Paper review: Sensitivity Analysis for Inverse Probability Weighting Estimators via the Percentile Bootstrap

A review of Zhao, Small, and Bhattacharya (2019) on conducting robust sensitivity analysis for IPW estimators using marginal sensitivity models and the percentile bootstrap.

  1. Sensitivity Analysis for Inverse Probability Weighting Estimators via the Percentile Bootstrap
    Qingyuan Zhao, Dylan S. Small, and Bhaswar B. Bhattacharya
    Journal of the Royal Statistical Society Series B: Statistical Methodology, Sep 2019

The Setup: The Data Generating Process

$$\mu := \mathbb{E}_0[Y]$$


The “Missing at Random” (MAR) Assumption

$$e_0(x, y) = \mathbb{P}_0(A = 1 \mid X = x, Y = y)$$

$$e_0(x, y) = e_0(x)$$


Sensitivity Model

Definition: The Marginal Sensitivity Model

Fix a parameter $\Lambda \ge 1$. We assume that our missingness mechanism $e(x, y)$ belongs to a set $\mathcal{E}(\Lambda)$, defined as:

$$\mathcal{E}(\Lambda) = \left\{ e(x, y) \in [0, 1] : \frac{1}{\Lambda} \le \text{OR}(e(x, y), e_0(x)) \le \Lambda, \text{ for all } x \in \mathcal{X}, y \in \mathbb{R} \right\}$$

Where $\text{OR}$ is the odds ratio, defined as:

$$\text{OR}(p_1, p_2) = \frac{p_1 / (1 - p_1)}{p_2 / (1 - p_2)}$$

By adjusting $\Lambda$, you control the “budget” of unmeasured confounding. If $\Lambda = 1$, you are strictly enforcing the MAR assumption. As $\Lambda$ grows, you allow for more severe violations of MAR.


The Logistic Representation

It is often much easier to write this model using the logistic (logit) scale.

Let’s define the logit functions for our probabilities:

$$g_0(x) = \text{logit}\{e_0(x)\} = \log \left( \frac{e_0(x)}{1 - e_0(x)} \right)$$

$$g_0(x, y) = \text{logit}\{e_0(x, y)\}$$

$$h_0(x, y) = g_0(x) - g_0(x, y)$$

So $h$ quantifies the degree of shift. We can now express our sensitivity model probability as a function of this shift, $h$:

$$e^{(h)}(x, y) = \left[ 1 + \exp\{h(x, y) - g_0(x)\} \right]^{-1}$$

Key Takeaway: The marginal sensitivity model simply puts a bound on the $L_\infty$-norm of $h$.

If we define $\lambda = \log(\Lambda)$, the sensitivity model translates to:

$$\mathcal{H}(\lambda) = \left\{ h : \mathcal{X} \times \mathbb{R} \to \mathbb{R} \text{ and } \|h\|_\infty \le \lambda \right\}$$

It is easy to see the model’s behavior: as $h(x, y) \to \pm\infty$, the selection probability $e^{(h)}(x, y)$ approaches $0$ or $1$.