Weekend Rain Probability with Dependence
The probability of rain on Saturday is $P(S) = 0.2$ and the probability of rain on Sunday is $P(U) = 0.3$.
- If Saturday rain and Sunday rain are independent, compute $P(\text{rain at least one day})$.
- Suppose Saturday rain makes Sunday rain more likely, i.e., $P(U \mid S) > P(U)$. Does $P(\text{rain at least one day})$ increase or decrease compared to the independent case?
- Without any assumption about the dependence between Saturday and Sunday rain, what is the tightest possible range for $P(\text{rain at least one day})$?
Hints
- Write out the inclusion-exclusion formula $P(S \cup U) = P(S) + P(U) - P(S \cap U)$ and notice that $P(S) + P(U)$ is fixed at $0.5$.
- For part (ii), positive dependence means $P(S \cap U) > P(S) P(U)$. How does increasing $P(S \cap U)$ affect $P(S \cup U)$?
- For part (iii), use the Frechet bounds: $P(S \cap U)$ ranges from $\max(0, P(S)+P(U)-1)$ to $\min(P(S), P(U))$. Substitute each extreme into the inclusion-exclusion formula.
Worked Solution
How to Think About It: This is a clean inclusion-exclusion exercise with a twist: you need to understand how dependence shifts the joint probability $P(S \cap U)$ and what that does to the union. The key identity is $P(S \cup U) = P(S) + P(U) - P(S \cap U) = 0.5 - P(S \cap U)$. So the union moves in the opposite direction of the intersection -- the more the rain events overlap, the less total "rain coverage" you get across the weekend. That is the central insight for all three parts.
Quick Estimate: Under independence, $P(S \cap U) = 0.06$, so $P(S \cup U) = 0.44$. If the events are perfectly bundled (always rain both days or neither), the intersection is as large as possible ($0.2$), and the union shrinks to $0.3$. If the events are perfectly anti-bundled (never rain both days), the intersection is $0$ and the union grows to $0.5$. So the answer to part (iii) should be $[0.3, 0.5]$, and positive dependence should push us below $0.44$.
Approach: Use inclusion-exclusion together with the Frechet bounds on $P(S \cap U)$.
Formal Solution:
Let $S$ = Saturday rain, $U$ = Sunday rain. Throughout, $P(S) = 0.2$, $P(U) = 0.3$.
The inclusion-exclusion identity gives:
$P(S \cup U) = P(S) + P(U) - P(S \cap U) = 0.5 - P(S \cap U)$
(i) Independent case:
$P(S \cap U) = P(S) \cdot P(U) = 0.06$
$P(S \cup U) = 0.5 - 0.06 = 0.44$
(ii) Positive dependence ($P(U \mid S) > P(U) = 0.3$):
Positive dependence means $P(S \cap U) = P(S) \cdot P(U \mid S) > 0.2 \times 0.3 = 0.06$. Since $P(S \cup U) = 0.5 - P(S \cap U)$, a larger intersection means a smaller union.
$P(S \cup U) < 0.5 - 0.06 = 0.44$
The probability of rain at least once decreases. Intuitively, the rain events "bunch together" -- when it rains Saturday, it is more likely to also rain Sunday, but when Saturday is dry, Sunday tends to be drier too. The clustering reduces total weekend rain coverage.
(iii) Range under arbitrary dependence:
The Frechet bounds constrain the joint probability:
$\max(0,\; P(S) + P(U) - 1) \leq P(S \cap U) \leq \min(P(S),\, P(U))$
$\max(0,\; -0.5) \leq P(S \cap U) \leq \min(0.2,\, 0.3)$
$0 \leq P(S \cap U) \leq 0.2$
Plugging into $P(S \cup U) = 0.5 - P(S \cap U)$:
- Maximum union: $P(S \cap U) = 0$ gives $P(S \cup U) = 0.5$ (the events never overlap).
- Minimum union: $P(S \cap U) = 0.2$ gives $P(S \cup U) = 0.3$ (Saturday rain is a subset of Sunday rain).
Answer:
- (i) $P(\text{rain at least once}) = 0.44$ under independence.
- (ii) Positive dependence decreases $P(\text{rain at least once})$ below $0.44$.
- (iii) $P(\text{rain at least once}) \in [0.3,\, 0.5]$.
Intuition
The core lesson is that inclusion-exclusion makes the union a decreasing function of the intersection: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. When two events are positively dependent, they cluster together, making the intersection larger and the union smaller. Think of it like umbrellas -- if rainy days tend to come in pairs, you might need your umbrella fewer distinct days, even though each rainy day is wetter. This exact reasoning appears constantly in portfolio risk: positively correlated losses "bunch up" into fewer but more severe drawdowns, while the probability of experiencing at least some loss can actually decrease relative to the independent case.
The Frechet bounds in part (iii) are a fundamental tool in quantitative risk. When you know marginal distributions but not the copula (dependence structure), these bounds give you the worst-case and best-case joint probabilities. In practice, regulators and risk managers use them to stress-test portfolios when the correlation structure is uncertain.