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Difference in means (DIM) estimand decomposition

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Apr 6, 2021

Unsectioned

(7 cards)

Difference in means (DIM) estimand decomposition

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

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

Difference in means (DIM)

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

Standardized mean difference (SMD)

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

where \(s^2\) is sample variance. Typically performed for each covariate.

Individual treatment effect

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

Average treatment effect among the treated (ATT)

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

Average treatment effect (ATE)

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

Fisher's exact test for the causal sharp null hypothesis

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

Let \(\Omega\) be set of possible ways to assign treatments

- Calculate \(\hat{\theta}_\text{true}\) (DIM)
- Calculate \(\hat{\theta}(\omega)\) for every \(\omega \in \Omega\)
- Compare \(\hat{\theta}_\text{true}\) to \(\hat{\theta}(\omega)\)'s to see how extreme

Chapter 1

(4 cards)

Sharp causal null hypothesis

There is no *individual* causal effect, \(Y^{a=1} = Y^{a=0}\) for all individuals

Causal null hypothesis

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

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

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

Average causal effect

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

or

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

Chapter 2

(4 cards)

Mean exchangeability

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

Standardization

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

Inverse probability (IP) weighting

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

Exchangeability

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

Chapter 3

(2 cards)

**Excess fraction** or **attributable fraction**

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

Identifiability conditions for nonparametric identification of ACEs

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

- Treatment