{ "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 47: Factorisation & the Markov Property\n", "\n", "*Beginner to Ninja: Math for Causal Inference*\n", "\n", "A DAG is also a statement about probability. Each variable depends only on its parents, so the joint distribution factorises into small pieces and implies conditional independences we can test.\n", "\n", "**Learning outcomes**\n", "\n", "- Write the Markov factorisation of a joint distribution for any DAG.\n", "- Count the parameters saved by the factorisation.\n", "- Compute joint, marginal and conditional probabilities from conditional probability tables.\n", "- State and prove the local Markov property and use it to list testable independences." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Theory\n", "\n", "### From the chain rule to the DAG\n", "\n", "For any ordering of variables, the **chain rule** of probability gives\n", "\n", "$$\n", "P(v_1, \\dots, v_p) = \\prod_{i=1}^p P(v_i \\mid v_1, \\dots, v_{i-1}).\n", "$$\n", "\n", "A DAG says more: in a topological order, each variable depends on its predecessors **only through its parents**. The joint distribution then **factorises**:\n", "\n", "$$\n", "P(v_1, \\dots, v_p) = \\prod_{i=1}^p P\\big(v_i \\mid \\operatorname{pa}(v_i)\\big).\n", "$$\n", "\n", "A distribution that factorises this way is **Markov** with respect to the DAG. Each factor $P(v_i \\mid \\operatorname{pa}(v_i))$ is a **conditional probability table (CPT)** for discrete variables, or a regression model for continuous ones.\n", "\n", "### Fewer parameters\n", "\n", "With $p$ binary variables, an unrestricted joint distribution needs $2^p - 1$ numbers. Under the factorisation, node $i$ needs one probability for each configuration of its parents, $2^{|\\operatorname{pa}(v_i)|}$, so the total is\n", "\n", "$$\n", "\\sum_{i=1}^p 2^{|\\operatorname{pa}(v_i)|}.\n", "$$\n", "\n", "With 20 variables and at most two parents each: at most 80 numbers instead of about a million.\n", "\n", "### The local Markov property\n", "\n", "Factorisation is equivalent to the **local Markov property**:\n", "\n", "$$\n", "V_i \\perp\\!\\!\\!\\perp \\operatorname{nd}(V_i) \\setminus \\operatorname{pa}(V_i) \\;\\big|\\; \\operatorname{pa}(V_i),\n", "$$\n", "\n", "where $\\operatorname{nd}(V_i)$ are the non-descendants of $V_i$. \"Once you know a variable's direct causes, its other non-effects tell you nothing more about it.\"\n", "\n", "### Three building blocks\n", "\n", "| Structure | Factorisation | Implied independence |\n", "|---|---|---|\n", "| Chain $Z \\to X \\to Y$ | $P(z)P(x \\mid z)P(y \\mid x)$ | $Y \\perp\\!\\!\\!\\perp Z \\mid X$ |\n", "| Fork $Z \\leftarrow X \\to Y$ | $P(x)P(z \\mid x)P(y \\mid x)$ | $Y \\perp\\!\\!\\!\\perp Z \\mid X$ |\n", "| Collider $Z \\to X \\leftarrow Y$ | $P(z)P(y)P(x \\mid z, y)$ | $Y \\perp\\!\\!\\!\\perp Z$ (but not given $X$) |\n", "\n", "Day 48 studies these in depth.\n", "\n", "### Faithfulness\n", "\n", "The Markov property says which independences *must* hold. A distribution may have *extra* independences, for instance when two causal channels cancel exactly. **Faithfulness** assumes it does not: every independence in the data comes from the graph. Causal discovery (Day 98) relies on it.\n", "\n", "### Testable implications\n", "\n", "Each implied independence is a **testable implication** of the DAG. If data reject $Y \\perp\\!\\!\\!\\perp Z \\mid X$, the graph is wrong. If data agree, the graph survives, but other graphs may imply the same independences.\n", "\n", "## Part 1 · Warm-up\n", "\n", "For Exercises 3–4 use the chain $Z \\to X \\to Y$ with binary variables, $P(Z = 1) = 0.4$, $P(X = 1 \\mid Z = 1) = 0.7$, $P(X = 1 \\mid Z = 0) = 0.2$, $P(Y = 1 \\mid X = 1) = 0.6$, $P(Y = 1 \\mid X = 0) = 0.1$." ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 1 · Writing the factorisation\n", "\n", "Write the factorisation for the DAG $A \\to B$, $A \\to C$, $B \\to D$, $C \\to D$, $D \\to E$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Parents: $A: \\varnothing$; $B: A$; $C: A$; $D: B, C$; $E: D$.\n", "\n", "**Step 2.** One factor per node, conditioning on its parents:\n", "\n", "$$\n", "P(a, b, c, d, e) = P(a)\\,P(b \\mid a)\\,P(c \\mid a)\\,P(d \\mid b, c)\\,P(e \\mid d).\n", "$$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 2 · Counting parameters\n", "\n", "All five variables in Exercise 1 are binary. How many parameters does the factorisation need, compared with an unrestricted joint distribution?\n", "\n", "
Show solution\n", "\n", "**Step 1.** Unrestricted: $2^5 - 1 = 31$.\n", "\n", "**Step 2.** Factorised: $2^0 + 2^1 + 2^1 + 2^2 + 2^1 = 1 + 2 + 2 + 4 + 2 = 11$.\n", "\n", "**Answer:** 11 versus 31.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 3 · A joint probability\n", "\n", "Compute $P(Z = 1, X = 1, Y = 1)$ and $P(Z = 0, X = 1, Y = 0)$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Factorisation: $P(z, x, y) = P(z)P(x \\mid z)P(y \\mid x)$.\n", "\n", "**Step 2.** $P(1, 1, 1) = 0.4 \\times 0.7 \\times 0.6 = 0.168$.\n", "\n", "**Step 3.** $P(0, 1, 0) = 0.6 \\times 0.2 \\times (1 - 0.6) = 0.6 \\times 0.2 \\times 0.4 = 0.048$.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 4 · Marginals by summing out\n", "\n", "Compute $P(X = 1)$ and $P(Y = 1)$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Sum over $Z$: $P(X = 1) = 0.4(0.7) + 0.6(0.2) = 0.28 + 0.12 = 0.40$.\n", "\n", "**Step 2.** Sum over $X$: $P(Y = 1) = P(X = 1)(0.6) + P(X = 0)(0.1)$.\n", "\n", "**Step 3.** $P(Y = 1) = 0.40(0.6) + 0.60(0.1) = 0.24 + 0.06 = 0.30$.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 5 · Local Markov statements\n", "\n", "For the DAG in Exercise 1, write the local Markov independence for $D$ and for $E$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** $D$: parents $\\{B, C\\}$; descendants $\\{D, E\\}$; so the non-descendants other than parents are $\\{A\\}$.\n", "\n", "**Step 2.** $D \\perp\\!\\!\\!\\perp A \\mid B, C$.\n", "\n", "**Step 3.** $E$: parent $\\{D\\}$; non-descendants other than the parent are $\\{A, B, C\\}$.\n", "\n", "**Step 4.** $E \\perp\\!\\!\\!\\perp \\{A, B, C\\} \\mid D$.\n", "\n", "
\n", "\n", "## Part 2 · Proofs" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 6 · The chain screens off\n", "\n", "For $Z \\to X \\to Y$, prove $P(y \\mid x, z) = P(y \\mid x)$ from the factorisation.\n", "\n", "
Show solution\n", "\n", "**Step 1.** By definition $P(y \\mid x, z) = \\dfrac{P(z, x, y)}{P(z, x)}$.\n", "\n", "**Step 2.** Numerator: $P(z)P(x \\mid z)P(y \\mid x)$.\n", "\n", "**Step 3.** Denominator: sum the numerator over $y$: $P(z)P(x \\mid z)\\sum_y P(y \\mid x) = P(z)P(x \\mid z)$.\n", "\n", "**Step 4.** The ratio is $P(y \\mid x)$, which does not involve $z$. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 7 · The fork screens off\n", "\n", "For $Z \\leftarrow X \\to Y$, prove $Y \\perp\\!\\!\\!\\perp Z \\mid X$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Factorisation: $P(x, y, z) = P(x)P(z \\mid x)P(y \\mid x)$.\n", "\n", "**Step 2.** Divide by $P(x)$: $P(y, z \\mid x) = P(z \\mid x)\\,P(y \\mid x)$.\n", "\n", "**Step 3.** That is the definition of conditional independence. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 8 · The collider separates its causes\n", "\n", "For $Z \\to X \\leftarrow Y$, prove $Z \\perp\\!\\!\\!\\perp Y$ marginally.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Factorisation: $P(z, x, y) = P(z)P(y)P(x \\mid z, y)$.\n", "\n", "**Step 2.** Sum over $x$: $P(z, y) = P(z)P(y)\\sum_x P(x \\mid z, y) = P(z)P(y)$.\n", "\n", "**Step 3.** So $Z$ and $Y$ are independent. $\\blacksquare$ (Conditioning on $X$ generally destroys this; see Exercise 14.)\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 9 · The factorisation is a valid distribution\n", "\n", "Prove that $\\prod_i P(v_i \\mid \\operatorname{pa}(v_i))$ sums to 1 over all values, for any choice of CPTs.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Order the nodes topologically, $V_1, \\dots, V_p$. The last node $V_p$ is a parent of no one, so it appears only in its own factor.\n", "\n", "**Step 2.** Summing over $v_p$ first: $\\sum_{v_p} P(v_p \\mid \\operatorname{pa}(v_p)) = 1$, which removes the last factor.\n", "\n", "**Step 3.** What remains is the same product for $V_1, \\dots, V_{p-1}$, in which $V_{p-1}$ is now last.\n", "\n", "**Step 4.** Repeat until no factors remain: the total is 1. $\\blacksquare$\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 10 · Local Markov implies factorisation\n", "\n", "Prove that if each $V_i$ satisfies the local Markov property, then the joint distribution factorises over the DAG.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Take a topological order and apply the chain rule: $P(v) = \\prod_i P(v_i \\mid v_1, \\dots, v_{i-1})$.\n", "\n", "**Step 2.** The predecessors $\\{V_1, \\dots, V_{i-1}\\}$ contain all parents of $V_i$ and no descendants of $V_i$ (descendants come later in the order).\n", "\n", "**Step 3.** So the predecessors are $\\operatorname{pa}(V_i)$ plus some other non-descendants.\n", "\n", "**Step 4.** By local Markov, those other non-descendants can be dropped: $P(v_i \\mid v_1, \\dots, v_{i-1}) = P(v_i \\mid \\operatorname{pa}(v_i))$.\n", "\n", "**Step 5.** Substituting gives the factorisation. $\\blacksquare$\n", "\n", "
\n", "\n", "## Part 3 · Applied" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 11 · Vaccination and age\n", "\n", "In a flu season, being older ($O = 1$, 30% of the population) affects both vaccination $V$ and infection $I$, and vaccination affects infection: $O \\to V$, $O \\to I$, $V \\to I$. $P(V = 1 \\mid O = 1) = 0.8$, $P(V = 1 \\mid O = 0) = 0.4$. Infection risks $P(I = 1 \\mid V, O)$: young unvaccinated 0.10, young vaccinated 0.04, old unvaccinated 0.30, old vaccinated 0.12. Compute $P(I = 1)$, $P(I = 1 \\mid V = 1)$ and $P(I = 1 \\mid V = 0)$. Compare the crude risk ratio with the age-specific ones.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Joint probabilities with $I = 1$, using $P(o)P(v \\mid o)P(I = 1 \\mid v, o)$:\n", "\n", "- young, vaccinated: $0.7 \\times 0.4 \\times 0.04 = 0.0112$;\n", "- young, unvaccinated: $0.7 \\times 0.6 \\times 0.10 = 0.0420$;\n", "- old, vaccinated: $0.3 \\times 0.8 \\times 0.12 = 0.0288$;\n", "- old, unvaccinated: $0.3 \\times 0.2 \\times 0.30 = 0.0180$.\n", "\n", "**Step 2.** $P(I = 1) = 0.0112 + 0.0420 + 0.0288 + 0.0180 = 0.100$.\n", "\n", "**Step 3.** $P(V = 1) = 0.7(0.4) + 0.3(0.8) = 0.52$, so $P(I = 1 \\mid V = 1) = (0.0112 + 0.0288)/0.52 = 0.04/0.52 \\approx 0.077$.\n", "\n", "**Step 4.** $P(I = 1 \\mid V = 0) = (0.0420 + 0.0180)/0.48 = 0.06/0.48 = 0.125$.\n", "\n", "**Step 5.** Crude risk ratio: $0.077/0.125 \\approx 0.62$. Age-specific: $0.04/0.10 = 0.4$ and $0.12/0.30 = 0.4$.\n", "\n", "**Answer:** the vaccine cuts risk by 60% in each age group, but the crude comparison suggests only 38%, because the vaccinated are disproportionately old (the fork $V \\leftarrow O \\to I$).\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 12 · A credit-scoring model\n", "\n", "A bank models 10 binary borrower characteristics with a DAG in which each node has at most two parents. Give an upper bound on the number of parameters and compare with the unrestricted joint distribution. What is gained, and what is risked?\n", "\n", "
Show solution\n", "\n", "**Step 1.** Each node needs at most $2^2 = 4$ parameters, so at most $10 \\times 4 = 40$.\n", "\n", "**Step 2.** Unrestricted: $2^{10} - 1 = 1023$.\n", "\n", "**Step 3.** Gain: each CPT is estimated from many borrowers, so estimates are precise even for rare combinations.\n", "\n", "**Answer:** at most 40 versus 1,023. The risk is that the missing arrows are wrong; the DAG's implied independences should then be tested.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 13 · Testing a chain with survey data\n", "\n", "A survey of 1,000 jobseekers records programme participation $P$, whether skills improved $S$ and employment $E$. Counts:\n", "\n", "| $P$ | $S$ | $E = 1$ | $E = 0$ |\n", "|---|---|---|---|\n", "| 1 | 1 | 210 | 90 |\n", "| 1 | 0 | 80 | 120 |\n", "| 0 | 1 | 105 | 45 |\n", "| 0 | 0 | 140 | 210 |\n", "\n", "Do the data agree with the chain $P \\to S \\to E$ (no direct effect of the programme on employment)?\n", "\n", "
Show solution\n", "\n", "**Step 1.** The chain implies $E \\perp\\!\\!\\!\\perp P \\mid S$.\n", "\n", "**Step 2.** With improved skills: $P(E = 1 \\mid S = 1, P = 1) = 210/300 = 0.70$ and $P(E = 1 \\mid S = 1, P = 0) = 105/150 = 0.70$.\n", "\n", "**Step 3.** Without: $P(E = 1 \\mid S = 0, P = 1) = 80/200 = 0.40$ and $P(E = 1 \\mid S = 0, P = 0) = 140/350 = 0.40$.\n", "\n", "**Step 4.** Marginally: $P(E = 1 \\mid P = 1) = 290/500 = 0.58$ versus $P(E = 1 \\mid P = 0) = 245/500 = 0.49$.\n", "\n", "**Answer:** consistent with the chain: the programme is associated with employment, but only through skills. (With sampling noise, a formal test such as a stratified chi-square would be used.)\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 14 · Hospital admissions (Berkson)\n", "\n", "Flu $F$ (10% of people) and injury $J$ (5%) are independent. Hospital admission $H$ depends on both: $P(H = 1 \\mid F, J)$ is 0.01 for neither, 0.3 for flu only, 0.5 for injury only and 0.8 for both. Among admitted patients, compute $P(J = 1)$, $P(J = 1 \\mid F = 1)$ and $P(J = 1 \\mid F = 0)$.\n", "\n", "
Show solution\n", "\n", "**Step 1.** Joint probabilities with $H = 1$:\n", "\n", "- neither: $0.9 \\times 0.95 \\times 0.01 = 0.00855$;\n", "- flu only: $0.1 \\times 0.95 \\times 0.3 = 0.0285$;\n", "- injury only: $0.9 \\times 0.05 \\times 0.5 = 0.0225$;\n", "- both: $0.1 \\times 0.05 \\times 0.8 = 0.004$.\n", "\n", "**Step 2.** $P(H = 1) = 0.06355$.\n", "\n", "**Step 3.** $P(J = 1 \\mid H = 1) = (0.0225 + 0.004)/0.06355 \\approx 0.417$.\n", "\n", "**Step 4.** $P(J = 1 \\mid H = 1, F = 1) = 0.004/(0.0285 + 0.004) \\approx 0.123$.\n", "\n", "**Step 5.** $P(J = 1 \\mid H = 1, F = 0) = 0.0225/(0.00855 + 0.0225) \\approx 0.725$.\n", "\n", "**Answer:** in the population flu and injury are independent, but among hospital patients those with flu are far *less* likely to be injured. Conditioning on the collider $H$ creates the association.\n", "\n", "
" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Exercise 15 · Microfinance testable implications\n", "\n", "A DAG for micro-entrepreneurs: education $E \\to$ loan $L$, $E \\to$ income $I$, $L \\to$ business start $B$, $B \\to I$, rainfall $R \\to I$. List the local Markov independences, and say which one you would test first with survey data.\n", "\n", "
Show solution\n", "\n", "**Step 1.** $E$: no parents; non-descendants other than itself: $R$. So $E \\perp\\!\\!\\!\\perp R$.\n", "\n", "**Step 2.** $R$: no parents; non-descendants: $E, L, B$. So $R \\perp\\!\\!\\!\\perp \\{E, L, B\\}$.\n", "\n", "**Step 3.** $L$: parent $E$; non-descendants other than $E$: $R$. So $L \\perp\\!\\!\\!\\perp R \\mid E$.\n", "\n", "**Step 4.** $B$: parent $L$; non-descendants: $E, R$. So $B \\perp\\!\\!\\!\\perp \\{E, R\\} \\mid L$.\n", "\n", "**Step 5.** $I$: parents $E, B, R$; non-descendant $L$. So $I \\perp\\!\\!\\!\\perp L \\mid E, B, R$.\n", "\n", "**Answer:** test $I \\perp\\!\\!\\!\\perp L \\mid E, B, R$ first: it says loans raise income *only* by starting businesses. Loans used for consumption smoothing or to pay off costlier debt would violate it.\n", "\n", "
\n", "\n", "## Verify in Python" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "*Your work:*" ] }, { "cell_type": "code", "metadata": {}, "execution_count": null, "outputs": [], "source": [ "import itertools\n", "import numpy as np\n", "\n", "# Exercises 3-4: chain Z -> X -> Y\n", "pZ = {1: .4, 0: .6}; pX1 = {1: .7, 0: .2}; pY1 = {1: .6, 0: .1}\n", "P = {(z, x, y): pZ[z] * (pX1[z] if x else 1 - pX1[z]) * (pY1[x] if y else 1 - pY1[x])\n", " for z, x, y in itertools.product([0, 1], repeat=3)}\n", "print(\"sums to\", round(sum(P.values()), 10), \" P(1,1,1) =\", round(P[1, 1, 1], 3), \" P(0,1,0) =\", round(P[0, 1, 0], 3))\n", "print(\"P(X=1) =\", round(sum(v for k, v in P.items() if k[1]), 3), \" P(Y=1) =\", round(sum(v for k, v in P.items() if k[2]), 3))\n", "for x in (0, 1): # Y independent of Z given X\n", " print(f\"P(Y=1|X={x},Z=z):\", [round(P[z, x, 1] / (P[z, x, 0] + P[z, x, 1]), 3) for z in (0, 1)])\n", "\n", "# Exercise 11: vaccination and age\n", "pO = {1: .3, 0: .7}; pV1 = {1: .8, 0: .4}; pI1 = {(0, 0): .10, (1, 0): .04, (0, 1): .30, (1, 1): .12} # key (v, o)\n", "J = {(o, v): pO[o] * (pV1[o] if v else 1 - pV1[o]) * pI1[v, o] for o in (0, 1) for v in (0, 1)}\n", "pV = {v: sum(pO[o] * (pV1[o] if v else 1 - pV1[o]) for o in (0, 1)) for v in (0, 1)}\n", "r1 = (J[0, 1] + J[1, 1]) / pV[1]; r0 = (J[0, 0] + J[1, 0]) / pV[0]\n", "print(\"P(I=1) =\", round(sum(J.values()), 4), \" P(I|V=1) =\", round(r1, 4), \" P(I|V=0) =\", round(r0, 4), \" crude RR =\", round(r1 / r0, 3))\n", "\n", "# Exercise 13: survey counts (P, S) -> (E=1, E=0)\n", "c = {(1, 1): (210, 90), (1, 0): (80, 120), (0, 1): (105, 45), (0, 0): (140, 210)}\n", "for (p, s), (a, b) in c.items():\n", " print(f\"P(E=1 | P={p}, S={s}) = {a / (a + b):.2f}\")\n", "\n", "# Exercise 14: Berkson's bias\n", "pF, pJ = .1, .05; pH = {(0, 0): .01, (1, 0): .3, (0, 1): .5, (1, 1): .8}\n", "H = {(f, j): (pF if f else 1 - pF) * (pJ if j else 1 - pJ) * pH[f, j] for f in (0, 1) for j in (0, 1)}\n", "tot = sum(H.values())\n", "print(\"P(H=1) =\", round(tot, 5), \" P(J|H) =\", round((H[0, 1] + H[1, 1]) / tot, 3),\n", " \" P(J|H,F=1) =\", round(H[1, 1] / (H[1, 0] + H[1, 1]), 3), \" P(J|H,F=0) =\", round(H[0, 1] / (H[0, 0] + H[0, 1]), 3))" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Key takeaways\n", "\n", "- A DAG factorises the joint distribution into one conditional distribution per node given its parents.\n", "- This is equivalent to the local Markov property: each variable is independent of its non-descendants given its parents.\n", "- Every implied independence is a testable implication; colliders are the exception that creates dependence when conditioned on." ] } ] }