Range of a Union Probability from Marginals

Worked Solution

How to Think About It: With the marginals fixed but the joint free, the union probability is pinned only by the overlap $P(I_1 \text{ and } I_2)$. Push the events to overlap as much as possible to make the union small; pull them apart to make it large. Those two extremes are exactly the Frechet bounds, and you can read them off without algebra. The correlation of the indicators is also an increasing function of the overlap, so the same two extremes bound it too.

Quick Estimate: Maximum overlap: the smaller event ($I_1$, prob $0.7$) sits entirely inside the larger ($I_2$, prob $0.8$), so the union is just $0.8$. Minimum overlap: $0.7 + 0.8 = 1.5 > 1$, so the events cannot be disjoint -- they must overlap by at least $0.5$, which forces the union up to its ceiling of $1$. Independence would give an overlap of $0.56$ and a union of $0.94$, with correlation $0$; because the marginals differ, the correlation cannot reach $\pm 1$ at either extreme.

Formal Solution:

Part (a): Union. Use inclusion-exclusion: $$P(I_1 \cup I_2) = P(I_1) + P(I_2) - P(I_1 \cap I_2) = 1.5 - P(I_1 \cap I_2).$$ The intersection is bounded by the Frechet inequalities: $$\max(0,\, 0.7 + 0.8 - 1) \le P(I_1 \cap I_2) \le \min(0.7, 0.8),$$ i.e. $0.5 \le P(I_1 \cap I_2) \le 0.7$. Substituting the endpoints: - Largest intersection $0.7$ gives the smallest union $1.5 - 0.7 = 0.8$. - Smallest intersection $0.5$ gives the largest union $1.5 - 0.5 = 1.0$.

Part (b): Default correlation. The same overlap $q = P(I_1 \cap I_2)$ drives the correlation. For indicators, $\text{Var}(I_1) = 0.7 \cdot 0.3 = 0.21$, $\text{Var}(I_2) = 0.8 \cdot 0.2 = 0.16$ and $\text{Cov}(I_1, I_2) = q - 0.7 \cdot 0.8 = q - 0.56$, so $$\rho = \frac{q - 0.56}{\sqrt{0.21 \cdot 0.16}} = \frac{q - 0.56}{\sqrt{0.0336}},$$ increasing in $q$. Map the Frechet endpoints $q \in [0.5, 0.7]$ through it: - $q = 0.7$ (nesting): $\rho = 0.14/\sqrt{0.0336} = \sqrt{7/12} \approx 0.764$. - $q = 0.5$ (least overlap): $\rho = -0.06/\sqrt{0.0336} = -\sqrt{3/28} \approx -0.327$.

So $\rho \in [-\sqrt{3/28},\ \sqrt{7/12}] \approx [-0.327,\ 0.764]$. The range is not $[-1, 1]$: correlation $+1$ would need $I_1 = I_2$, impossible with different marginals, and $-1$ would need $I_2 = 1 - I_1$, impossible since $0.8 \ne 0.3$. It is also lopsided, because both names are likely to default, so they are forced to overlap and cannot be strongly negatively correlated. Independence ($q = 0.56$, union $0.94$) sits inside both ranges.

Answer: (a) $P(I_1 \text{ or } I_2) \in [0.8,\ 1.0]$. The lower end is a perfect nesting ($I_1 \subseteq I_2$); the upper end is the least-overlapping configuration the marginals allow, with overlap $0.5$. (b) $\rho \in [-\sqrt{3/28},\ \sqrt{7/12}] \approx [-0.327,\ 0.764]$, with the same two configurations at the endpoints.