A nonparametric method for producing isolines of bivariate exceedance probabilities

Abstract

We present a method for drawing isolines indicating regions of equal joint exceedance probability for bivariate data. The method relies on bivariate regular variation, a dependence framework widely used for extremes. The method we utilize for characterizing dependence in the tail is largely nonparametric. The extremes framework enables drawing isolines corresponding to very low exceedance probabilities and may even lie beyond the range of the data; such cases would be problematic for standard nonparametric methods. Furthermore, we extend this method to the case of asymptotic independence and propose a procedure which smooths the transition from hidden regular variation in the interior to the first-order behavior on the axes. We propose a diagnostic plot for assessing the isoline estimate and choice of smoothing, and a bootstrap procedure to visually assess uncertainty.

Appendix

We show that the smoothed scaling procedure in Section 4 preserves the requirement that isolines of exceedance probabilities must have negative slopes.

Assume the estimated survival function is strictly decreasing, i.e., if x₁ = (x_1,1,x_1,2)≠x₂ = (x_2,1,x_2,2), x_1,1 ≤ x_2,1, and x_1,2 ≤ x_2,2, then $\hat {\bar F}_{\boldsymbol X}(\boldsymbol x_{1}) > \hat {\bar F}_{\boldsymbol X}(\boldsymbol x_{2})$. We first show that it follows that any isoline $\hat \ell _{\boldsymbol X}(p)$ must have a negative slope. Let x₁ and x₂ be two distinct locations on $\hat \ell _{\boldsymbol X}(p)$. WLOG, assume x_2,2 − x_1,2 ≥ 0 and x_2,1 − x_1,1 ≥ 0, implying the slope is not negative. This implies x_2,2 ≥ x_1,2 and x_2,1 ≥ x_1,1, but $\hat {\bar F}_{\boldsymbol X}(\boldsymbol x_{1}) = \hat {\bar F}_{\boldsymbol X}(\boldsymbol x_{2})$, which is a contradiction.

As the transformation to Fréchet scale is monotonic, $\hat \ell _{\boldsymbol Z}(p)$ has negative slopes.

Let $\boldsymbol z_{1}^{(proj)}, \boldsymbol z_{2}^{(proj)}$ be any two points in $\hat \ell _{\boldsymbol Z}(p_{proj})$. Let s = p_base/p_proj. Let $\boldsymbol z_{1}^{(base)}, \boldsymbol z_{2}^{(base)}$ be the points in $\hat \ell _{\boldsymbol Z}(p_{base})$ such that $\boldsymbol z_{i}^{(proj)} = (s^{\eta _{1}(\boldsymbol z_{i}^{(base)})} z^{(base)}_{i,1},$$s^{\eta _{2}(\boldsymbol z_{i}^{(base)})} z^{(base)}_{i,2})$ for i = 1, 2. WLOG assume $z^{(base)}_{2,2} < z^{(base)}_{1,2}$ and $z^{(base)}_{2,1} > z^{(base)}_{1,1}$.

Note that since $z^{(base)}_{2,1} > z^{(base)}_{1,1}$, $\eta _{1}(\boldsymbol z^{(base)}_{2}) > \eta _{1}(\boldsymbol z^{(base)}_{1})$, and likewise since $z^{(base)}_{2,2} < z^{(base)}_{1,2}$, $\eta _{2}(\boldsymbol z^{(base)}_{2}) < \eta _{2}(\boldsymbol z^{(base)}_{1})$. Hence,

$$ z^{(proj)}_{2,2} \!- z^{(proj)}_{1,2} = s^{\eta_{2}(\boldsymbol z_{2}^{(base)})} z^{(base)}_{2,2} \!- s^{\eta_{2}(\boldsymbol z_{1}^{(base)})} z^{(base)}_{1,2} \!<\! s^{\eta_{2}(\boldsymbol z_{1}^{(base)})} (z^{(base)\!}_{2,2} - z^{(base)}_{1,2}) \!<\! 0, $$

and

$$ z^{(proj)}_{2,1} - z^{(proj)}_{1,1} = s^{\eta_{1}(\boldsymbol z_{2}^{(base)})} z^{(base)}_{2,1} \!- s^{\eta_{1}(\boldsymbol z_{1}^{(base)})} z^{(base)}_{1,1} \!>\! s^{\eta_{1}(\boldsymbol z_{1}^{(base)})} (z^{(base)}_{2,1} \!- z^{(base)}_{1,1}) \!>\! 0. $$

Thus, the slope between any two points on $\hat \ell _{\boldsymbol Z}(p_{proj})$ is negative, and since the marginal transformation is monotonic, the slope between any two points on $\hat \ell _{\boldsymbol X}(p_{proj})$ is negative.