{ "nbformat": 4, "nbformat_minor": 5, "metadata": { "kernelspec": { "name": "python3", "display_name": "Python 3" }, "language_info": { "name": "python" }, "colab": { "provenance": [] } }, "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Day 53: Adjustment Sets & Bad Controls\n", "\n", "*Beginner to Ninja: Math for Causal Inference*\n", "\n", "Not every covariate belongs in the regression. Confounders must be in; outcome predictors help precision; instruments hurt precision and amplify bias; mediators, colliders, M-structures and descendants of the outcome bias the estimate. Among valid sets, some are more precise than others.\n", "\n", "**Learning outcomes**\n", "\n", "- Classify a candidate control as good, neutral (precision-helping or precision-hurting) or bad, from the DAG.\n", "- Quantify the precision effects of outcome predictors and instruments in a linear model.\n", "- Derive the biases from controlling for a confounded mediator, an M-structure collider and an instrument with unobserved confounding.\n", "- Choose a precise admissible set and avoid the \"Table 2 fallacy\"." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Theory\n", "\n", "### A catalogue of controls\n", "\n", "Let $X$ be the treatment and $Y$ the outcome. For a candidate control $Z$ (and possibly unobserved $U$):\n", "\n", "| Structure | Verdict | Why |\n", "|---|---|---|\n", "| $X \\leftarrow Z \\to Y$ | **good** | blocks a back-door path |\n", "| $X \\leftarrow Z \\leftarrow U \\to Y$ | **good** | blocks the path through $U$ (a proxy) |\n", "| $Z \\to Y$ only | **neutral, helps precision** | explains outcome variance |\n", "| $Z \\to X$ only | **neutral, hurts precision** | removes treatment variation |\n", "| $Z \\to X$, $X \\leftarrow U \\to Y$ | **bad: bias amplification** | removes clean variation, keeps confounded variation |\n", "| $X \\to Z \\to Y$ | **bad** (for the total effect) | blocks part of the effect |\n", "| $X \\to Z \\leftarrow Y$, or $X \\to Z \\leftarrow U \\to Y$ | **bad** | conditioning on a collider |\n", "| $X \\leftarrow U_1 \\to Z \\leftarrow U_2 \\to Y$ | **bad: M-bias** | pre-treatment collider |\n", "| $Y \\to Z$ | **bad** | selection on the outcome |\n", "\n", "\"Pre-treatment\" is not the same as \"safe\": the M-structure collider can be measured before treatment.\n", "\n", "### Precision in a linear model\n", "\n", "In $Y = \\tau X + \\boldsymbol\\gamma^\\top\\mathbf Z + \\varepsilon$ estimated by OLS, Frisch–Waugh–Lovell gives\n", "\n", "$$\n", "\\operatorname{Var}(\\hat\\tau) \\approx \\frac{\\sigma^2_{Y \\mid X, \\mathbf Z}}{n\\,\\operatorname{Var}(X \\mid \\mathbf Z)}.\n", "$$\n", "\n", "- An outcome predictor shrinks the numerator: more precision.\n", "- A treatment predictor shrinks the denominator: less precision (the variance inflation factor $1/(1 - R^2_{X \\mid \\mathbf Z})$).\n", "\n", "### Choosing among admissible sets\n", "\n", "Several sets may satisfy the back-door criterion. In linear models the asymptotically most precise one is the **optimal adjustment set**: roughly, the parents of $Y$ (and of mediators on the causal path) that are not themselves descendants of $X$, excluding $X$. Practical rule: prefer confounders close to the outcome over confounders close to the treatment.\n", "\n", "### The Table 2 fallacy\n", "\n", "A regression designed to estimate the effect of $X$ has coefficients on the controls too. They are generally **not** causal effects of the controls: $X$ may be a mediator or collider for them. Report only the coefficient the design was built for.\n", "\n", "## Part 1 · Warm-up" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 1 · Classify the controls\n", "\n", "For the effect of $X$ on $Y$, classify $Z$: (a) $Z \\to X$, $Z \\to Y$; (b) $X \\to Z$, $Y \\to Z$; (c) $Z \\to Y$ only; (d) $Z \\to X$ only, no confounding; (e) $X \\to Z \\to Y$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** (a) Confounder: good.\n", "\n", "**Step 2.** (b) Collider: bad.\n", "\n", "**Step 3.** (c) Outcome predictor: neutral, improves precision.\n", "\n", "**Step 4.** (d) Instrument-like: neutral for bias, reduces precision.\n", "\n", "**Step 5.** (e) Mediator: bad for the total effect.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 2 · Precision from an outcome predictor\n", "\n", "$Y = \\tau X + 2Z + \\varepsilon$, with $Z \\perp\\!\\!\\!\\perp X$, $\\operatorname{Var}(Z) = 1$, $\\operatorname{Var}(\\varepsilon) = 1$. By what factor does including $Z$ reduce the standard error of $\\hat\\tau$?\n", "\n", "
Show solution\n", "\n", "**Step 1.** Without $Z$: residual variance $4\\operatorname{Var}(Z) + 1 = 5$.\n", "\n", "**Step 2.** With $Z$: residual variance 1.\n", "\n", "**Step 3.** $\\operatorname{Var}(X \\mid Z) = \\operatorname{Var}(X)$ since they are independent, so the denominator is unchanged.\n", "\n", "**Answer:** variance falls by a factor of 5; the SE by $\\sqrt5 \\approx 2.24$.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 3 · Precision lost to an instrument\n", "\n", "$X = Z + u$ with $\\operatorname{Var}(Z) = \\operatorname{Var}(u) = 1$, and $Z$ affects $Y$ only through $X$ (no confounding). By what factor does controlling for $Z$ raise $\\operatorname{Var}(\\hat\\tau)$?\n", "\n", "
Show solution\n", "\n", "**Step 1.** $\\operatorname{Var}(X) = 2$ and $\\operatorname{Var}(X \\mid Z) = \\operatorname{Var}(u) = 1$.\n", "\n", "**Step 2.** The residual variance of $Y$ is unchanged (Z has no direct effect on $Y$).\n", "\n", "**Answer:** the variance doubles ($R^2_{X \\mid Z} = 0.5$); the SE rises by $\\sqrt2$.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 4 · M-bias\n", "\n", "In $U_1 \\to X$, $U_1 \\to M$, $U_2 \\to M$, $U_2 \\to Y$, $X \\to Y$, with $U_1, U_2$ unobserved, is $\\varnothing$ admissible? Is $\\{M\\}$?\n", "\n", "
Show solution\n", "\n", "**Step 1.** The only back-door path is $X \\leftarrow U_1 \\to M \\leftarrow U_2 \\to Y$, with a collider at $M$.\n", "\n", "**Step 2.** Given $\\varnothing$ it is blocked: $\\varnothing$ is admissible.\n", "\n", "**Step 3.** Given $M$ it opens: $\\{M\\}$ is not admissible, although $M$ is pre-treatment and correlated with both $X$ and $Y$.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 5 · Controlling for a consequence of the outcome\n", "\n", "Why is it a mistake to control for $D$ when $Y \\to D$ (for example, $D$ = hospital discharge diagnosis, $Y$ = disease severity)?\n", "\n", "
Show solution\n", "\n", "**Step 1.** $X \\to Y \\to D$: $D$ is a descendant of $X$, violating the back-door condition.\n", "\n", "**Step 2.** Conditioning on $D$ partially conditions on $Y$ itself: among units with the same $D$, variation in $Y$ is restricted, which attenuates (or distorts) the estimated effect.\n", "\n", "**Answer:** a descendant of the outcome is a bad control.\n", "\n", "
\n", "\n", "## Part 2 · Proofs" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 6 · The variance formula\n", "\n", "In $Y = \\tau X + \\boldsymbol\\gamma^\\top\\mathbf Z + \\varepsilon$ with homoskedastic $\\varepsilon$, show $\\operatorname{Var}(\\hat\\tau \\mid X, \\mathbf Z) = \\sigma^2/\\sum_i \\tilde X_i^2$, where $\\tilde X$ is $X$ residualised on $\\mathbf Z$, and conclude the approximation in the theory.\n", "\n", "
Show solution\n", "\n", "**Step 1.** FWL: $\\hat\\tau = \\sum_i \\tilde X_i Y_i/\\sum_i \\tilde X_i^2$.\n", "\n", "**Step 2.** Substitute $Y_i = \\tau X_i + \\boldsymbol\\gamma^\\top\\mathbf Z_i + \\varepsilon_i$; since $\\tilde X$ is orthogonal to $\\mathbf Z$ and $\\sum \\tilde X_i X_i = \\sum \\tilde X_i^2$, $\\hat\\tau = \\tau + \\sum \\tilde X_i\\varepsilon_i/\\sum \\tilde X_i^2$.\n", "\n", "**Step 3.** $\\operatorname{Var}(\\hat\\tau \\mid X, \\mathbf Z) = \\sigma^2\\sum \\tilde X_i^2/(\\sum \\tilde X_i^2)^2 = \\sigma^2/\\sum \\tilde X_i^2$.\n", "\n", "**Step 4.** $\\sum \\tilde X_i^2 \\approx n\\operatorname{Var}(X \\mid \\mathbf Z)$ for large $n$. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 7 · Bias amplification\n", "\n", "Let $X = aZ + bU + e_X$ and $Y = \\tau X + cU + e_Y$, with $Z, U, e_X, e_Y$ independent, $\\operatorname{Var}(Z) = \\operatorname{Var}(U) = 1$, $\\operatorname{Var}(e_X) = s^2$, and $U$ unobserved. Show the bias of $\\hat\\tau$ is $cb/(a^2 + b^2 + s^2)$ without $Z$ and $cb/(b^2 + s^2)$ with $Z$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Without $Z$: the slope is $\\tau + c\\operatorname{Cov}(X, U)/\\operatorname{Var}(X) = \\tau + cb/(a^2 + b^2 + s^2)$ (Day 50, Exercise 8).\n", "\n", "**Step 2.** With $Z$: by FWL, use $\\tilde X = X - aZ = bU + e_X$.\n", "\n", "**Step 3.** Slope $= \\operatorname{Cov}(\\tilde X, Y)/\\operatorname{Var}(\\tilde X)$, and $\\operatorname{Cov}(\\tilde X, Y) = \\tau\\operatorname{Var}(\\tilde X) + cb$.\n", "\n", "**Step 4.** So the bias is $cb/(b^2 + s^2)$, which is larger: controlling for $Z$ removed clean variation ($a^2$) from the denominator. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 8 · Outcome predictors do not change the estimand\n", "\n", "Assume $\\mathbf Z$ is admissible and $W$ is independent of $(X, \\mathbf Z)$ and affects only $Y$. Prove that $\\mathbf Z \\cup \\{W\\}$ is admissible.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Condition 1: $W$ has no parents among the descendants of $X$ (it is independent of $X$ with arrows only into $Y$), so it is not a descendant of $X$.\n", "\n", "**Step 2.** Condition 2: every path through $W$ is $\\cdots Y \\leftarrow W$ with $W$ as an endpoint-adjacent node; $W$ has no other neighbours, so no back-door path from $X$ passes through $W$.\n", "\n", "**Step 3.** Conditioning on $W$ cannot open a path: $W$ is not a collider or a descendant of one on any $X$–$Y$ path.\n", "\n", "**Step 4.** Paths blocked by $\\mathbf Z$ remain blocked. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 9 · Controlling for a confounded mediator\n", "\n", "Let $M = aX + U + e_M$ and $Y = dX + bM + cU + e_Y$, with $X, U, e_M, e_Y$ independent, $\\operatorname{Var}(U) = v_U$, $\\operatorname{Var}(e_M) = v_M$. Show that the coefficient on $X$ in the regression of $Y$ on $(X, M)$ is $d - cak$, with $k = v_U/(v_U + v_M)$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Given $X$ and $M$, we know $U + e_M = M - aX$. By normal (or linear) projection, $E[U \\mid X, M] = k(M - aX)$.\n", "\n", "**Step 2.** $E[Y \\mid X, M] = dX + bM + cE[U \\mid X, M] = dX + bM + ck(M - aX)$.\n", "\n", "**Step 3.** Collect terms: $(d - cak)X + (b + ck)M$.\n", "\n", "**Answer:** the coefficient is the direct effect $d$ only if $c = 0$ (no mediator–outcome confounding). Otherwise $M$ is a collider on $X \\to M \\leftarrow U \\to Y$. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 10 · The size of M-bias\n", "\n", "In Exercise 4 with linear equations $X = \\alpha U_1 + e_X$, $M = U_1 + U_2 + e_M$, $Y = \\tau X + \\beta U_2 + e_Y$ (all disturbances independent with variance 1), show that controlling for $M$ adds bias $-\\alpha\\beta/(2\\alpha^2 + 3)$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** $E[Y \\mid X, M] = \\tau X + \\beta E[U_2 \\mid X, M]$.\n", "\n", "**Step 2.** $\\operatorname{Var}(X) = \\alpha^2 + 1$, $\\operatorname{Var}(M) = 3$, $\\operatorname{Cov}(X, M) = \\alpha$; $\\operatorname{Cov}(U_2, X) = 0$, $\\operatorname{Cov}(U_2, M) = 1$.\n", "\n", "**Step 3.** Projection coefficients: $(0, 1)\\begin{pmatrix} \\alpha^2 + 1 & \\alpha \\\\ \\alpha & 3 \\end{pmatrix}^{-1} = \\dfrac{(-\\alpha,\\; \\alpha^2 + 1)}{3(\\alpha^2 + 1) - \\alpha^2}$.\n", "\n", "**Step 4.** The coefficient on $X$ in $E[U_2 \\mid X, M]$ is $-\\alpha/(2\\alpha^2 + 3)$.\n", "\n", "**Answer:** bias $-\\alpha\\beta/(2\\alpha^2 + 3)$; with $\\alpha = \\beta = 1$ it is −0.2. Usually modest, but real. $\\blacksquare$\n", "\n", "
\n", "\n", "## Part 3 · Applied" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 11 · Medicaid expansion and mortality\n", "\n", "You study the effect of a state's Medicaid expansion $X$ on county mortality $Y$. Classify each candidate control: (a) pre-expansion county poverty rate; (b) post-expansion insurance coverage; (c) post-expansion number of hospital closures, affected by both expansion and by local death rates through hospital finances; (d) county rainfall; (e) pre-expansion mortality trend.\n", "\n", "
Show solution\n", "\n", "**Step 1.** (a) Affects expansion decisions and mortality: confounder, include.\n", "\n", "**Step 2.** (b) Mediator: expansion works through coverage. Exclude for the total effect.\n", "\n", "**Step 3.** (c) Descendant of $X$ and of $Y$: collider-like and a descendant of the outcome. Exclude.\n", "\n", "**Step 4.** (d) Unrelated to expansion; weakly predicts mortality at most. Harmless; include only if it explains outcome variance.\n", "\n", "**Step 5.** (e) Strong predictor of mortality and plausibly of expansion: include (good for bias and precision).\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 12 · A cash-transfer RCT\n", "\n", "In a randomised cash-transfer trial with 2,000 households, baseline consumption explains 60% of the variance of endline consumption. A field officer also proposes controlling for the household's distance to the payment point, which affected take-up but not consumption. What happens to the standard error of the effect of being offered the transfer with each control?\n", "\n", "
Show solution\n", "\n", "**Step 1.** Offer is randomised, so both controls leave the estimand unchanged.\n", "\n", "**Step 2.** Baseline consumption: residual variance falls to 40%; SE falls by $\\sqrt{0.4} \\approx 0.63$, i.e. 37% smaller.\n", "\n", "**Step 3.** Distance: it predicts take-up, not the random offer itself, and not consumption; it neither helps nor hurts the offer coefficient (beyond a tiny degrees-of-freedom cost).\n", "\n", "**Answer:** control for baseline consumption; distance is irrelevant for the intention-to-treat effect. (If the treatment were take-up rather than the offer, distance would be an instrument and should not be a control.)\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 13 · Bias amplification in returns to schooling\n", "\n", "Schooling $S = 0.8D + 1.0A + e_S$, where $D$ is distance to the nearest college (standardised, negative effect coded positive), $A$ unobserved ability (variance 1), $\\operatorname{Var}(e_S) = 1$; log wage $Y = 0.08S + 0.10A + e_Y$. Compute the ability bias with and without controlling for $D$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Use Exercise 7 with $a = 0.8$, $b = 1$, $s^2 = 1$, $c = 0.10$.\n", "\n", "**Step 2.** Without $D$: $0.10 \\times 1/(0.64 + 1 + 1) = 0.10/2.64 \\approx 0.038$.\n", "\n", "**Step 3.** With $D$: $0.10/(1 + 1) = 0.050$.\n", "\n", "**Answer:** controlling for distance raises the bias from 3.8 to 5.0 points (estimated return 11.8% to 13.0% instead of 8%). Distance belongs in an IV analysis (Phase 5), not in the control list.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 14 · Choosing the most precise admissible set\n", "\n", "In a smoking-cessation study, $A$ (age) $\\to X$ (counselling), $A \\to B$ (baseline lung function), $B \\to Y$ (lung function at one year), $X \\to Y$. Both $\\{A\\}$ and $\\{B\\}$ are admissible. In the simulation below, which gives the smaller standard error, and why?\n", "\n", "
Show solution\n", "\n", "**Step 1.** Back-door path: $X \\leftarrow A \\to B \\to Y$, blocked by $A$ (fork) or by $B$ (chain).\n", "\n", "**Step 2.** $B$ is a parent of $Y$: controlling for it removes the large outcome variance it explains, shrinking the numerator of the variance formula.\n", "\n", "**Step 3.** $B$ is only a noisy proxy for $A$, so $\\operatorname{Var}(X \\mid B) > \\operatorname{Var}(X \\mid A)$: more treatment variation is kept, enlarging the denominator.\n", "\n", "**Step 4.** $A$ does the opposite on both counts: it predicts $X$ strongly and $Y$ only through $B$.\n", "\n", "**Answer:** $\\{B\\}$ is more precise; the simulation shows an SE ratio of about 0.38. Prefer confounders near the outcome.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 15 · The Table 2 fallacy\n", "\n", "A paper estimates the effect of air pollution $P$ on asthma admissions $Y$, adjusting for temperature $T$ and income $I$ (both confounders; temperature also affects pollution). Its table reports coefficients for $P$, $T$ and $I$. A journalist writes \"temperature has no effect on asthma once pollution is accounted for\". What is wrong?\n", "\n", "
Show solution\n", "\n", "**Step 1.** The regression was designed so that $\\{T, I\\}$ is an admissible set for $P$.\n", "\n", "**Step 2.** For temperature, pollution is a **mediator** ($T \\to P \\to Y$). Conditioning on $P$ blocks that channel, so the $T$ coefficient is at most a direct effect, and possibly confounded by variables not needed for $P$.\n", "\n", "**Answer:** the $T$ coefficient is not the total effect of temperature; only the coefficient on $P$ has the causal interpretation the design supports.\n", "\n", "
\n", "\n", "## Verify in Python" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "code", "metadata": {}, "execution_count": null, "outputs": [], "source": [ "import numpy as np\n", "\n", "rng = np.random.default_rng(53)\n", "def coef_se(y, X):\n", " X = np.column_stack([np.ones(len(y))] + X)\n", " b, *_ = np.linalg.lstsq(X, y, rcond=None); r = y - X @ b\n", " V = r @ r / (len(y) - X.shape[1]) * np.linalg.inv(X.T @ X)\n", " return b[1], np.sqrt(V[1, 1])\n", "\n", "n = 200_000\n", "# Exercise 7 / 13: bias amplification\n", "D = rng.normal(size=n); A = rng.normal(size=n)\n", "S = .8 * D + A + rng.normal(size=n); Y = .08 * S + .10 * A + rng.normal(size=n)\n", "print(\"schooling: no control %.4f, control for distance %.4f (truth 0.08)\" % (coef_se(Y, [S])[0], coef_se(Y, [S, D])[0]))\n", "\n", "# Exercise 9: confounded mediator (a=1, b=1, c=1, d=0.5, vU=vM=1 -> coef 0.5 - 0.5 = 0)\n", "X = rng.normal(size=n); U = rng.normal(size=n)\n", "M = X + U + rng.normal(size=n); Y = .5 * X + M + U + rng.normal(size=n)\n", "print(\"coef on X given M: %.3f (direct effect 0.5, formula 0.0)\" % coef_se(Y, [X, M])[0])\n", "\n", "# Exercise 10: M-bias with alpha = beta = 1 (tau = 1)\n", "U1, U2 = rng.normal(size=n), rng.normal(size=n)\n", "X = U1 + rng.normal(size=n); M = U1 + U2 + rng.normal(size=n); Y = X + U2 + rng.normal(size=n)\n", "print(\"M-bias: without M %.3f, with M %.3f (bias -0.2)\" % (coef_se(Y, [X])[0], coef_se(Y, [X, M])[0]))\n", "\n", "# Exercise 14: precision of two admissible sets, 400 patients, 2000 replications\n", "est = {\"A\": [], \"B\": []}\n", "for _ in range(2000):\n", " m = 400\n", " A = rng.normal(size=m); X = (1.5 * A + rng.normal(size=m) > 0).astype(float)\n", " B = A + rng.normal(size=m); Y = 1.0 * X + 2 * B + rng.normal(size=m)\n", " est[\"A\"].append(coef_se(Y, [X, A])[0]); est[\"B\"].append(coef_se(Y, [X, B])[0])\n", "sa, sb = np.std(est[\"A\"]), np.std(est[\"B\"])\n", "print(f\"mean A {np.mean(est['A']):.3f}, B {np.mean(est['B']):.3f}; SD A {sa:.3f}, B {sb:.3f}, ratio {sb / sa:.2f}\")" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Key takeaways\n", "\n", "- Include confounders (or their proxies); add strong outcome predictors for precision.\n", "- Leave out instruments, mediators, colliders (including pre-treatment M-colliders) and descendants of the outcome.\n", "- Among admissible sets, those closer to the outcome are more precise; coefficients on controls are not causal by default." ] } ] }