-1} \oint_{C_j}\frac{G_n(z)}{F_n(z)}\,dz .$$

The integral accumulates the observable ratio around the circle; $\Theta_{C_j}$ divides it by the enclosed count and $2\pi i$.

• The integral of $F_n'/F_n$ equals $2\pi i$ times the zero count.
• The contamination $B_n/H_n$ is analytic—complex differentiable throughout the interior—so its integral is zero.

Each enclosed zero contributes the same coefficient. Dividing by their count recovers $\theta_0$.

What guarantees a usable circle?

The first requirement is that regression error contributes no zeros:

$$\sup_{|z|\le R_1}|H_n(z)-1| \le R_1 e^{2C_gR_1}\varepsilon_{1,n}.$$

This bounds how far the nuisance transform can move from one across the search disk.

Bounded treatment regressions give $|D_n|\le2C_g$. Under the main result’s error gate, the right side is at most $1/4$, so $|H_n|\ge3/4$.

Consequently, $F_n=MH_n$ and $M$ have the same zeros there, including multiplicities.

The mixture makes the same protection concrete:

$$E_P\!\left[ Z_n\sin\!\left(\frac{\pi Z_n}{2}\right) \right] = e^{-\pi^2/8} E_P\!\left[ \cos\!\left(\frac{\pi D_n}{2}\right) \right],$$

This is its population denominator. When $E_P|D_n|\le1/\pi$, it is at least $e^{-\pi^2/8}/2$.

Regression error changes instrument strength while preserving exact cancellation.

Suppose the innovation transform had no zeros throughout a large disk.

• Then $\log M$ would be analytic throughout that disk, with value zero at the origin because $M(0)=1$.
• The sub-Gaussian tail restriction bounds transform growth: the real part of $\log M$ has a quadratic upper envelope.
• Analytic derivative bounds then force derivatives of order $k\ge3$ at the origin to become small as the zero-free disk grows.
• But that derivative is $\kappa_k(\eta)$, whose magnitude must be at least the fixed $\delta$.

These restrictions cannot coexist on an arbitrarily large zero-free disk.

$$A_k=(2^{k+4}k!)^{1/(k-2)},\qquad R_0=A_k(\psi_{\eta}^2/\delta)^{1/(k-2)},\qquad M(z)=E_P[e^{z\eta}].$$

This computes the guaranteed zero radius: $A_k$ depends only on order, $\psi_{\eta}$ controls growth, and $\delta$ prevents higher derivatives from being too small.

Set $R_1=R_0+1$. Every circle with radius strictly between $R_0$ and $R_1$ encloses the guaranteed zero.

Avoiding boundary zeros alone is insufficient: estimation also needs the transform bounded away from zero.

• The fixed tail envelope and normalization $M(0)=1$ bound how many zeros can occur in a slightly larger disk.
• Place finitely many candidate radii between $R_0$ and $R_1$, with enough candidates to avoid small neighborhoods of every zero’s distance from the origin.
• After accounting for these zeros, the same analytic growth control bounds the remaining factor away from zero on a protected circle.

Thus one bank, fixed before seeing the distribution, guarantees a circle with positive count and a fixed positive boundary lower bound $a_{\star}$.

Since $|H_n|\ge3/4$, this protection transfers to $F_n$. The protected index may vary across distributions; the bank and lower bound remain fixed.

Split the sample: a pilot fold selects the circle and count; an independent evaluation fold estimates the coefficient.

For fold $a$, with observation indices $I_a$, compute

$$\widehat F_{a,n}(z)=|I_a|^{-1}\sum_{i\in I_a}\exp(zZ_{n,i}), \qquad \widehat G_{a,n}(z)=|I_a|^{-1}\sum_{i\in I_a}Y_i\exp(zZ_{n,i}),$$

Here $Z_{n,i}=T_i-\bar g_n(X_i)$. These sample averages estimate $F_n$ and $G_n$.

For each pilot circle with a positive boundary denominator, compute its empirical zero count:

$$N_j= \frac{1}{2\pi i}\oint_{C_j} \frac{\widehat F'_{0,n}(z)} {\widehat F_{0,n}(z)}\,dz$$

The logarithmic-derivative integral counts zeros of the pilot transform. Numerical integration isolates its unique integer value.

Retain circles with boundary lower bound at least $a_{\star}/2$ and $N_j\ge1$. Select the retained circle with the strongest lower bound; call its index $j_{\star}$.

Why does its pilot count equal the population count?

A zero count changes only when a zero crosses the contour boundary.

If $\sup_{z\in C_j}|\widehat F_{0,n}(z)-F_n(z)|<\inf_{z\in C_j}|F_n(z)|$, continuously moving from $F_n$ to $\widehat F_{0,n}$ never makes a boundary value zero.

Therefore their enclosed counts agree, even if individual zeros move or split.

Under our model, range, tail and sampling assumptions,

$$E_{P^{\otimes n}}\!\left[\sup_{|z|\le R_1} |\widehat F_{a,n}(z)-F_n(z)|^2\right] + E_{P^{\otimes n}}\!\left[\sup_{|z|\le R_1} |\widehat G_{a,n}(z)-G_n(z)|^2\right] \le \frac{K}{|I_a|}.$$

Here $K$ is fixed. The bound controls both transforms uniformly across the whole search disk, hence every candidate circle.

• Small pilot error makes the guaranteed protected circle eligible.
• Every retained circle then has a population boundary margin larger than that error, so $N_{j_{\star}}=N_{C_{j_{\star}}}(F_n)$.
• Crossing a fixed error threshold has probability at most a constant divided by $n$, by the squared-error bound.

Thus an incorrect normalization is confined to an event whose contribution to MSE can be controlled.

On the independent evaluation fold, check the selected boundary denominator again and approximate

$$\frac{1}{N_{j_{\star}}2\pi i} \oint_{C_{j_{\star}}} \frac{\widehat G_{1,n}(z)} {\widehat F_{1,n}(z)}\,dz$$

This is the empirical identifying integral on the selected circle, normalized by its pilot count.

• Correct normalization: on the small-error event, the pilot count equals the population count.
• Stable division: protected boundary denominators turn transform errors into coefficient errors with a fixed amplification bound.
• Bounded failures: clip the real output to $[-C_{\theta},C_{\theta}]$; failed checks return zero. The squared loss is bounded, so events with probability at most a constant divided by $n$ contribute at most that order to MSE.
• Numerical accuracy: integration error is at most $1/n$.

Exact population cancellation and uniform transform variance yield the $C/n$ upper bound.

For the matching lower bound, keep treatment, regressions and supplied inputs fixed; vary the coefficient with independent Gaussian outcome noise. Coefficients separated by $h/\sqrt n$, for a fixed small $h$, remain difficult to distinguish.

What happens when non-Gaussian separation weakens with sample size?

$$|\kappa_k(\eta_n)|\ge\delta_n \qquad\text{with}\qquad \delta_n\downarrow0$$

Here $\eta_n$ is a changing treatment-noise law and $\delta_n$ its shrinking separation threshold.

• Determine sharp minimax MSE as a function of $(n,\varepsilon_{1,n},\varepsilon_{2,n},\delta_n)$.
• Select from data among double/debiased machine learning, finite-order corrections, and contour procedures.
• Establish uniformly valid inference and matching local minimax lower bounds.

• Transform zeros give weights that cancel every covariate-dependent treatment-regression shift; contour integration removes the remaining covariate contamination exactly.
• Fixed cumulant separation and tail bounds guarantee a localized zero and a protected circle in a finite bank. Small mean absolute treatment error preserves that geometry.
• Uniform transform control preserves the pilot zero count and stabilizes evaluation. Under the model, independence, fixed bounds, i.i.d. sampling and explicit treatment-error gate, the resulting fixed-input minimax MSE is of order $n^{-1}$.

A known innovation-transform zero yields a shift-annihilating instrument and, when its denominator is nonzero, an identifying moment ratio.

A contour with a zero-free boundary, zero-free nuisance factor inside, and positive zero count identifies the coefficient exactly.

With fixed cumulant separation and the treatment-error gate, the finite contour statistic attains matched fixed-code MSE bounds and the stated generalized-quantile bound.