 # Causal Inference

What If

Scott Mueller (lvl 17)
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Difference in means (DIM) estimand decomposition

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Date created

Apr 6, 2021

## Cards(17)

Unsectioned

(7 cards)

Difference in means (DIM) estimand decomposition

Front

\begin{aligned}E[Y_i|D_i=1] &- E[Y_i|D_i=0]\\&= E[Y_{1i}|D_i=1] - E[Y_{0i}|D_i=1]\\&=E[Y_{1i} - Y_{0i}|D_i=1]\\&+ (E[Y_{0i}|D_i=1] - E[Y_{0i}|D_i=0])\end{aligned}

where the first term in the last expression is the ATT and the second term is DIM had everyone been untreated

Back

Difference in means (DIM)

Front

$$E[Y_i|D_i=1] - E[Y_i|D_i=0]$$

Back

Standardized mean difference (SMD)

Front

$$\frac{\overline{X}_\text{treatment} - \overline{X}_\text{control}}{\sqrt{\frac{s^2_\text{treatment} + s^2_\text{control}}2}}$$

where $$s^2$$ is sample variance. Typically performed for each covariate.

Back

Individual treatment effect

Front

$$\tau_i = Y_{1i} - Y_{0i}$$

Back

Average treatment effect among the treated (ATT)

Front

$$E[\tau_i|D_i = 1] = E[Y_{1i} - Y_{0i}|D_i = 1]$$

Back

Average treatment effect (ATE)

Front

$$E[\tau_i] = E[Y_{1i} - Y_{0i}]$$

Back

Fisher's exact test for the causal sharp null hypothesis

Front

$$H_0: Y_{1i} = Y_{0i}, \forall i$$

Let $$\Omega$$ be set of possible ways to assign treatments

1. Calculate $$\hat{\theta}_\text{true}$$ (DIM)
2. Calculate $$\hat{\theta}(\omega)$$ for every $$\omega \in \Omega$$
3. Compare $$\hat{\theta}_\text{true}$$ to $$\hat{\theta}(\omega)$$'s to see how extreme
Back

Chapter 1

(4 cards)

Sharp causal null hypothesis

Front

There is no individual causal effect, $$Y^{a=1} = Y^{a=0}$$ for all individuals

Back

Causal null hypothesis

Front

There is no causal effect,

$$P(Y^{a=1} = 1) = P(Y^{a=0} = 1)$$

or

$$E\left[Y^{a=1}\right] = E\left[Y^{a=0}\right]$$ for nondichotomous outcomes

Back

Number needed to treat (NNT) for treatments that reduce the average number of cases (negative causal risk difference)

Front

$$\frac{-1}{P\left(Y^{a=1} = 1\right) - P\left(Y^{a=0} = 1\right)}$$

Back

Average causal effect

Front

$$P\left(Y^{a=1} = 1\right) - P\left(Y^{a=0} = 1\right)$$

or

$$E\left[Y^{a=1}\right] - E\left[Y^{a=0}\right]$$

Back

Chapter 2

(4 cards)

Mean exchangeability

Front

\begin{aligned}E\left(Y^a|A=0\right) &= E\left(Y^a|A=1\right)\\&= E\left(Y^a\right)\end{aligned}

Back

Standardization

Front

Weighted average of the stratum-specific risks $$P\left(Y^a=1|L=0\right)$$ and $$P\left(Y^a=1|L=1\right)$$ with weights equal to the proportion of individuals in the population with $$L = 0$$ and $$L = 1$$, respectively

Back

Inverse probability (IP) weighting

Front

Construct pseudo-population by weighting every individual with $$P(A = a|L = \ell)^{-1}$$

Back

Exchangeability

Front

\begin{aligned}P\left(Y^a=1|A=0\right) &= P\left(Y^a=1|A=1\right)\\&= P\left(Y^a=1\right)\end{aligned}

Back

Chapter 3

(2 cards)

Excess fraction or attributable fraction

Front

$$\frac{P(Y = 1) - P\left(Y^{a=0} = 1\right)}{P(Y = 1)}$$

Back

Identifiability conditions for nonparametric identification of ACEs

Front
• Exchangeability
• Positivity (aka experimental treatment assumption)
• Consistency
• Treatment sufficiently well-defined
• Sufficiently well-defined treatments must be present in the data
Back