The world of quantitative finance is filled with challenging probability puzzles that reveal your analytical reasoning and mathematical depth. One classic example often encountered in interviews for systematic trading roles, such as at Susquehanna International Group (SIG), is the overlapping time interval problem. This problem tests your ability to model random events, apply geometric probability, and communicate a clear, logical solution. In this article, we will thoroughly analyze and solve the following SIG Quantitative Systematic Trader Interview Question:

SIG Quantitative Systematic Trader Interview Question: Overlapping Time Intervals Probability Problem

Problem Statement

Question: Two delivery drivers arrive uniformly at random between 1 pm and 2 pm. Each driver stays exactly 10 minutes. What is the probability that their time intervals overlap?

Table of Contents

• Problem Clarification
• Key Concepts Involved
• Mathematical Formulation
• Visualizing the Problem
• Geometric Probability Approach
• Step-by-Step Solution
• Generalization and Extensions
• Python Simulation
• Interview Tips and Insights
• Summary

Problem Clarification

Let's start by parsing exactly what is being asked:

• Both drivers arrive at a random time, uniformly distributed between 1:00 pm and 2:00 pm.
• Upon arrival, each stays for exactly 10 minutes.
• What is the probability that their time intervals (the 10-minute periods) overlap?

This is a classic interval overlap problem that can be solved using geometric probability. The uniformity in arrival times suggests a continuous uniform distribution.

Key Concepts Involved

Before diving into the solution, let's review the concepts that underpin this problem:

• Uniform Distribution: Each arrival time is equally likely anywhere in the interval [0, 60] minutes past 1 pm.
• Geometric Probability: Probability as a ratio of areas (or volumes) in the sample space.
• Intervals Overlap: Two intervals [A, A+10] and [B, B+10] overlap if and only if their intersection is non-empty.

Mathematical Formulation

Let’s define variables:

• Let X be the arrival time (in minutes after 1 pm) of the first driver.
• Let Y be the arrival time (in minutes after 1 pm) of the second driver.

Both X and Y are independent and uniformly distributed random variables on [0, 60].

Each driver stays for 10 minutes. So, driver 1 is present from X to X+10, and driver 2 is present from Y to Y+10.

The question is: What is the probability that these two intervals overlap?

Visualizing the Problem

Let’s represent all possible pairs of arrival times on a 2D square:

• The x-axis is X (0 to 60).
• The y-axis is Y (0 to 60).
• The sample space is the 60 x 60 square, with area 3600.

For which points (X, Y) do the intervals [X, X+10] and [Y, Y+10] overlap? This will be a region inside the square.

Geometric Probability Approach

This is a textbook case for geometric probability: the probability is the ratio of the area of the "overlap" region to the total area of the square.

Let’s formalize when the intervals overlap.

• The intervals [X, X+10] and [Y, Y+10] overlap if and only if:
  • \([X, X+10] \cap [Y, Y+10] \neq \emptyset\)
• Equivalently, the two intervals overlap unless one of them starts after the other ends. That is:
  • \(X \geq Y+10\) (driver 1 arrives after driver 2 has left), or
  • \(Y \geq X+10\) (driver 2 arrives after driver 1 has left)
• Therefore, the intervals overlap if and only if:
  • \(|X-Y| < 10\)

Thus, the overlap region in the (X, Y) plane is the area where \(|X-Y| < 10\).

Step-by-Step Solution

Step 1: Express the Overlap Condition Mathematically

We’ve established that overlap occurs when:

\[ |X - Y| < 10 \]

The total sample space is the square \(0 \leq X \leq 60\), \(0 \leq Y \leq 60\). Its area is \(60 \times 60 = 3600\).

Step 2: Find the Area Where |X – Y| < 10

The region \(|X - Y| < 10\) is the set of points within 10 units of the line X=Y.

Visually, this is the region between the lines \(Y = X + 10\) and \(Y = X - 10\).

Let’s compute the area of this region inside the 60x60 square.

Step 3: Compute the Area of the Overlap Region

We want the area where \(|X - Y| < 10\) and \(0 \leq X, Y \leq 60\).

Let’s do this by integrating over X:

For each fixed X in [0, 60], Y must satisfy:

• \(X - 10 < Y < X + 10\)
• But Y must also be in [0, 60]

For each X,

• Lower limit: \(L = \max(0, X - 10)\)
• Upper limit: \(U = \min(60, X + 10)\)

So for each X, the length of Y-interval is \(U - L\).

The area is:

\[ \text{Area} = \int_{X=0}^{60} \left[ \min(60, X+10) - \max(0, X-10) \right] dX \]

Step 4: Break the Integration into Regions

There are three cases for X:

1. Case 1: \(0 \leq X < 10\)
   • Lower limit: \(0\)
   • Upper limit: \(X + 10\)
   • Interval length: \(X + 10 - 0 = X + 10\)
2. Case 2: \(10 \leq X \leq 50\)
   • Lower limit: \(X - 10\)
   • Upper limit: \(X + 10\)
   • Interval length: \(X + 10 - (X - 10) = 20\)
3. Case 3: \(50 < X \leq 60\)
   • Lower limit: \(X - 10\)
   • Upper limit: \(60\)
   • Interval length: \(60 - (X - 10) = 70 - X\)

Step 5: Write the Full Integral

So, the total area is:

\[ \begin{align*} \text{Area} &= \int_{0}^{10} (X + 10) dX + \int_{10}^{50} 20\, dX + \int_{50}^{60} (70 - X) dX \\ \end{align*} \]

Step 6: Evaluate Each Integral

Let’s compute each term:

• First integral:
  \[ \int_{0}^{10} (X + 10) dX = \int_{0}^{10} X dX + \int_{0}^{10} 10 dX = \left[ \frac{1}{2} X^2 \right]_0^{10} + [10X]_0^{10} \]
  \[ = \frac{1}{2}(10^2) + 10(10) = 50 + 100 = 150 \]
• Second integral:
  \[ \int_{10}^{50} 20\, dX = 20 \times (50 - 10) = 20 \times 40 = 800 \]
• Third integral:
  \[ \int_{50}^{60} (70 - X) dX = \int_{50}^{60} 70 dX - \int_{50}^{60} X dX = [70X]_{50}^{60} - [\frac{1}{2} X^2]_{50}^{60} \]
  \[ = 70 \times (60 - 50) - \frac{1}{2}(60^2 - 50^2) \]
  \(60^2 = 3600,\ 50^2 = 2500\)
  \[ = 70 \times 10 - \frac{1}{2}(3600 - 2500) = 700 - \frac{1}{2}(1100) = 700 - 550 = 150 \]

Adding up:

\(150 + 800 + 150 = 1100\)

Step 7: Compute the Probability

Recall that the total area is 3600.

\[ P = \frac{\text{Area where intervals overlap}}{\text{Total area}} = \frac{1100}{3600} = \frac{11}{36} \approx 0.3056 \]

Final Answer

The probability that their intervals overlap is:

\[ \boxed{\frac{11}{36} \approx 30.56\%} \]

Generalization and Extensions

If Each Driver Stays for t Minutes (t ≤ 60)

Let’s generalize. Suppose each driver stays for \(t\) minutes, and arrival window is \(T\) minutes.

Sample space: area \(T^2\).

Overlap occurs if \(|X - Y| < t\).

For each \(X\), the length of Y-interval is:

• For \(0 \leq X < t\): \(X + t\)
• For \(t \leq X \leq T - t\): \(2t\)
• For \(T - t < X \leq T\): \(T - (X - t)\)

So, the area becomes:

\[ \int_{0}^{t} (X + t) dX + \int_{t}^{T-t} 2t\, dX + \int_{T-t}^{T} (T - (X - t)) dX \]

Compute:

\[ \int_{0}^{t} (X + t) dX = \frac{1}{2} t^2 + t^2 = \frac{3}{2} t^2 \] \[ \int_{t}^{T-t} 2t\, dX = 2t (T - 2t) \] \[ \int_{T-t}^{T} (T - (X - t)) dX = \int_{T-t}^{T} (T - X + t) dX = \left[ (T + t)X - \frac{1}{2} X^2 \right]_{T-t}^{T} \]

But you can show that this last integral equals \(\frac{3}{2} t^2\) as well. So, total area:

\[ \text{Area} = 2 \times \frac{3}{2} t^2 + 2t(T - 2t) = 3 t^2 + 2t(T - 2t) \] \[ = 3t^2 +2tT - 4t^2 = 2tT - t^2 \] So, the total area is: \[ \text{Area}_{\text{overlap}} = 2tT - t^2 \] Therefore, for general \( t \leq T \): \[ P_{\text{overlap}} = \frac{2tT - t^2}{T^2} \] For our specific problem, \( t = 10,\, T = 60 \): \[ P_{\text{overlap}} = \frac{2 \times 10 \times 60 - 10^2}{60^2} = \frac{1200 - 100}{3600} = \frac{1100}{3600} = \frac{11}{36} \] This matches our previous calculation.

Further Generalization: Drivers Stay Different Amounts of Time

You might be asked, "Suppose the first driver stays for \( t_1 \) minutes and the second for \( t_2 \) minutes, both arriving uniformly at random between 0 and \( T \) minutes. What is the probability that their intervals overlap?"

In this case, for overlap, the condition is:

\[ |X - Y| < t_1 + t_2 \]

The formula for the probability becomes:

\[ P_{\text{overlap}} = \frac{2T(t_1 + t_2) - (t_1 + t_2)^2}{T^2} \]

If \( t_1 + t_2 > T \), the probability is 1, since their intervals must overlap.

Python Simulation

In interviews, it's often helpful to validate your result with a quick simulation. Here's a Python snippet to estimate the probability empirically:

import numpy as np

def overlap_probability(num_trials=100000, t=10, T=60):
    count = 0
    for _ in range(num_trials):
        x = np.random.uniform(0, T)
        y = np.random.uniform(0, T)
        if abs(x - y) < t:
            count += 1
    return count / num_trials

prob = overlap_probability()
print(f"Simulated probability: {prob:.4f}")

Running this simulation should yield a value close to 0.3056 (i.e., \( \frac{11}{36} \)).

Interview Tips and Insights

• State all assumptions clearly. For example, clarify that arrivals are independent and uniformly distributed, and that staying time is fixed.
• Use geometric probability for continuous uniform cases.
• Draw the region. Visualizing the 2D sample space and the overlap region is often key to understanding and explaining your reasoning.
• Break the integration into cases. When integrating over a region with changing boundaries, split into manageable intervals.
• Generalize your answer. Interviewers often want to see if you can extend your solution to more general cases.
• Check by simulation. If time permits, mention that you would validate the result with a simple Monte Carlo simulation.

Summary

The "overlapping time intervals" problem is a classic in quantitative interviews. Here’s a recap of the solution process:

Final Answer: The probability that the two drivers' time intervals overlap is:

\[ \boxed{\frac{11}{36} \approx 30.56\%} \]

This classic problem not only tests your probability and integration skills but also your ability to communicate logic and generalize patterns—essential skills for a successful quantitative systematic trader at firms like SIG.

References & Further Reading

• Feller, W. (1968). An Introduction to Probability Theory and Its Applications, Vol. 1.
• Geometric Probability (Wikipedia)
• Quant StackExchange: Overlapping Time Intervals Probability