Difference in means (DIM) estimand decomposition
Front
Active users
1
All-time users
1
Favorites
0
Last updated
4 years ago
Date created
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
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