Quantitative Researcher interviews at top firms like Two Sigma, WorldQuant, Jane Street, and Citadel are renowned for their rigor and depth in testing probability, statistics, combinatorics, programming, and critical thinking. Understanding the solutions and the reasoning behind these problems is essential for any aspiring quant. In this article, we walk through some of the most frequently asked and challenging quant interview questions, providing detailed solutions, mathematical explanations, and code where required.

Quant Researcher Interview Questions - Two Sigma & Other Top Firms

2. Probability & Combinatorics: Bear and the Fish (Jane Street)

Question

A bear wants to catch 3 fish from a river. When he has caught 3 fish, he'll leave. When a fish comes, there is a 1/2 chance he'll catch it. What is the probability that the 5th fish will not be caught?

Detailed Solution

Let’s define the process step by step.

• The bear leaves after catching 3 fish.
• Each fish has a 0.5 chance of being caught.
• We are interested in the probability that the 5th fish is not caught.

Step 1: For the bear to be present at the 5th fish

The bear leaves as soon as he catches the 3rd fish. For the bear to be present when the 5th fish arrives, he must have caught at most 2 fish in the first 4.

Let \( X \) be the number of fish caught in the first 4.

• If \( X = 0, 1, 2 \): The bear is still present for the 5th fish.
• If \( X \geq 3 \): The bear has left before the 5th fish.

Let’s compute \( P(X \leq 2) \), where \( X \sim \text{Binomial}(4, 0.5) \).

\[ P(X \leq 2) = \sum_{k=0}^2 \binom{4}{k} \left(\frac{1}{2}\right)^4 \]

• \( \binom{4}{0} = 1 \)
• \( \binom{4}{1} = 4 \)
• \( \binom{4}{2} = 6 \)

\[ P(X \leq 2) = \frac{1 + 4 + 6}{16} = \frac{11}{16} \]

Step 2: Probability the bear catches the 5th fish

Given the bear is present, probability he catches 5th fish = 0.5.

Step 3: Probability the 5th fish is not caught

The probability the bear is present is \( \frac{11}{16} \), and then the chance of not catching is 0.5. If the bear is gone (with probability \( 1 - \frac{11}{16} = \frac{5}{16} \)), the fish is, by default, not caught.

So, total probability: \[ P(\text{5th not caught}) = P(\text{bear gone}) + P(\text{bear present}) \times P(\text{not caught} | \text{present}) \] \[ = \frac{5}{16} + \frac{11}{16} \times 0.5 \] \[ = \frac{5}{16} + \frac{11}{32} \] \[ = \frac{10}{32} + \frac{11}{32} = \frac{21}{32} \]

Final Answer

The probability that the 5th fish is not caught is \( \frac{21}{32} \).

3. Probability: Students Passing Test (Citadel)

Question

Chance that a student passes the test is 10%. What is the chance that out of 400 students AT LEAST 50 pass the test? The offered answers were 5%, 10%, 15%, 20%, 25%.

Detailed Solution

This is a classic binomial probability approximation problem.

• Number of students: \( n = 400 \)
• Probability of success: \( p = 0.1 \)
• Let \( X \) be the number of students passing. \( X \sim \mathrm{Binomial}(400, 0.1) \).
• We want \( P(X \geq 50) \).

Step 1: Calculate Mean and Standard Deviation

\[ \mu = np = 400 \times 0.1 = 40 \] \[ \sigma = \sqrt{np(1-p)} = \sqrt{400 \times 0.1 \times 0.9} = \sqrt{36} = 6 \]

Step 2: Normal Approximation

We use the normal approximation to the binomial: \[ P(X \geq 50) \approx P\left( Z \geq \frac{49.5 - 40}{6} \right) \] Where 49.5 is the continuity correction. \[ \frac{49.5 - 40}{6} = \frac{9.5}{6} \approx 1.58 \]

\[ P(Z \geq 1.58) \] From standard normal tables, \[ P(Z \geq 1.58) \approx 0.057 \] That is, about 5.7%.

Step 3: Closest Multiple Choice Answer

Closest to 5%, so the answer is 5%.

4. Two Sigma Coding & Probability Problems

4.1 Coding Fibonacci

Code the n-th term of Fibonacci, using iteration, then recursion only, then recursion and memoization. What is the time complexity and space complexity for all three cases?

Iterative Solution

def fib_iterative(n):
    if n <= 1:
        return n
    a, b = 0, 1
    for _ in range(2, n+1):
        a, b = b, a + b
    return b

• Time complexity: \( O(n) \)
• Space complexity: \( O(1) \)

Recursive Solution

def fib_recursive(n):
    if n <= 1:
        return n
    return fib_recursive(n-1) + fib_recursive(n-2)

• Time complexity: \( O(2^n) \)
• Space complexity: \( O(n) \) (call stack)

Recursive with Memoization

def fib_memo(n, memo={}):
    if n in memo:
        return memo[n]
    if n <= 1:
        memo[n] = n
        return n
    memo[n] = fib_memo(n-1, memo) + fib_memo(n-2, memo)
    return memo[n]

• Time complexity: \( O(n) \)
• Space complexity: \( O(n) \) (for memo dictionary)

4.2 Linear Regression Data Duplication Intuition

Linear regression of X and y, X is n-dimensional and y is 1-dimensional, how would beta, standard error of beta, r², and t-stat of beta change if you duplicate the data? Can you explain the results intuitively?

Mathematical Explanation

• Suppose original data is D with n samples.
• Duplicate to form D' = D + D (2n samples, but identical data).
• Let’s consider the effects on OLS regression:

Summary

• Beta: Same
• Standard Error: Decreases by \( 1/\sqrt{2} \)
• r²: Same
• t-stat: Increases by \( \sqrt{2} \)

Intuition

Duplicating data does not add new information, it just increases apparent sample size, reducing standard error, but the fit quality does not change.

4.3 Bayesian Probability: Friends Telling the Weather

London has probability \( p \) of rain. You have 3friends, each of whom has a \( \frac{2}{3} \) probability of telling the truth and \( \frac{1}{3} \) probability of lying. Suppose all friends tell you "rain". What is the probability it actually rains?

Step 1: Define the Events

• Let \( R \): It rains
• Let \( T_i \): Friend \( i \) says "rain"
• Each friend tells truth with probability \( \frac{2}{3} \), lies with \( \frac{1}{3} \)

We are asked to find \( P(R | T_1 \cap T_2 \cap T_3) \).

Step 2: Use Bayes’ Theorem

By Bayes' Theorem: \[ P(R | T_1 T_2 T_3) = \frac{P(T_1 T_2 T_3 | R) P(R)}{P(T_1 T_2 T_3)} \]

Step 3: Compute Likelihoods

• If it rains, each friend says "rain" with probability \( \frac{2}{3} \); all friends: \( \left(\frac{2}{3}\right)^3 = \frac{8}{27} \)
• If it does not rain, each friend lies and says "rain" with probability \( \frac{1}{3} \); all friends: \( \left(\frac{1}{3}\right)^3 = \frac{1}{27} \)

Step 4: Total Probability

\[ P(T_1 T_2 T_3) = P(T_1 T_2 T_3 | R)P(R) + P(T_1 T_2 T_3 | \neg R)P(\neg R) \] \[ = \frac{8}{27}p + \frac{1}{27}(1-p) \]

Step 5: Plug Back into Bayes’ Formula

\[ P(R | T_1 T_2 T_3) = \frac{\frac{8}{27}p}{\frac{8}{27}p + \frac{1}{27}(1-p)} = \frac{8p}{8p + (1-p)} = \frac{8p}{7p + 1} \]

Step 6: Intuitive Explanation

If \( p \) is small (i.e., rain is rare), hearing "rain" from all friends is much more likely due to lying, but as \( p \) increases, the chance that they’re telling the truth increases.

Example: If \( p = 0.1 \)

\[ P(R | T_1 T_2 T_3) = \frac{8 \times 0.1}{7 \times 0.1 + 1} = \frac{0.8}{0.7 + 1} = \frac{0.8}{1.7} \approx 0.4706 \]

So, even if rain is unlikely, if all friends say "rain", the probability it actually rains increases significantly.

4.4 Leetcode Problems: DSA and DP

Two Sigma and other quant firms frequently ask algorithm and data structure questions, focusing on dynamic programming (DP), searching, sorting, and problem decomposition. Let's discuss a classic DP problem and its solution.

Example Problem: Longest Increasing Subsequence (LIS)

Given an unsorted array of integers, find the length of the longest increasing subsequence.

DP Solution (O(n^2))

def lengthOfLIS(nums):
    n = len(nums)
    if n == 0:
        return 0
    dp = [1] * n
    for i in range(n):
        for j in range(i):
            if nums[i] > nums[j]:
                dp[i] = max(dp[i], dp[j] + 1)
    return max(dp)

• Time Complexity: \( O(n^2) \)
• Space Complexity: \( O(n) \)

Explanation

For each index \( i \), \( dp[i] \) stores the length of the longest increasing subsequence ending at \( i \). For each pair \( (j, i) \), if \( nums[i] > nums[j] \), update \( dp[i] \).

Optimized Solution (O(n log n))

import bisect
def lengthOfLIS(nums):
    sub = []
    for num in nums:
        idx = bisect.bisect_left(sub, num)
        if idx == len(sub):
            sub.append(num)
        else:
            sub[idx] = num
    return len(sub)

• Time Complexity: \( O(n \log n) \)
• Space Complexity: \( O(n) \)

Explanation

Maintain a list \( sub \) representing the smallest tail of all increasing subsequences of different lengths. Use binary search to maintain \( sub \).

Summary Table: Key Interview Problem Types and Concepts

Conclusion

Quantitative researcher interviews at firms like WorldQuant, Jane Street, Citadel, and Two Sigma test not just mathematical and programming skills, but also your ability to reason under uncertainty, spot edge cases, and explain concepts clearly. Mastering these problems requires not just rote memorization, but deep understanding and practice. Use the step-by-step solutions and explanations above as a template for approaching similar interview questions. Good luck in your quant journey!

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