A review of Zhao, Small, and Bhattacharya (2019) on conducting robust sensitivity analysis for IPW estimators using marginal sensitivity models and the percentile bootstrap.
We assume that $(A_1, X_1, Y_1), \dots, (A_n, X_n, Y_n)$ are independent and identically distributed (IID) from a joint true data-generating distribution, $F_0$.
Crucially, $Y_i$ is only observed if $A_i = 1$. In reality, the data we actually get to see is $(A_i, X_i, A_i Y_i)$.
$$\mu := \mathbb{E}_0[Y]$$
$$e_0(x, y) = \mathbb{P}_0(A = 1 \mid X = x, Y = y)$$
$$e_0(x, y) = e_0(x)$$
MAR assumes that the probability of missingness depends only on the observed covariates, not the unobserved response.
However, MAR is a strong assumption, and in many real-world scenarios, it is easily violated.
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.
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 $1$.