Algorithm 1:

The sampling algorithm.

Input: A longitudinal dataset $\mathbb{D}$ consisting of visit-level records for $\mathit{N}$ participants. For each participant $\mathit{i}$, a visit set is denoted as ${\mathit{V}}_{\mathit{i}}=\left\{{\mathit{v}}_{\mathit{i}\mathbf{1}},{\mathit{v}}_{\mathit{i}\mathbf{2}},\dots ,{\mathit{v}}_{\mathit{i}{\mathit{T}}_{\mathit{i}}}\right\}$, where ${\mathit{T}}_{\mathit{i}}$ indicates a number of total visits, including: participant ID and baseline demographics at $\mathit{T}=\mathbf{1}$, and cognitive impairment status, comorbidities and NPS indicators over time
Output: Set of bootstrap samples $𝓑=\left\{{\mathit{B}}_{\mathbf{1}},{\mathit{B}}_{\mathbf{2}},\dots ,{\mathit{B}}_{\mathbf{100}}\right\}$ with each dataset ${\mathit{B}}_{\mathit{b}}\in {ℝ}^{\mathit{N}\times \mathit{d}}$
1	Initialize $ℬ\leftarrow \varnothing$
2	for $b=1$ to 100 do
	Initialize $B_b\leftarrow \varnothing$
	for $i=1$ to $N$ do
	Let $V_i={\left\{v_{ij}\right\}}_{j=1}^{T_i}$, sorted by time
	Partition $V_i=V^n(\text{normal}\text{cognition})\cup V^c(\text{cognitively impaired})$
	Let status $s_{i{T}_i}\in \{\mathit{\text{normal}}\mathit{\text{cognition}},\mathit{\text{cognitively}}\mathit{\text{impaired}}\}$
	Extract ${\mathit{x}}_i^0\leftarrow$ baseline demographics from $v_{i1}$
	if $s_{i{T}_i}=\mathit{\text{cognitively}}\mathit{\text{impaired}}$ then
	Random sample $S_i^{\prime }\subset V^n\cup V^{\text{c}},\left|S_i^{\prime }\right|=3,s.t.$
	$S_i^{\prime }=\left\{\begin{array}{l}\left\{v_{i{t}_1}^n,v_{i{t}_2}^c,v_{i{t}_3}^c\right\}\hfill \\ \left\{v_{i{t}_1}^n,v_{i{t}_2}^n,v_{i{t}_3}^c\right\}\hfill \end{array}\right.$
	Random sample $S_i^c\subset V_i^c,\left|S_i^c\right|=3$
	else
	Random sample $S_i^{\prime }\subset V_i^n,\left|S_i^{\prime }\right|=3$
	end if
	Let ${\mathit{x}}_i^{\prime }$ disease and NPS statuses from $S_i^{\prime }$
	Let ${\mathit{x}}_i\in {ℝ}^d\leftarrow \text{concatenate}{\mathit{x}}_i^0\text{and}{\mathit{x}}_i^{\prime }$
	Append ${\mathit{x}}_i$ to $B_b$
	end for
	Append $B_b$ to $ℬ$
	end for
3	Return $ℬ$