You are given:

\[ P(A \text{ defaults}) = 0.50, \quad P(B \text{ defaults}) = 0.30 \]

Let’s denote:

  • \( 1_A \): Indicator for A defaulting
  • \( 1_B \): Indicator for B defaulting

Part 1: Bounds on \( P(A \cup B) \)

By the inclusion-exclusion principle:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Since \( P(A) = 0.5 \) and \( P(B) = 0.3 \), we have:

\[ P(A \cup B) = 0.5 + 0.3 - P(A \cap B) \]

To bound this, we need bounds on \( P(A \cap B) \).

  • Minimum of \( P(A \cap B) \): \( \max(0, P(A) + P(B) - 1) = \max(0, 0.5 + 0.3 - 1) = \boxed{0.0} \)
  • Maximum of \( P(A \cap B) \): \( \min(P(A), P(B)) = \boxed{0.3} \)

Therefore:

  • Max \( P(A \cup B) \) = \( 0.5 + 0.3 - 0 = \boxed{0.8} \)
  • Min \( P(A \cup B) \) = \( 0.5 + 0.3 - 0.3 = \boxed{0.5} \)

Part 2: Bounds on Correlation \( \mathrm{Corr}(1_A, 1_B) \)

We compute the correlation of the binary indicators:

\[ \mathrm{Corr}(1_A, 1_B) = \frac{\mathrm{Cov}(1_A, 1_B)}{\sqrt{\mathrm{Var}(1_A) \cdot \mathrm{Var}(1_B)}} \]

  • \( \mathrm{Cov}(1_A, 1_B) = P(A \cap B) - P(A)P(B) \)
  • \( \mathrm{Var}(1_A) = 0.5(1 - 0.5) = 0.25 \)
  • \( \mathrm{Var}(1_B) = 0.3(1 - 0.3) = 0.21 \)

\[ \mathrm{Corr}(1_A, 1_B) = \frac{P(A \cap B) - 0.15}{\sqrt{0.25 \cdot 0.21}} = \frac{P(A \cap B) - 0.15}{\sqrt{0.0525}} \]

  • Max \( P(A \cap B) = 0.3 \): Correlation = \( \frac{0.3 - 0.15}{\sqrt{0.0525}} \approx \boxed{0.6547} \)
  • Min \( P(A \cap B) = 0.0 \): Correlation = \( \frac{0 - 0.15}{\sqrt{0.0525}} \approx \boxed{-0.6547} \)

Final Answers

(1) Bounds on \( P(A \cup B) \):

\[ \boxed{0.5 \le P(A \cup B) \le 0.8} \]

(2) Corresponding bounds on \( \mathrm{Corr}(1_A, 1_B) \):

\[ \boxed{-0.6547 \le \mathrm{Corr}(1_A, 1_B) \le 0.6547} \]

Reference