Weekend Rain Bounds via Frechet-Hoeffding

Worked Solution

How to Think About It: This is a classic question about what you can say about a joint probability when you only know the marginals. In practice, this comes up constantly -- you know the probability of two individual events but the dependence structure is unknown. The Frechet-Hoeffding bounds give you the tightest possible interval for $P(A \cap B)$ over all possible couplings of $A$ and $B$. The qualitative parts test whether you understand what positive and negative dependence mean in plain terms.

Quick Estimate: Take $p = 0.4$, $q = 0.3$. Independence gives $P(A \cap B) = 0.12$. The maximum possible overlap is $\min(0.4, 0.3) = 0.3$ (Sunday rain implies Saturday rain). The minimum is $\max(0.4 + 0.3 - 1, 0) = \max(-0.3, 0) = 0$. So for these marginals, $P(A \cap B) \in [0, 0.3]$, with independence sitting at $0.12$. Positive dependence pushes toward $0.3$; negative dependence pushes toward $0$.

Approach: Apply inclusion-exclusion and the constraint that all probabilities lie in $[0,1]$.

Formal Solution:

Part 1: Frechet-Hoeffding bounds.

We need two constraints on $P(A \cap B)$:

• *Upper bound:* Since $A \cap B \subseteq A$ and $A \cap B \subseteq B$, we have $P(A \cap B) \leq \min(p, q)$.

• *Lower bound:* By inclusion-exclusion, $P(A \cup B) = p + q - P(A \cap B)$. Since $P(A \cup B) \leq 1$, we get $P(A \cap B) \geq p + q - 1$. Also $P(A \cap B) \geq 0$. Combining: $P(A \cap B) \geq \max(p + q - 1, 0)$.

Both bounds are tight -- they are achieved by specific joint distributions:

• The upper bound $\min(p, q)$ is achieved when one event is a subset of the other (maximal positive dependence).
• The lower bound $\max(p + q - 1, 0)$ is achieved when the events are as "anti-aligned" as possible (maximal negative dependence, or mutual exclusivity when $p + q \leq 1$).

So the tight range is:

$$\max(p + q - 1, \, 0) \;\leq\; P(A \cap B) \;\leq\; \min(p, q)$$

Part 2: "Rain usually lasts at least two days."

This statement means that if it rains on one day, it very likely rains on the adjacent day as well. In probabilistic terms, this is strong positive dependence: $P(B \mid A)$ is high (close to 1), and similarly $P(A \mid B)$ is high. Since $P(A \cap B) = P(B \mid A) \cdot p$, a high conditional probability pushes $P(A \cap B)$ upward.

In the extreme case, rain on one day guarantees rain on the other, giving $P(A \cap B) = \min(p, q)$, the upper Frechet bound.

So $P(A \cap B)$ tends toward the upper end of the range, near $\min(p, q)$.

Part 3: "Rain rarely lasts more than one day."

This means rain events tend to be isolated -- if it rains on Saturday, it is unlikely to rain on Sunday, and vice versa. This is negative dependence: $P(B \mid A)$ is small relative to $P(B)$. This pushes $P(A \cap B)$ downward.

In the extreme case where rain never lasts two consecutive days, $A$ and $B$ are mutually exclusive (if $p + q \leq 1$), giving $P(A \cap B) = 0$. More generally, $P(A \cap B) \approx \max(p + q - 1, 0)$.

So $P(A \cap B)$ tends toward the lower end of the range, near $\max(p + q - 1, 0)$.

Answer:

$$\max(p + q - 1, \, 0) \;\leq\; P(A \cap B) \;\leq\; \min(p, q)$$

"Rain lasts two days" implies strong positive dependence, pushing $P(A \cap B)$ toward the upper bound $\min(p, q)$. "Rain rarely lasts more than one day" implies negative dependence, pushing $P(A \cap B)$ toward the lower bound $\max(p + q - 1, 0)$.