← Jane Street Interview Insights
This is where inclusion-exclusion saves you.
Use the inclusion-exclusion principle to express P(A∪B) = p + q - P(A∩B), then bound P(A∩B) using the constraints 0 ≤ P(A∩B) ≤ min(p, q) and also P(A∩B) ≥ max(0, p+q-1) from the union bound. The range is [max(p,q), min(1, p+q)], and you must construct explicit probability spaces (e.g., using a uniform distribution over outcomes) to show both endpoints are achievable.
Pro tip: When explaining achievability, explicitly construct simple probability spaces (like a 4-outcome sample space with probabilities) to demonstrate the extreme cases. This shows you can translate abstract bounds into concrete examples, a skill valued in quantitative roles.
Let A be rain on Saturday and B be rain on Sunday. Then P(A∪B) = P(A) + P(B) - P(A∩B) = p + q - P(A∩B).
Since P(A∩B) ≤ P(A) = p and P(A∩B) ≤ P(B) = q, we have P(A∩B) ≤ min(p, q). Also, by the union bound, P(A∩B) ≥ P(A) + P(B) - 1 = p + q - 1, and trivially P(A∩B) ≥ 0, so P(A∩B) ≥ max(0, p+q-1).
Substitute the bounds into P(A∪B) = p + q - P(A∩B). The maximum occurs when P(A∩B) is minimized: P(A∪B) ≤ p + q - max(0, p+q-1) = min(1, p+q). The minimum occurs when P(A∩B) is maximized: P(A∪B) ≥ p + q - min(p, q) = max(p, q). Thus the range is [max(p,q), min(1, p+q)].
To achieve P(A∪B) = max(p,q), assume without loss of generality that p ≥ q. Then set P(A∩B) = q (so B is a subset of A). This is possible if q ≤ p. Construct a probability space where B implies A, e.g., outcomes: A only with prob p-q, both with prob q, neither with prob 1-p.
To achieve P(A∪B) = min(1, p+q), we need P(A∩B) = max(0, p+q-1). If p+q ≤ 1, set P(A∩B)=0 (disjoint events). If p+q > 1, set P(A∩B)=p+q-1, which is feasible since p+q-1 ≤ min(p,q). Construct explicit probability spaces accordingly.
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The short-duration case means rain on Saturday makes rain on Sunday less likely, so negative dependence, so P(A intersect B) is smaller than pq, so the union is pushed toward the upper end of the range.
Clarify the underlying probability model and assumptions, then reason about how the duration of rain events affects the probability of at least one rainy day within a given range. Use conditional probability and independence assumptions to derive bounds or exact probabilities, and discuss how the answer changes when events persist for multiple days.
Pro tip: Explicitly state your assumptions about the distribution of rain events and the independence of days, as Jane Street values clear probabilistic reasoning and the ability to handle ambiguity. Acknowledge that the true probability depends on the correlation between days, and provide a range or bounds rather than a single number.
Restate the problem in your own words and ask clarifying questions about the range, the definition of a rain event, and whether events are independent. Assume a simple model if no additional information is given.
If rain events last exactly one day, then each day is either rainy or not, and the probability of at least one rainy day in a range of N days is 1 - (1 - p)^N, where p is the daily probability of rain. Discuss how p might be estimated from historical data.
If rain events persist for two or more days, the days are not independent. Model the probability of at least one rainy day as 1 minus the probability that no rain event starts within the range, considering that an event could start before the range and continue into it.
Compare the probabilities under the two scenarios and explain how the persistence of rain events increases the chance of at least one rainy day for a given range, especially for shorter ranges. Discuss the impact of the range length and the event duration.
Conclude with a summary of how the true probability falls within a range depending on the assumptions, and provide bounds or a sensitivity analysis. Emphasize that without additional data, the exact probability cannot be determined.
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Once r is given the union is just p + q minus r, which is clean.
First, clarify the problem setup: define events A and B as rainy on day 1 and day 2, with P(A)=p, P(B)=q, and P(A∩B)=r. Then use the inclusion-exclusion principle to compute P(A∪B) = p + q - r, and determine feasible r by ensuring all probabilities are between 0 and 1 and consistent with the individual marginals (i.e., max(0, p+q-1) ≤ r ≤ min(p, q)).
Pro tip: Emphasize that r is not arbitrary; it must satisfy the Fréchet–Hoeffding bounds, which reflect the possible dependence between the two rainy days. This shows you understand the constraints beyond just the formula.
Let A be the event that day 1 is rainy and B be the event that day 2 is rainy. Then P(A)=p, P(B)=q, and P(A∩B)=r.
The probability of at least one rainy day is P(A∪B) = P(A) + P(B) - P(A∩B) = p + q - r.
Since probabilities must be between 0 and 1, we need 0 ≤ r ≤ 1, r ≤ p, r ≤ q, and also p + q - r ≤ 1 (from P(A∪B) ≤ 1). These yield max(0, p+q-1) ≤ r ≤ min(p, q).
Check that the bounds ensure P(A|B) and P(B|A) are valid probabilities, i.e., r/q ≤ 1 and r/p ≤ 1, which are already covered by r ≤ p and r ≤ q.
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Under independence P(A intersect B) = pq, so the union is p + q minus pq.
First, clarify the setup: two days with rain probabilities p and q, both in (0,1). The range for P(at least one rainy day) is [max(p,q), min(1, p+q)] by the union bound and monotonicity. Under independence, P = p + q - pq. Show that this value lies strictly inside the range when p and q are strictly between 0 and 1, and explain why it cannot equal either endpoint.
Pro tip: Emphasize that independence is a specific dependence structure that yields a value strictly between the Fréchet bounds. This demonstrates understanding of how dependence affects joint probabilities, a key concept in probability and statistics.
Confirm that p and q are the marginal probabilities of rain on each day, both strictly between 0 and 1, and that the two days are independent.
State that for any joint distribution, P(A∪B) is bounded below by max(p,q) and above by min(1, p+q). These are the Fréchet bounds.
Under independence, P(A∪B) = p + q - pq. Show that this expression is strictly greater than max(p,q) and strictly less than min(1, p+q) when 0 < p,q < 1.
Demonstrate algebraically: p + q - pq > p because q(1-p) > 0, and similarly > q. Also p + q - pq < p + q because pq > 0, and < 1 because (1-p)(1-q) > 0.
Conclude that the independent value lies strictly inside the range and cannot touch either endpoint unless p or q is 0 or 1.
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I did not see this coming and kind of stalled.
First, clarify that the marginal probabilities alone do not determine the joint distribution, so the probability of at least one rainy day can vary. Then, derive the minimum and maximum possible values by considering extreme dependence structures: for the minimum, make rainy days as mutually exclusive as possible; for the maximum, make them as positively dependent as possible (e.g., one event implies the others).
Pro tip: Explicitly state that without additional assumptions (like independence), the probability is not uniquely determined—this demonstrates statistical maturity and avoids the common trap of assuming independence. Then, show the bounds using the union bound and the fact that the maximum probability is capped at 1.
State that the marginal probabilities are given for each day, but the joint distribution is unspecified. Emphasize that the answer is a range, not a single value.
Use the union bound (Boole's inequality): P(at least one) ≤ sum of marginals. For the lower bound, consider the maximum possible overlap of the complement events (no rain). The minimum occurs when rainy days are as mutually exclusive as possible, but if the sum of marginals exceeds 1, the minimum is sum - 2 (by inclusion-exclusion for three events). More precisely, the minimum is max(0, sum of marginals - 2) when considering three events? Actually, for three events, the minimum of P(A∪B∪C) given marginals is max(0, sum - 2) if sum > 2? Wait, let's derive: The minimum of P(A∪B∪C) is max(0, sum - 2) because the maximum overlap of complements is limited by 1. But careful: For three events, the minimum is max(0, sum - 2) only if sum ≤ 3? Actually, the minimum is max(0, sum - 2) because the sum of probabilities of the complements is 3 - sum, and the intersection of complements has probability at most 1, so P(no rain) ≤ 1, but we want to maximize P(no rain) to minimize P(at least one). The maximum possible P(no rain) is min(1, 3 - sum) because the sum of the probabilities of the complements is 3 - sum, and the intersection of complements is at most the minimum of the complements' probabilities, but more simply, P(no rain) ≤ 1 and also P(no rain) ≤ 3 - sum? Actually, P(no rain) = 1 - P(at least one). The sum of marginals = sum P(A_i). We have P(at least one) ≥ sum P(A_i) - sum_{i<j} P(A_i ∩ A_j) + ... but that's not directly helpful. The standard result: For n events, the minimum of P(∪ A_i) given marginals p_i is max(0, sum p_i - (n-1)). For n=3, min = max(0, sum p_i - 2). The maximum is min(1, sum p_i). So the range is [max(0, sum p_i - 2), min(1, sum p_i)].
The maximum probability of at least one rainy day is min(1, sum of marginals). This occurs when the events are as positively dependent as possible, e.g., if one event occurs then all occur, or by making the events nested.
Construct simple examples to illustrate the bounds. For instance, if all marginals are 0.5, the range is [max(0, 1.5-2)=0, min(1,1.5)=1], so any value between 0 and 1 is possible. If marginals are 0.2, 0.3, 0.4, sum=0.9, range is [0, 0.9].
Conclude that the probability of rain on at least one of the three days lies in the interval [max(0, p1+p2+p3 - 2), min(1, p1+p2+p3)].
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P(A intersect B) = p + q minus P(A union B), so that's determined exactly.
Start by recalling the inclusion-exclusion principle: P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Since only P(A ∪ B) is given, express P(A ∩ B) in terms of the unknown P(A) and P(B), then use probability bounds to derive the range of possible values. For the conditional probability, use the definition P(B|A) = P(A ∩ B) / P(A) and analyze how it varies with P(A) and P(B) within the constraints.
Pro tip: Emphasize that without additional information, the intersection and conditional probability are not uniquely determined; demonstrating the range of possibilities with concrete numerical examples shows rigor and practical insight.
Write the formula P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Since P(A ∪ B) is known, rearrange to P(A ∩ B) = P(A) + P(B) - P(A ∪ B).
Use the constraints 0 ≤ P(A), P(B) ≤ 1 and P(A ∩ B) ≤ min(P(A), P(B)) to bound P(A ∩ B). Also note P(A ∩ B) ≥ max(0, P(A) + P(B) - 1) by subadditivity.
Combine the bounds to find the minimum and maximum possible values of P(A ∩ B) given only P(A ∪ B). The minimum is max(0, 2P(A ∪ B) - 1) if P(A) = P(B) = P(A ∪ B), and the maximum is P(A ∪ B) (when one event is a subset of the other).
Express P(B|A) = P(A ∩ B) / P(A). Since P(A) is unknown, P(B|A) can range from 0 to 1, but with constraints: if P(A) is large, P(B|A) may be small; if P(A) is small, P(B|A) may be large. Provide examples to illustrate.
Summarize that without additional information about P(A) or P(B), the intersection and conditional probability are not uniquely determined; only bounds can be provided.
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