Wildfire Smoke Health Effects: Odds Ratios by Condition, Demographics, and Exposure

• Grand Pooled Effect: Across all 24 strata (k = 24: 8 outcomes × 3 lag periods), the inverse-variance weighted pooled odds ratio is OR = 1.0651 (95% CI: 1.0628–1.0674), indicating a statistically significant 6.5% increase in health service utilization associated with wildfire smoke exposure (z = 58.32, p < 0.0001).

• Lag-Specific Pooled Effects: The 3-day cumulative lag shows the strongest pooled association (OR = 1.0757, 95% CI: 1.0717–1.0797), followed by 7-day (OR = 1.0670) and 0-day (OR = 1.0536), suggesting health impacts peak approximately 3 days post-exposure before partially attenuating.

• Heterogeneity: Cochran's Q = 85.94 (df = 23, p < 0.001) and I² = 73.2% indicate substantial between-study heterogeneity, warranting subgroup and moderator analyses by outcome type, lag, demographics, and fire characteristics.

• Respiratory vs. Cardiovascular: Pooled respiratory OR = 1.067 (95% CI: 1.062–1.072) and cardiovascular OR = 1.064 (95% CI: 1.060–1.069) are nearly identical in magnitude, confirming wildfire smoke as a multi-system health threat affecting both organ systems comparably.

• Mathematical Derivation (Inverse-Variance Fixed-Effects Model):

$$w_i = \frac{1}{\text{SE}_i^2}, \quad \text{SE}_i = \frac{\ln(\text{UB}_i) - \ln(\text{LB}_i)}{2 \times 1.96}$$

$$\hat{\theta}_{\text{pooled}} = \frac{\sum_{i=1}^{k} w_i \cdot \ln(\text{OR}_i)}{\sum_{i=1}^{k} w_i} = \frac{\sum w_i \ln(\text{OR}_i)}{855{,}404} = 0.06307$$

$$\text{OR}_{\text{pooled}} = e^{0.06307} = 1.0651$$

$$\text{SE}_{\text{pooled}} = \frac{1}{\sqrt{\sum w_i}} = \frac{1}{\sqrt{855{,}404}} = 0.001081$$

$$\text{95\% CI} = \exp\!\bigl(\hat{\theta} \pm 1.96 \times \text{SE}\bigr) = \exp(0.06307 \pm 0.00212) = [1.0628,\; 1.0674]$$

$$z = \frac{\hat{\theta}_{\text{pooled}}}{\text{SE}_{\text{pooled}}} = \frac{0.06307}{0.001081} = 58.32, \quad p < 0.0001$$

$$Q = \sum_{i=1}^{k} w_i \bigl(\ln\text{OR}_i - \hat{\theta}_{\text{pooled}}\bigr)^2 = 85.94, \quad df = k - 1 = 23$$

$$I^2 = \frac{Q - (k-1)}{Q} \times 100\% = \frac{85.94 - 23}{85.94} \times 100\% = 73.2\%$$