{ "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 50: Structural Causal Models\n", "\n", "*Beginner to Ninja: Math for Causal Inference*\n", "\n", "A structural causal model writes each variable as a function of its direct causes and an exogenous disturbance. It generates the DAG, the joint distribution and, unlike a regression, answers questions about interventions.\n", "\n", "**Learning outcomes**\n", "\n", "- Define a structural causal model and draw the DAG it induces, including bidirected edges for dependent disturbances.\n", "- Compute means, variances and covariances implied by a linear SCM, using Σ = (I − B)⁻¹ Ω (I − B)⁻ᵀ.\n", "- Distinguish structural coefficients from regression coefficients and compute omitted-variable bias.\n", "- Explain observational equivalence, where different causal models produce the same data." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Theory\n", "\n", "### Definition\n", "\n", "A **structural causal model (SCM)** $\\mathcal M = (\\mathbf U, \\mathbf V, \\mathbf F, P(\\mathbf U))$ has:\n", "\n", "- **exogenous** variables $\\mathbf U$, determined outside the model, with distribution $P(\\mathbf U)$;\n", "- **endogenous** variables $\\mathbf V = (V_1, \\dots, V_p)$;\n", "- one **structural equation** per endogenous variable,\n", "\n", "$$\n", "V_i := f_i\\big(\\operatorname{pa}(V_i), U_i\\big).\n", "$$\n", "\n", "The symbol $:=$ is an *assignment*: $V_i$ is produced by its causes. Unlike an algebraic equation, it cannot be rearranged to \"solve for\" a cause.\n", "\n", "### The induced graph\n", "\n", "Draw $W \\to V_i$ for each $W \\in \\operatorname{pa}(V_i)$. If the $U_i$ are mutually independent the model is **Markovian**, the graph is a DAG, and the distribution of $\\mathbf V$ factorises over it (Day 47). If some $U_i, U_j$ are dependent (an omitted common cause), draw a **bidirected edge** $V_i \\leftrightarrow V_j$: the model is **semi-Markovian**. A bidirected edge is shorthand for $V_i \\leftarrow U \\to V_j$ with $U$ unobserved.\n", "\n", "### Linear SCMs\n", "\n", "Stack the equations: $\\mathbf V = \\mathbf B\\mathbf V + \\boldsymbol\\varepsilon$, where $B_{ij}$ is the coefficient of $V_j$ in the equation for $V_i$ and $\\operatorname{Var}(\\boldsymbol\\varepsilon) = \\boldsymbol\\Omega$ (diagonal if Markovian). Solving,\n", "\n", "$$\n", "\\mathbf V = (\\mathbf I - \\mathbf B)^{-1}\\boldsymbol\\varepsilon, \\qquad \\boldsymbol\\Sigma = \\operatorname{Var}(\\mathbf V) = (\\mathbf I - \\mathbf B)^{-1}\\boldsymbol\\Omega(\\mathbf I - \\mathbf B)^{-\\top}.\n", "$$\n", "\n", "Because the graph is acyclic, $\\mathbf B$ is nilpotent and $(\\mathbf I - \\mathbf B)^{-1} = \\mathbf I + \\mathbf B + \\mathbf B^2 + \\dots$: entry $(i, j)$ is the **total effect** of $V_j$ on $V_i$, the sum over directed paths of the products of their coefficients.\n", "\n", "### Structural versus regression coefficients\n", "\n", "In $Y := \\beta X + \\gamma A + U_Y$, $\\beta$ is causal: change $X$ by one unit, holding $A$ and $U_Y$ fixed, and $Y$ moves by $\\beta$. The regression slope of $Y$ on $X$ alone is\n", "\n", "$$\n", "\\frac{\\operatorname{Cov}(X, Y)}{\\operatorname{Var}(X)} = \\beta + \\gamma\\,\\frac{\\operatorname{Cov}(X, A)}{\\operatorname{Var}(X)},\n", "$$\n", "\n", "which equals $\\beta$ only if the omitted cause $A$ is uncorrelated with $X$. The SCM tells you which regression (if any) recovers $\\beta$.\n", "\n", "### Observational equivalence\n", "\n", "Different SCMs can generate the same distribution. With Gaussian variables, $X \\to Y$ and $Y \\to X$ fit any correlation equally well. Data alone cannot decide causal direction there; assumptions (time order, experiments, non-Gaussianity, Day 98) must.\n", "\n", "### What an SCM adds\n", "\n", "An SCM supports three kinds of questions (Pearl's ladder, Day 59): **association** (from $P(\\mathbf V)$), **intervention** (replace an equation, Day 51) and **counterfactuals** (keep each unit's $U$, Day 59).\n", "\n", "## Part 1 · Warm-up" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 1 · From equations to graph\n", "\n", "Draw the graph for $A := U_A$, $S := f_S(A, U_S)$, $O := f_O(S, U_O)$, $W := f_W(A, S, O, U_W)$, where $U_S$ and $U_W$ are correlated and the others independent.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Directed edges from the arguments: $A \\to S$, $S \\to O$, $A \\to W$, $S \\to W$, $O \\to W$.\n", "\n", "**Step 2.** Correlated disturbances $U_S, U_W$: bidirected edge $S \\leftrightarrow W$.\n", "\n", "**Answer:** a semi-Markovian graph; the bidirected edge says an unobserved factor (say, motivation) affects both schooling and wages.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 2 · Mean and variance\n", "\n", "$X := U_X$ with $U_X \\sim N(2, 1)$ and $Y := 3 + 0.5X + U_Y$ with $U_Y \\sim N(0, 0.75)$ independent. Find $E[Y]$, $\\operatorname{Var}(Y)$ and $\\operatorname{corr}(X, Y)$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** $E[Y] = 3 + 0.5(2) = 4$.\n", "\n", "**Step 2.** $\\operatorname{Var}(Y) = 0.5^2(1) + 0.75 = 1$.\n", "\n", "**Step 3.** $\\operatorname{Cov}(X, Y) = 0.5\\operatorname{Var}(X) = 0.5$, so $\\operatorname{corr} = 0.5/\\sqrt{1 \\times 1} = 0.5$.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 3 · Total effect by paths\n", "\n", "In a linear SCM, $X \\xrightarrow{2} M \\xrightarrow{3} Y$ and $X \\xrightarrow{-1} Y$. Find the total effect of $X$ on $Y$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Directed paths: $X \\to M \\to Y$ with product $2 \\times 3 = 6$; $X \\to Y$ with $-1$.\n", "\n", "**Step 2.** Sum: $6 - 1 = 5$.\n", "\n", "**Answer:** 5.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 4 · The covariance matrix\n", "\n", "For $X := \\varepsilon_X$, $Y := 2X + \\varepsilon_Y$ with $\\operatorname{Var}(\\varepsilon_X) = 1$ and $\\operatorname{Var}(\\varepsilon_Y) = 3$, compute $\\boldsymbol\\Sigma$ using $(\\mathbf I - \\mathbf B)^{-1}$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** With order $(X, Y)$: $\\mathbf B = \\begin{pmatrix} 0 & 0 \\\\ 2 & 0 \\end{pmatrix}$, so $(\\mathbf I - \\mathbf B)^{-1} = \\mathbf I + \\mathbf B = \\begin{pmatrix} 1 & 0 \\\\ 2 & 1 \\end{pmatrix}$.\n", "\n", "**Step 2.** $\\boldsymbol\\Omega = \\operatorname{diag}(1, 3)$.\n", "\n", "**Step 3.**\n", "\n", "$$\n", "\\boldsymbol\\Sigma = \\begin{pmatrix} 1 & 0 \\\\ 2 & 1 \\end{pmatrix}\\begin{pmatrix} 1 & 0 \\\\ 0 & 3 \\end{pmatrix}\\begin{pmatrix} 1 & 2 \\\\ 0 & 1 \\end{pmatrix} = \\begin{pmatrix} 1 & 2 \\\\ 2 & 7 \\end{pmatrix}.\n", "$$\n", "\n", "**Answer:** $\\operatorname{Var}(X) = 1$, $\\operatorname{Cov}(X, Y) = 2$, $\\operatorname{Var}(Y) = 4 + 3 = 7$.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 5 · Assignment, not equation\n", "\n", "From $Y := 2X + U_Y$, a student writes $X = (Y - U_Y)/2$ and concludes that raising $Y$ raises $X$. What is wrong?\n", "\n", "
Show solution\n", "\n", "**Step 1.** The structural equation describes how $Y$ is produced; $X$ has its own equation, $X := f_X(\\operatorname{pa}(X), U_X)$, which does not involve $Y$.\n", "\n", "**Step 2.** Intervening on $Y$ replaces $Y$'s equation and leaves $X$'s untouched (Day 51).\n", "\n", "**Answer:** algebraic rearrangement is valid for the observed values but not for causal claims; $X$ does not respond to changes in $Y$.\n", "\n", "
\n", "\n", "## Part 2 · Proofs" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 6 · Markovian SCMs factorise\n", "\n", "Prove that if the $U_i$ are independent and the graph is acyclic, then $P(\\mathbf v) = \\prod_i P(v_i \\mid \\operatorname{pa}(v_i))$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Order the variables topologically. By recursive substitution, each predecessor $V_j$ ($j < i$) is a function of $U_1, \\dots, U_{i-1}$ only.\n", "\n", "**Step 2.** $U_i$ is independent of $(U_1, \\dots, U_{i-1})$, hence of the predecessors $(V_1, \\dots, V_{i-1})$.\n", "\n", "**Step 3.** So given the predecessors, $V_i = f_i(\\operatorname{pa}(V_i), U_i)$ has a distribution that depends only on $\\operatorname{pa}(V_i)$: $P(v_i \\mid v_1, \\dots, v_{i-1}) = P(v_i \\mid \\operatorname{pa}(v_i))$.\n", "\n", "**Step 4.** Substitute into the chain rule. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 7 · Total effects from (I − B)⁻¹\n", "\n", "Prove that for an acyclic linear SCM, $(\\mathbf I - \\mathbf B)^{-1} = \\sum_{k=0}^{p-1}\\mathbf B^k$ and that entry $(i, j)$ for $i \\ne j$ is the sum over directed paths from $V_j$ to $V_i$ of the products of coefficients.\n", "\n", "
Show solution\n", "\n", "**Step 1.** $\\mathbf B$ is nilpotent ($\\mathbf B^p = 0$, Day 46), so $(\\mathbf I - \\mathbf B)\\sum_{k=0}^{p-1}\\mathbf B^k = \\mathbf I - \\mathbf B^p = \\mathbf I$.\n", "\n", "**Step 2.** $(\\mathbf B^k)_{ij} = \\sum B_{i m_1}B_{m_1 m_2}\\cdots B_{m_{k-1} j}$ over intermediate nodes: each non-zero term is a directed path $V_j \\to V_{m_{k-1}} \\to \\dots \\to V_i$ of length $k$, weighted by its coefficients.\n", "\n", "**Step 3.** Summing over $k$ adds all directed paths. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 8 · Omitted-variable bias\n", "\n", "Let $Y := \\beta X + \\gamma A + U_Y$ with $U_Y$ independent of $(X, A)$. Prove $\\operatorname{Cov}(X, Y)/\\operatorname{Var}(X) = \\beta + \\gamma\\operatorname{Cov}(X, A)/\\operatorname{Var}(X)$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** $\\operatorname{Cov}(X, Y) = \\beta\\operatorname{Var}(X) + \\gamma\\operatorname{Cov}(X, A) + \\operatorname{Cov}(X, U_Y)$.\n", "\n", "**Step 2.** The last term is 0 by independence.\n", "\n", "**Step 3.** Divide by $\\operatorname{Var}(X)$. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 9 · Reversing a Gaussian arrow\n", "\n", "Let $X \\sim N(0, 1)$ and $Y := bX + U_Y$, $U_Y \\sim N(0, 1 - b^2)$, $|b| < 1$. Construct an SCM with $Y \\to X$ that generates the same joint distribution.\n", "\n", "
Show solution\n", "\n", "**Step 1.** The original model gives $\\operatorname{Var}(Y) = b^2 + (1 - b^2) = 1$ and $\\operatorname{Cov}(X, Y) = b$.\n", "\n", "**Step 2.** Try $Y := U'_Y \\sim N(0, 1)$ and $X := bY + U'_X$ with $U'_X \\sim N(0, 1 - b^2)$ independent.\n", "\n", "**Step 3.** Then $\\operatorname{Var}(X) = b^2 + 1 - b^2 = 1$ and $\\operatorname{Cov}(X, Y) = b$.\n", "\n", "**Step 4.** Both are bivariate normal with mean 0 and the same covariance matrix, so the distributions coincide. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 10 · When regression is structural\n", "\n", "In a Markovian linear SCM, prove that regressing $V_i$ on all its parents recovers the structural coefficients.\n", "\n", "
Show solution\n", "\n", "**Step 1.** $V_i = \\sum_{j \\in \\operatorname{pa}(i)} B_{ij}V_j + \\varepsilon_i$.\n", "\n", "**Step 2.** Each parent is a function of the disturbances of its own ancestors, and $V_i$ is not among them (acyclicity). With independent disturbances, $\\varepsilon_i$ is independent of all parents.\n", "\n", "**Step 3.** So $E[\\varepsilon_i \\mid \\operatorname{pa}(V_i)] = 0$: the OLS moment condition holds at the true coefficients, and the population regression coefficients equal $B_{ij}$. $\\blacksquare$\n", "\n", "
\n", "\n", "## Part 3 · Applied" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 11 · Returns to schooling with ability\n", "\n", "An SCM for log wages: ability $A \\sim N(0, 1)$; years of schooling $S := 12 + 2A + U_S$ with $\\operatorname{Var}(U_S) = 4$; $\\log W := 1.5 + 0.08S + 0.10A + U_W$, all disturbances independent. What does a regression of $\\log W$ on $S$ alone estimate? What about on $S$ and $A$?\n", "\n", "
Show solution\n", "\n", "**Step 1.** $\\operatorname{Var}(S) = 2^2(1) + 4 = 8$ and $\\operatorname{Cov}(S, A) = 2$.\n", "\n", "**Step 2.** By Exercise 8, the short-regression slope is $0.08 + 0.10 \\times 2/8 = 0.08 + 0.025 = 0.105$.\n", "\n", "**Step 3.** With $A$ included, the regression is on all parents of $\\log W$, so it recovers $0.08$ (Exercise 10).\n", "\n", "**Answer:** 10.5% per year without ability, 8% with it. The 2.5-point gap is ability bias; in practice $A$ is unobserved, which motivates instruments (Phase 5).\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 12 · A binary vaccine model\n", "\n", "$U_I, U_V, U_E$ are independent uniform on $(0, 1)$. Exposure $I := \\mathbf 1[U_I < 0.1]$, vaccination $V := \\mathbf 1[U_V < 0.6]$, protection $E := \\mathbf 1[U_E < 0.9]$ (the vaccine works for 90%), disease $D := I\\,(1 - VE)$. Compute $P(D = 1)$, $P(D = 1 \\mid V = 1)$ and $P(D = 1 \\mid V = 0)$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** $D = 1$ requires exposure and no protection: $P(D = 1) = P(I = 1)\\big(1 - P(V = 1)P(E = 1)\\big) = 0.1(1 - 0.6 \\times 0.9) = 0.046$.\n", "\n", "**Step 2.** Given $V = 1$: $P(D = 1 \\mid V = 1) = 0.1 \\times (1 - 0.9) = 0.01$.\n", "\n", "**Step 3.** Given $V = 0$: $P(D = 1 \\mid V = 0) = 0.1$.\n", "\n", "**Answer:** 4.6%, 1% and 10%. Because $V$ is independent of $U_I$ and $U_E$ here, the observed risk ratio 0.1 equals the causal one.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 13 · Monetary policy multipliers\n", "\n", "A linear macro SCM: interest rate $R$ affects investment ($R \\xrightarrow{-2} I$) and consumption ($R \\xrightarrow{-0.5} C$); investment raises consumption ($I \\xrightarrow{0.3} C$, via income); output $Y := I + C + G$. Compute the total effect of $R$ on $Y$ by paths and by $(\\mathbf I - \\mathbf B)^{-1}$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Directed paths $R \\to Y$: $R \\to I \\to Y$: $-2 \\times 1 = -2$; $R \\to C \\to Y$: $-0.5 \\times 1 = -0.5$; $R \\to I \\to C \\to Y$: $-2 \\times 0.3 \\times 1 = -0.6$.\n", "\n", "**Step 2.** Total: $-2 - 0.5 - 0.6 = -3.1$.\n", "\n", "**Step 3.** With order $(R, I, C, Y)$, the matrix computation (code below) gives the entry $[(\\mathbf I - \\mathbf B)^{-1}]_{Y, R} = -3.1$.\n", "\n", "**Answer:** a one-point rise in the policy rate lowers output by 3.1 units; the regression of $Y$ on $R$ alone recovers this only if $R$ is set independently of the disturbances.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 14 · Exercise and fitness\n", "\n", "A health survey finds $\\operatorname{corr}(\\text{exercise hours}, \\text{fitness score}) = 0.6$, both standardised and normal. Write two SCMs, one with exercise causing fitness and one with fitness causing exercise, that both fit the data exactly. What evidence could separate them?\n", "\n", "
Show solution\n", "\n", "**Step 1.** Model 1: $X := U_X \\sim N(0, 1)$, $Y := 0.6X + U_Y$ with $\\operatorname{Var}(U_Y) = 1 - 0.36 = 0.64$.\n", "\n", "**Step 2.** Model 2: $Y := U'_Y \\sim N(0, 1)$, $X := 0.6Y + U'_X$ with $\\operatorname{Var}(U'_X) = 0.64$.\n", "\n", "**Step 3.** By Exercise 9 both give the same bivariate normal distribution.\n", "\n", "**Answer:** only extra information can separate them: an experiment (randomise an exercise programme), time order in panel data (exercise this month, fitness next month), or non-Gaussian features (Day 98).\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 15 · Making a bidirected edge explicit\n", "\n", "A health economist models hospital quality $Q$ and patient survival $Y$ with $Q \\to Y$, and notes that wealthier areas have both better hospitals and healthier patients, but area wealth is not measured. Write a semi-Markovian SCM and show what the regression of $Y$ on $Q$ estimates in the linear case $Q := aH + U_Q$, $Y := \\beta Q + cH + U_Y$, with unobserved wealth $H \\sim N(0, 1)$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Graph: $Q \\to Y$ and $Q \\leftrightarrow Y$ (shorthand for $Q \\leftarrow H \\to Y$ with $H$ unobserved).\n", "\n", "**Step 2.** $\\operatorname{Var}(Q) = a^2 + \\operatorname{Var}(U_Q)$ and $\\operatorname{Cov}(Q, H) = a$.\n", "\n", "**Step 3.** By Exercise 8 the slope is $\\beta + c\\,a/(a^2 + \\operatorname{Var}(U_Q))$.\n", "\n", "**Answer:** biased upward when wealth raises both quality and survival ($a, c > 0$). Days 52–56 ask when such bias can be removed with the variables you do observe.\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", "def implied_cov(B, Omega):\n", " \"\"\"Covariance of V = B V + eps for an acyclic linear SCM.\"\"\"\n", " T = np.linalg.inv(np.eye(len(B)) - B)\n", " return T @ Omega @ T.T, T\n", "\n", "# Exercise 4\n", "S, _ = implied_cov(np.array([[0, 0], [2, 0.]]), np.diag([1, 3.]))\n", "print(\"Sigma:\\n\", S)\n", "\n", "# Exercise 11: ability bias (order A, S, logW)\n", "B = np.zeros((3, 3)); B[1, 0] = 2; B[2, 1] = .08; B[2, 0] = .10\n", "S, _ = implied_cov(B, np.diag([1, 4, .25]))\n", "print(\"short slope:\", round(S[1, 2] / S[1, 1], 4))\n", "coef = np.linalg.solve(S[:2, :2], S[:2, 2])\n", "print(\"long regression (A, S):\", coef.round(4))\n", "\n", "# Exercise 13: total effects, order R, I, C, Y\n", "B = np.zeros((4, 4)); B[1, 0] = -2; B[2, 0] = -.5; B[2, 1] = .3; B[3, 1] = 1; B[3, 2] = 1\n", "_, T = implied_cov(B, np.eye(4))\n", "print(\"total effect R -> Y:\", T[3, 0].round(3))\n", "\n", "# Exercise 12: binary vaccine model by simulation\n", "rng = np.random.default_rng(50); n = 2_000_000\n", "I = rng.random(n) < .1; V = rng.random(n) < .6; E = rng.random(n) < .9\n", "D = I & ~(V & E)\n", "print(f\"P(D) {D.mean():.4f} P(D|V=1) {D[V].mean():.4f} P(D|V=0) {D[~V].mean():.4f}\")\n", "\n", "# Exercise 14: two directions, one distribution\n", "X = rng.normal(size=n); Y = .6 * X + np.sqrt(.64) * rng.normal(size=n)\n", "Y2 = rng.normal(size=n); X2 = .6 * Y2 + np.sqrt(.64) * rng.normal(size=n)\n", "print(\"cov X->Y model:\", np.cov(X, Y).round(3).tolist(), \" cov Y->X model:\", np.cov(X2, Y2).round(3).tolist())" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Key takeaways\n", "\n", "- An SCM assigns each variable a mechanism $V_i := f_i(\\operatorname{pa}(V_i), U_i)$; independent disturbances give a DAG, dependent ones bidirected edges.\n", "- In linear SCMs, $(\\mathbf I - \\mathbf B)^{-1}$ collects total effects and $\\boldsymbol\\Sigma$ follows from it; regression recovers a structural coefficient only when the regressors block every non-causal path.\n", "- Data can be silent about direction: different SCMs can be observationally equivalent." ] } ] }