1.1.1 Motivation

The intervention based motivation for understanding causal effects:

• We want to know if a particular intervention (like aid) caused a particular outcome (like reduced corruption).
• We need to know:
  1. What happened?
  2. What would the outcome have been if there were no intervention?
• The problem:
  1. … this is hard
  2. … this is impossible

The problem in 2 is that you need to know what would have happened if things were different. You need information on a counterfactual.

1.1.2 Potential Outcomes

• For each unit, we assume that there are two post-treatment outcomes: \(Y_i(1)\) and \(Y_i(0)\).
  • \(Y(1)\) is the outcome that would obtain if the unit received the treatment.
  • \(Y(0)\) is the outcome that would obtain if it did not.
• The causal effect of Treatment (relative to Control) is: \(\tau_i = Y_i(1) - Y_i(0)\)
• Note:
  • The causal effect is defined at the individual level.
  • There is no “data generating process” or functional form.
  • The causal effect is defined relative to something else, so a counterfactual must be conceivable (did Germany cause the second world war?).
  • Are there any substantive assumptions made here so far?

1.1.10 Causal claims: Contribution or attribution?

The counterfactual model is about attribution in a very conditional sense.

• This is problem for research programs that define “explanation” in terms of figuring out the things that cause \(Y\)

• Real difficulties conceptualizing what it means to say one cause is more important than another cause. What does that mean?

1.1.19 Causal claims: The rub and the solution

• Now for some magic. We really want to estimate: \[ \tau_i = Y_i(1) - Y_i(0)\]

• BUT: We never can observe both \(Y_i(1)\) and \(Y_i(0)\)

• Say we lower our sights and try to estimate an average treatment effect: \[ \tau = \mathbb{E} [Y(1)-Y(0)]\]

• Now make use of the fact that \[\mathbb E[Y(1)-Y(0)] = \mathbb E[Y(1)]- \mathbb E [Y(0)] \]

• In words: The average of differences is equal to the difference of averages; here, the average treatment effect is equal to the difference in average outcomes in treatment and control units.

• The magic is that while we can’t hope to measure the differences; we are good at measuring averages.

1.1.20 Causal claims: The rub and the solution

• So we want to estimate \(\mathbb{E} [Y(1)]\) and \(\mathbb{E} [Y(0)]\).
• We know that we can estimate averages of a quantity by taking the average value from a random sample of units
• To do this here we need to select a random sample of the \(Y(1)\) values and a random sample of the \(Y(0)\) values, in other words, we randomly assign subjects to treatment and control conditions.
• When we do that we can in fact estimate: \[ \mathbb {E}_N[Y_i(1) | Z_i = 1) - \mathbb {E}_N(Y_i(0) | Z_i = 0]\] which in expectation equals: \[ \mathbb{E} [Y_i(1) | Z_i = 1 \text{ or } Z_i = 0] - \mathbb{E} [Y_i(0) | Z_i = 1 \text{ or } Z_i = 0]\]
• This highlights a deep connection between random assignment and random sampling: when we do random assignment we are in fact randomly sampling from different possible worlds.

1.1.27 Exercise your potential outcomes 1

Consider the following potential outcomes table:

Questions for us: What are the unit level treatment effects? What is the average treatment effect?

1.1.28 Exercise your potential outcomes 2

Consider the following potential outcomes table:

Questions for us: Fill in the blanks.

• Assuming a constant treatment effect of \(+1\)
• Assuming a constant treatment effect of \(-1\)
• Assuming an average treatment effect of \(0\)

What is the actual treatment effect?

1.3.3 Heterogeneous Effects with Endogeneous Categories

• Violence(Treatment)
• Reporting(Treatment, Violence)

Expected reporting given violence in control = Pr(Type 1)

Expected reporting given violence in treatment = 100%

Question: What is the actual effect of treatment on the propensity to report violence?

1.3.9 Missing data can create an endogeneous subgroup problem

• It is well known that missing data can undo the magic of random assignment.
• One seemingly promising approach is to match into pairs ex ante and drop pairs together ex post.
• Say potential outcomes looked like this (2 pairs of 2 units):

1.3.10 Missing data

• Say though that treated cases are likely to drop out of the sample if things go badly (e.g. they get a negative score or die)
• Then you might see no attrition if those would-be attritors are not treated.
• You might assume you have no problem (after all, no attrition).
• No missing data when the normal cases happens to be selected

1.3.11 Missing data

• But in cases in which you have attrition, dropping the pair doesn’t necessarily help.
• The problem is potential missingness still depends on potential outcomes
• The kicker is that the method can produce bias even if (in fact) there is no attrition!
• But missing data when the vulnerable cases happens to be selected

1.3.14 Multistage games

Question: Does visibility alter the extent to which subjects follow norms to punish antisocial behavior (and reward prosocial behavior)? Consider a trust game in which we are interested in how information on receivers affects their actions

What do we think? Does visibility make people react more to investments?

1.3.18 Multistage games

What to do?

Solutions?

1. Analysis could focus on the effect of treatment on respondent behavior, directly.
   • This would get the correct answer but to a different question (Does information affect the share of contributions returned by subjects on average?)
2. Strategy method can sometimes help address the problem, but note that that is (a) changing the question and (b) putting demands on respondent imagination and honesty
3. First mover action could be directly manipulated, but unless deception is used that is also changing the question
4. First movers could be selected because they act in predictable ways (bordering on deception?)

Take away: Proceed with extreme caution when estimating effects beyond the first stage.

1.5.1 Key insight

The most powerful results from the study of DAGs give procedures for figuring out when conditioning aids or hinders causal identification.

• You can read off a confounding variable from a DAG.
  • You figure out what to condition on for causal identification.
• You can read off “colliders” from a DAG
  • Sometimes you have to avoid conditioning on these
• Sometimes a variable might be both, so
  • you have to condition on it
  • you have to avoid conditioning on it
  • Ouch.

1.7.2 Conditional distributions: Factorization

• Consider a situation with variables \(X_1, X_2, \dots X_n\)
• The probability of outcome \(x\) can always be written in the form \(P(X_1 = x_1)P(X_2 = x_2|X_1=x_1)(X_3 = x_3|X_1=x_1, X_2 = x_2)\dots\).
• This can be done with any ordering of variables.
• However the representation can be greatly simplified if you can make use of a set of “parentage” relationships
• Given an ordering of variables, the Markovian parents of variable \(X_j\) are the minimal set of variables such that when you condition on these, \(X_j\) is independent of all other prior variables in the ordering
• We want to get to the point where we can can write: \(P(x) = \prod_j(x_j | pa_j)\)

1.7.3 Conditional distributions: Factorization from graphs

• We want to use causal graphs to represent these relations of conditional independence.
• Informally, an arrow, \(A \rightarrow B\) means that \(A\) is a cause of \(B\): that is, under some conditions, a change in \(A\) produces a change in \(B\).
  • Arrows carry no information about the type of effect; e.g. sign, size, or whether different causes are complements or substitutes
• We say that arrows point from parents to children, and by extension from ancestors to descendants.
• These are parents on the graph; but we will connect them to Markovian parents in a probability distribution \(P\).

1.7.4 Conditional distributions: Markov condition

• A DAG is just a graph in which some or all nodes are connected by directed edges (arrows) and there are no cyclical paths along these directed edges.
• Consider a DAG, \(G\), and consider the ancestry relations implied by \(G\): the distribution \(P\) is Markov relative to the graph \(G\) if every variable is independent of its nondescendants (in \(G\)) conditional on its parents (in \(G\)).
  • Markov condition: conditional on its parents, a variable is independent of its non-descendants.

Now we have what we need to simplify: if the Markov condition is satisfied, then instead of writing the full probability as \(P(x) = P(x_1)P(x_2|x_1)P(x_3|x_1, x_2)\) we can write \(P(x) = \prod_i P(x_i |pa_i)\).

1.7.5 Illustration

If \(P(a,b,c)\) is Markov relative to this graph then: \(C\) is independent of \(A\) given \(B\)

And instead of

\[\Pr(a,b,c) = \Pr(a)\Pr(a|b)\Pr(c|a, b)\]

we could now write:

\[\Pr(a,b,c) = \Pr(a)\Pr(a|b)\Pr(c|b)\]

1.7.6 Conditional distributions and interventions

We want the graphs to be able to represent the effects of interventions.

Pearl uses do notation to capture this idea.

\[\Pr(X_1, X_2,\dots | do(X_j = x_j))\] or

\[\Pr(X_1, X_2,\dots | \hat{x}_j)\]

denotes the distribution of \(X\) when a particular node (or set of nodes) is intervened upon and forced to a particular level, \(x_j\).

1.7.7 Conditional distributions given do operations

Note, in general: \[\Pr(X_1, X_2,\dots | do(X_j = x_j')) \neq \Pr(X_1, X_2,\dots | X_j = x_j')\] as an example we might imagine a situation where:

• for men, binary \(X\) always causes \(Y=1\)
• for women, \(Y=0\) regardless of \(X\)
• \(X=1\) for men only.

In that case \(\Pr(Y=1 | X = 1) = 1\) but \(\Pr(Y=1 | do(X = 1)) = .5\)

1.7.8 Conditional distributions given do operations

A DAG is “causal Bayesian network” or “Causal DAG” if (and only if) the probability distribution resulting from setting some set \(X_i\) to \(\hat{x'}_i\) (i.e. do(X=x')) is:

\[P_{\hat{x}_i}: P(x_1,x_2,\dots x_n|\hat{x}_i) = \mathbb{I}(x_i = x_i')\prod_{-i}P(x_j|pa_j)\]

This means that there is only probability mass on vectors in which \(x_i = x_i'\) (reflecting the success of control) and all other variables are determined by their parents, given the values that have been set for \(x_i\).

1.7.9 Conditional distributions given do operations

Illustration, say we have binary \(X\) causes binary \(M\) which cases binary \(Y\); say we intervene and set \(M=1\). Then what is the distribution of \((x,m,y)\)?

It is:

\[\Pr(x,m,y) = \Pr(x)\mathbb I(M = 1)\Pr(y|m)\]

1.8.3 Conditional independence from graphs

\(A\) and \(B\) are conditionally independent, given \(C\) if on every path between \(A\) and \(B\):

• there is some chain (\(\bullet\rightarrow \bullet\rightarrow\bullet\) or \(\bullet\leftarrow \bullet\leftarrow\bullet\)) or fork (\(\bullet\leftarrow \bullet\rightarrow\bullet\)) with the central element in \(C\),

or

• there is an inverted fork (\(\bullet\rightarrow \bullet\leftarrow\bullet\)) with the central element (and its descendants) not in \(C\)

Notes:

• In this case we say that \(A\) and \(B\) are d-separated by \(C\).
• \(A\), \(B\), and \(C\) can all be sets
• Note that a path can involve arrows pointing any direction \(\bullet\rightarrow \bullet\rightarrow \bullet\leftarrow \bullet\rightarrow\bullet\)

1.8.4 Test yourself

Are A and D unconditionally independent:

• if you do not condition on anything?
• if you condition on B?
• if you condition on C?
• if you condition on B and C?

1.8.5 Back to this example

• \(Z = f_1(U_1, U_2)\)
• \(X = f_2(U_2)\)
• \(Y = f_3(X, U_1)\)

1. Let’s graph this
2. Now: say we removed the arrow from \(X\) to \(Y\)
   • Would you expect to see a correlation between \(X\) and \(Y\) if you did not control for \(Z\)
   • Would you expect to see a correlation between \(X\) and \(Y\) if you did control for \(Z\)

1.8.6 Back to this example

Now: say we removed the arrow from \(X\) to \(Y\) - Would you expect to see a correlation between \(X\) and \(Y\) if you did not control for \(Z\) - Would you expect to see a correlation between \(X\) and \(Y\) if you did control for \(Z\)

1.9.1 From graphs to Causal Models

A “causal model” is:

1.Variables

• An ordered list of \(n\) endogenous nodes, \(\mathcal{V}= (V^1, V^2,\dots, V^n)\), with a specification of a range for each of them
• A list of \(n\) exogenous nodes, \(\Theta = (\theta^1, \theta^2,\dots , \theta^n)\)

2. A list of \(n\) functions \(\mathcal{F}= (f^1, f^2,\dots, f^n)\), one for each element of \(\mathcal{V}\) such that each \(f^i\) takes as arguments \(\theta^i\) as well as elements of \(\mathcal{V}\) that are prior to \(V^i\) in the ordering

3. A probability distribution over \(\Theta\)

1.9.4 Effects on a DAG

For instance the simplest model consistent with \(X \rightarrow Y\):

• Endogenous Nodes = \(\{X, Y\}\), both with range \(\{0,1\}\)

• Exogenous Nodes = \(\{\theta^X, \theta^Y\}\), with ranges \(\{\theta^X_0, \theta^X_1\}\) and \(\{\theta^Y_{00}\theta^Y_{01}, \theta^Y_{10}, \theta^Y_{11}\}\)

• Functional equations:

  • \(f_Y\): \(\theta^Y =\theta^Y_{ij} \rightarrow \{Y = i \text{ if } X=0; Y = j \text{ if } X=1\}\)
  • \(f_X\): \(\theta^X =\theta^X_{i} \rightarrow \{X = i\}\)

• Distributions on \(\Theta\): \(\Pr(\theta^i = \theta^i_k) = \lambda^i_k\)