Are you preparing for a Jane Street quantitative trader interview? One of the classic probability puzzles you may encounter deals with random intervals and their intersection properties. Understanding these questions not only tests your mathematical rigor but also your ability to model and analyze stochastic processes—a skill highly valued at quantitative trading firms like Jane Street. In this article, we’ll break down the famous "Probability of Overlapping Random Intervals" problem, explain key mathematical concepts involved, and guide you step by step through its solution.

Jane Street Quantitative Trader Interview Question: Probability of Overlapping Random Intervals

Table of Contents

• Problem Statement
• Key Concepts Involved
• Restating the Problem Mathematically
• How Are the Random Intervals Generated?
• Condition for Non-Empty Intersection
• Mathematical Formulation
• Step-by-Step Detailed Solution
• Generalization to n Intervals
• Python Simulation
• Jane Street Interview Tips
• Conclusion

Problem Statement

Question: Five intervals are independently generated by sampling two points uniformly from [0,1] and taking them as endpoints. What is the probability that all five intervals share a common non-empty intersection?

Key Concepts Involved

• Uniform Distribution
• Random Intervals
• Intersection of Intervals
• Order Statistics
• Probability Calculation

Restating the Problem Mathematically

Let’s formalize the question. For each interval \(I_i\) for \(i = 1, 2, 3, 4, 5\):

• Pick \(X_i, Y_i\) independently and uniformly at random from [0, 1].
• Let \(A_i = \min(X_i, Y_i)\), \(B_i = \max(X_i, Y_i)\), so \(I_i = [A_i, B_i]\).

How Are the Random Intervals Generated?

For each interval:

• Pick two numbers \(X\) and \(Y\) independently and uniformly distributed on [0, 1].
• Define the interval as \([A, B]\), where \(A = \min(X, Y)\) and \(B = \max(X, Y)\).
• The random interval is thus a segment of [0, 1], possibly the entire [0, 1] if \(X = 0\) and \(Y = 1\).

Condition for Non-Empty Intersection

For the intersection of all intervals to be non-empty, there must be at least one point \(x\) in [0, 1] that lies inside every interval \(I_i\). In other words: \[ \exists x \in [0, 1] \text{ such that } \forall i,\, x \in [A_i, B_i] \] This is equivalent to: \[ \max(A_1, A_2, ..., A_5) \leq \min(B_1, B_2, ..., B_5) \] If the maximum of the left endpoints is less than or equal to the minimum of the right endpoints, the intervals overlap.

Mathematical Formulation

Let’s introduce:

• \(M = \max(A_1, ..., A_5)\): the largest of the five lower endpoints.
• \(m = \min(B_1, ..., B_5)\): the smallest of the five upper endpoints.

Step-by-Step Detailed Solution

Step 1: Distribution of Endpoints

Given \(X\) and \(Y\) are both uniform on [0, 1], and independent, what is the joint distribution of \(A = \min(X, Y)\) and \(B = \max(X, Y)\)?

• The pair \((A, B)\) is uniformly distributed over the triangle \(0 \leq A \leq B \leq 1\).
• This is because for each pair \((X, Y)\), either \(X \leq Y\) or \(Y < X\), but both are symmetric.

Therefore, the joint PDF of \((A, B)\) is: \[ f_{A,B}(a, b) = 2, \quad 0 \leq a \leq b \leq 1 \] The factor 2 comes from the Jacobian of the transformation and the symmetry.

Step 2: Probability of Intersection

We want: \[ P(\max(A_1,...,A_5) \leq \min(B_1,...,B_5)) \] Let’s denote \(M_A = \max(A_1,...,A_5)\), \(m_B = \min(B_1,...,B_5)\).

The event that the intervals overlap is exactly \(M_A \leq m_B\). Let’s compute: \[ P(M_A \leq m_B) \] This is the probability that there exists some \(x \in [0, 1]\) such that \(x\) is in all five intervals.

Step 3: Reformulation

Let’s integrate over all possible values of \(M_A = m\) and \(m_B = n\), and account for the joint distribution.

• For the intersection to be non-empty, \(m \leq n\).
• Each interval’s \((A, B)\) is independently uniform over the triangle \(0 \leq A \leq B \leq 1\).

Step 4: Calculating the Probability for n Intervals

Let’s consider the general case of \(n\) intervals, then plug in \(n=5\).

Let’s fix \(m\) and \(n\) such that \(0 \leq m \leq n \leq 1\). For all intervals to overlap, all \(A_i \leq n\) and all \(B_i \geq m\), and at least one \(A_i = m\) and at least one \(B_i = n\).

For a single interval, the probability that \(A \leq n\) and \(B \geq m\) (with \(A \leq B\)) is the probability that the interval covers the point \((m, n)\), i.e., \(A \leq m \leq n \leq B\). But since the interval endpoints are random, we need to compute the chance that for all intervals, their overlap includes the interval \([m, n]\).

Step 5: Direct Probability Calculation

For each interval, the probability that it covers the point \(x\) is the probability that \(A \leq x \leq B\).

But we want the intersection to be non-empty, that is, there is some \(x\) such that for all \(i\), \(A_i \leq x \leq B_i\), i.e., \(M_A \leq x \leq m_B\). So, \(M_A \leq m_B\) is the only condition; for any \(x\) in \([M_A, m_B]\), all intervals cover \(x\).

Thus, the probability is: \[ P(M_A \leq m_B) \] But since all \((A_i, B_i)\) are independent, and the joint PDF is \(2\) over the triangle \(0 \leq A \leq B \leq 1\), we can proceed as follows.

Step 6: Calculating the Distribution of \(A\) and \(B\)

First, compute the marginal PDFs:

• For \(A\): \(P(A \leq a) = 1 - (1-a)^2\), so the PDF is \(f_A(a) = 2(1-a)\), for \(a \in [0,1]\).
• For \(B\): \(P(B \leq b) = b^2\), so the PDF is \(f_B(b) = 2b\), for \(b \in [0,1]\).

Step 7: Calculating the Final Probability

Let’s fix \(x \in [0,1]\). For each interval, the probability that it covers \(x\) is: \[ P(A \leq x \leq B) \] This is the probability that both \(A \leq x\) and \(B \geq x\), and \(A \leq B\).

For a single interval: \[ P(A \leq x, B \geq x) = P(\min(X, Y) \leq x, \max(X, Y) \geq x) \] But for \(X, Y\) uniform and independent on [0,1], this is the probability that one is at most \(x\) and the other is at least \(x\).

The probability that both \(X\) and \(Y\) are \(\leq x\) is \(x^2\).

The probability that both \(X\) and \(Y\) are \(\geq x\) is \((1-x)^2\).

Thus, the probability that one is at most \(x\) and the other at least \(x\) is: \[ 1 - x^2 - (1-x)^2 = 1 - (x^2 + (1-x)^2) \] \[ = 1 - (x^2 + 1 - 2x + x^2) = 1 - (1 - 2x + 2x^2) \] \[ = 2x - 2x^2 \] So, \[ P(x \in [A, B]) = 2x(1-x) \]

Step 8: Probability All Intervals Overlap at Some Point

For all five intervals to have a non-empty intersection, there must exist \(x \in [0,1]\) such that all five intervals contain \(x\). The probability that a specific \(x\) is contained in a particular interval is \(2x(1-x)\), as above. For five independent intervals, the probability that all five include \(x\) is \((2x(1-x))^5\).

Therefore, by the union bound over all possible \(x\), the probability that there exists at least one such \(x\) is: \[ P = \int_0^1 (2x(1-x))^5 dx \] This integral gives the probability that all five intervals overlap at some point.

Step 9: Calculating the Integral

Let’s compute: \[ P = \int_0^1 (2x(1-x))^5 dx = 2^5 \int_0^1 x^5(1-x)^5 dx = 32 \int_0^1 x^5(1-x)^5 dx \] The integral \(\int_0^1 x^m (1-x)^n dx\) is the Beta function \(B(m+1, n+1)\).

Recall: \[ B(m+1,n+1) = \frac{m!n!}{(m+n+1)!} \] For \(m = 5, n = 5\): \[ B(6,6) = \frac{5! \cdot 5!}{11!} = \frac{120 \cdot 120}{39916800} = \frac{14400}{39916800} = \frac{1}{2772} \] So, \[ P = 32 \cdot \frac{1}{2772} = \frac{32}{2772} = \frac{8}{693} \]

Final Answer: \[ \boxed{\frac{8}{693}} \]

Generalization to n Intervals

For \(n\) intervals, the probability is: \[ P_n = 2^n \int_0^1 x^n (1-x)^n dx = 2^n \cdot B(n+1, n+1) \] \[ = 2^n \cdot \frac{n!n!}{(2n+1)!} \] So, for \(n\) intervals, the probability that all intervals share a common intersection is: \[ P_n = \frac{2^n n! n!}{(2n+1)!} \]

Python Simulation

Let’s verify the result with

import numpy as np

def random_interval():
    x, y = np.random.uniform(0, 1, 2)
    return min(x, y), max(x, y)

def all_overlap(n_intervals=5):
    intervals = [random_interval() for _ in range(n_intervals)]
    left = max(interval[0] for interval in intervals)
    right = min(interval[1] for interval in intervals)
    return left <= right

# Run the simulation
trials = 1_000_000
successes = sum(all_overlap(5) for _ in range(trials))
probability = successes / trials

print(f"Estimated probability: {probability:.6f}")

Running this code (with a sufficient number of trials, e.g., 1,000,000), you should observe an estimated probability close to the exact value \( \frac{8}{693} \approx 0.01154 \).

Jane Street Interview Tips

Quantitative interviews at Jane Street and similar trading firms often feature probability and combinatorial puzzles like this. Here are some tips for tackling these:

• Clarify and formalize the problem. Restate the question in precise mathematical terms.
• Draw diagrams. Visualizing intervals and their overlap can help you find the right approach.
• Look for symmetry and invariance. Many intervals problems are greatly simplified by symmetry arguments.
• Identify relevant distributions. In this problem, understanding uniform and order statistics is essential.
• Generalize. Try to solve for \( n \) intervals, not just the specific value given. This shows depth of understanding.
• Check your answer numerically. Simulations are a great way to verify analytic results.

Conclusion

The "Probability of Overlapping Random Intervals" problem is a beautiful illustration of how combinatorial geometry and probability theory intersect. For five random intervals generated by selecting two independent uniform points on [0, 1] and using them as endpoints, the probability that all intervals overlap is:

\[ \boxed{\frac{8}{693}} \]

This result follows from understanding the distribution of random intervals, formulating the intersection condition, and calculating the appropriate integral using the Beta function. The general formula for \( n \) intervals is:

\[ P_n = \frac{2^n \cdot (n!)^2}{(2n+1)!} \]

Problems like these are common in Jane Street quantitative interviews because they test your ability to model random processes, think abstractly, and communicate mathematical arguments clearly. Practice with similar questions, verify your answers computationally, and approach every step methodically. Good luck with your interviews!

References

• Feller, W. An Introduction to Probability Theory and Its Applications, Vol. 1.
• Jane Street Sample Interview Questions: Official Jane Street Careers
• Probability Stack Exchange: Probability of overlap of n random intervals
• Wikipedia: Beta function