Quant Research Interview Questions - Jump Trading
Preparing for a quant interview at elite trading firms like Jump Trading, WorldQuant, or TGS Management Company requires a robust understanding of probability, statistics, linear algebra, and algorithmic thinking. In this guide, we solve and explain in detail some of the most challenging quant interview questions, focusing on Jump Trading and similar firms. These questions test not just mathematical knowledge but also creativity and problem-solving skills. Below, we walk through classic interview problems, offering thorough explanations, and, where appropriate, code snippets to help you master the concepts.
Quant Interview Questions – Jump Trading and Beyond
1. Expected Rolls for Consecutive Six 6s on a Fair Die
Question:
You keep rolling a fair die with 6 sides. What is the average number of rolls required to obtain six consecutive sixes for the first time?
Solution:
This is a variant of the classic problem of waiting time for a specific pattern in a sequence of independent trials. Here, we are interested in the expected number of rolls to get the pattern "666666" (six consecutive sixes).
Step 1: Defining States
Let’s define states based on the number of consecutive sixes observed at the end of the current sequence:
- State 0: No consecutive sixes.
- State 1: Last roll is one six, but not two in a row, and so on, up to...
- State 6: Last six rolls are all sixes (i.e., the target pattern). This is the absorbing state.
Let \( E_k \) denote the expected number of additional rolls needed to reach state 6, given that we are currently in state \( k \) (i.e., we have just rolled \( k \) consecutive sixes).
Step 2: Setting Up Recurrence Relations
We want to find \( E_0 \) (starting from scratch).
Let’s write the recurrence for \( 0 \leq k \leq 5 \):
At state \( k \) (with \( k \) consecutive sixes), on the next roll:
- With probability \( \frac{1}{6} \), you roll a six and move to state \( k+1 \).
- With probability \( \frac{5}{6} \), you roll something else and return to state 0 (since the consecutive streak is broken).
Therefore, the expected number of steps from state \( k \) is:
\[ E_k = 1 + \frac{1}{6}E_{k+1} + \frac{5}{6}E_0 \quad \text{for } k=0,1,2,3,4,5 \]
The base case: \[ E_6 = 0 \]
Step 3: Solving the Recurrence
Let’s solve backwards from \( E_5 \) to \( E_0 \).
- For \( k=5 \):
\[ E_5 = 1 + \frac{1}{6}E_6 + \frac{5}{6}E_0 = 1 + \frac{5}{6}E_0 \] (Since \( E_6 = 0 \)) - For \( k=4 \):
\[ E_4 = 1 + \frac{1}{6}E_5 + \frac{5}{6}E_0 \] - For \( k=3 \):
\[ E_3 = 1 + \frac{1}{6}E_4 + \frac{5}{6}E_0 \] - For \( k=2 \):
\[ E_2 = 1 + \frac{1}{6}E_3 + \frac{5}{6}E_0 \] - For \( k=1 \):
\[ E_1 = 1 + \frac{1}{6}E_2 + \frac{5}{6}E_0 \] - For \( k=0 \):
\[ E_0 = 1 + \frac{1}{6}E_1 + \frac{5}{6}E_0 \]
Let’s solve these equations step by step.
Step 3.1: Express Each \( E_k \) in Terms of \( E_0 \)
- \( E_5 = 1 + \frac{5}{6}E_0 \)
- \( E_4 = 1 + \frac{1}{6}E_5 + \frac{5}{6}E_0 \)
Substituting \( E_5 \): \[ E_4 = 1 + \frac{1}{6}(1 + \frac{5}{6}E_0) + \frac{5}{6}E_0 = 1 + \frac{1}{6} + \frac{5}{36}E_0 + \frac{5}{6}E_0 = \frac{7}{6} + (\frac{5}{36} + \frac{30}{36})E_0 = \frac{7}{6} + \frac{35}{36}E_0 \] - \( E_3 = 1 + \frac{1}{6}E_4 + \frac{5}{6}E_0 \)
Substitute \( E_4 \): \[ E_3 = 1 + \frac{1}{6}(\frac{7}{6} + \frac{35}{36}E_0) + \frac{5}{6}E_0 = 1 + \frac{7}{36} + \frac{35}{216}E_0 + \frac{5}{6}E_0 = \frac{43}{36} + (\frac{35}{216} + \frac{180}{216})E_0 = \frac{43}{36} + \frac{215}{216}E_0 \] - \( E_2 = 1 + \frac{1}{6}E_3 + \frac{5}{6}E_0 \)
Substitute \( E_3 \): \[ E_2 = 1 + \frac{1}{6}(\frac{43}{36} + \frac{215}{216}E_0) + \frac{5}{6}E_0 = 1 + \frac{43}{216} + \frac{215}{1296}E_0 + \frac{5}{6}E_0 = \frac{259}{216} + (\frac{215}{1296} + \frac{1080}{1296})E_0 = \frac{259}{216} + \frac{1295}{1296}E_0 \] - \( E_1 = 1 + \frac{1}{6}E_2 + \frac{5}{6}E_0 \)
Substitute \( E_2 \): \[ E_1 = 1 + \frac{1}{6}(\frac{259}{216} + \frac{1295}{1296}E_0) + \frac{5}{6}E_0 = 1 + \frac{259}{1296} + \frac{1295}{7776}E_0 + \frac{5}{6}E_0 = \frac{1555}{1296} + (\frac{1295}{7776} + \frac{6480}{7776})E_0 = \frac{1555}{1296} + \frac{7775}{7776}E_0 \] - \( E_0 = 1 + \frac{1}{6}E_1 + \frac{5}{6}E_0 \)
Rearranged: \[ E_0 - \frac{5}{6}E_0 = 1 + \frac{1}{6}E_1 \] \[ \frac{1}{6}E_0 = 1 + \frac{1}{6}E_1 \] \[ E_0 = 6 + E_1 \]
Step 3.2: Substitute \( E_1 \) in Terms of \( E_0 \)
Recall: \[ E_1 = \frac{1555}{1296} + \frac{7775}{7776}E_0 \]
Then: \[ E_0 = 6 + E_1 = 6 + \frac{1555}{1296} + \frac{7775}{7776}E_0 \]
Let’s isolate \( E_0 \): \[ E_0 - \frac{7775}{7776}E_0 = 6 + \frac{1555}{1296} \] \[ \frac{1}{7776}E_0 = 6 + \frac{1555}{1296} \] \[ E_0 = 7776 \times \left(6 + \frac{1555}{1296}\right) \]
Calculate \( 6 + \frac{1555}{1296} \): \[ 6 + \frac{1555}{1296} = \frac{6 \times 1296 + 1555}{1296} = \frac{7776 + 1555}{1296} = \frac{9331}{1296} \]
Therefore, \[ E_0 = 7776 \times \frac{9331}{1296} = \frac{7776 \times 9331}{1296} \] \[ = 6 \times 9331 = 55,986 \]
So, the expected number of rolls to get six consecutive sixes is 55,986.
Step 4: Generalization and Intuition
For \( n \) consecutive sixes, the expected number of rolls is \( 6^n \times (1 + \frac{1}{6} + \frac{1}{6^2} + ... + \frac{1}{6^{n-1}}) \). For \( n=6 \), this formula gives the same result.
2. Correlation Matrix and Positive Semi-Definiteness
Question:
A, B, and C are time series and the following rhos represent their correlation coefficients. \( \rho(A,B) = 0.7 \), \( \rho(B,C) = 0.8 \). What is \( \rho(A,C) \)? Why is the correlation matrix positive semi-definite?
Solution:
Step 1: Boundaries for \( \rho(A,C) \)
Given only the pairwise correlations \( \rho(A,B) \) and \( \rho(B,C) \), the value of \( \rho(A,C) \) is not uniquely determined but is bounded by the requirement that the correlation matrix remains positive semi-definite (PSD).
The correlation matrix is:
| A | B | C | |
|---|---|---|---|
| A | 1 | 0.7 | ? |
| B | 0.7 | 1 | 0.8 |
| C | ? | 0.8 | 1 |
Let \( x = \rho(A,C) \). The matrix is: \[ \begin{pmatrix} 1 & 0.7 & x \\ 0.7 & 1 & 0.8 \\ x & 0.8 & 1 \end{pmatrix} \]
For this matrix to be PSD, all its principal minors must be non-negative.
Step 2: Determinant Condition
The determinant of the correlation matrix must be \( \geq 0 \).
\[ \det = 1(1 \times 1 - 0.8 \times 0.8) - 0.7(0.7 \times 1 - 0.8x) + x(0.7 \times 0.8 - 1 \times x) \]
Let’s expand: \begin{align*} \det &= 1 \cdot (1 - 0.64) - 0.7 \cdot (0.7 - 0.8x) + x \cdot (0.56 - x) \\ &= 0.36 - 0.7 \cdot 0.7 + 0.7 \cdot 0.8x + 0.56x - x^2 \\ &= 0.36 - 0.49 + 0.56x + 0.56x - x^2 \\ &= -0.13 + 1.12x - x^2 \end{align*}
Set \( \det \geq 0 \): \[ -x^2 + 1.12x - 0.13 \geq 0 \] \[ x^2 - 1.12x + 0.13 \leq 0 \]
Solve for \( x \): \[ x = \frac{1.12 \pm \sqrt{1.12^2 - 4 \times 0.13}}{2} \] \[ = \frac{1.12 \pm \sqrt{1.2544 - 0.52}}{2} = \frac{1.12 \pm \sqrt{0.7344}}{2} \] \[ \sqrt{0.7344} \approx 0.857 \] \[ x_1 = \frac{1.12 + 0.857
\[ x_1 = \frac{1.12 + 0.857}{2} \approx \frac{1.977}{2} \approx 0.9885 \] \[ x_2 = \frac{1.12 - 0.857}{2} \approx \frac{0.263}{2} \approx 0.1315 \]
Therefore, for the correlation matrix to be positive semi-definite, \(\rho(A,C)\) must satisfy:
\[ 0.1315 \leq \rho(A,C) \leq 0.9885 \]
Step 3: Intuition and Generalization
This means that if two variables (A and C) are both highly correlated to a third variable (B), then their own correlation must be within a constrained range for the correlation matrix to remain valid (PSD). This is a common scenario in portfolio construction and risk management.
Step 4: Why is the Correlation Matrix Positive Semi-Definite?
By definition, a correlation matrix is positive semi-definite because for any real vector \( x \): \[ x^T C x \geq 0 \] where \( C \) is the correlation matrix.
- Geometric Meaning: This property ensures that the variance (which is always non-negative) of any linear combination of variables is also non-negative.
- Mathematical Explanation: Since a correlation matrix is a special case of a covariance matrix (normalized to variances of 1), and covariance matrices are always positive semi-definite, so are correlation matrices.
- Practical Reason: If a matrix were not PSD, you could find a portfolio (linear combination) with negative variance, which is not possible in the real world.
3. Generating a Gaussian Distribution from rand()
Question:
How can you generate a Gaussian (normal) distribution using only a uniform random number generator rand()?
Solution:
The rand() function typically gives a pseudo-random number uniformly distributed between 0 and 1. To generate a normal (Gaussian) distribution, two popular methods are widely used:
Method 1: Box-Muller Transform
The Box-Muller transform converts two independent uniform random variables into two independent standard normal random variables.
- Let \( U_1, U_2 \) be independent random variables uniformly distributed in (0,1).
- Then, set: \[ Z_0 = \sqrt{-2 \ln U_1} \cdot \cos(2\pi U_2) \] \[ Z_1 = \sqrt{-2 \ln U_1} \cdot \sin(2\pi U_2) \] Both \( Z_0 \) and \( Z_1 \) are independent standard normal variables (\( N(0,1) \)).
Sample Code (C/C++):
#include <math.h>
#include <stdlib.h>
double gaussian_rand() {
double U1 = (rand() + 1.0) / (RAND_MAX + 2.0);
double U2 = (rand() + 1.0) / (RAND_MAX + 2.0);
return sqrt(-2.0 * log(U1)) * cos(2.0 * M_PI * U2);
}
Method 2: Central Limit Theorem Approximation
Sum a large number of independent uniform random variables and normalize.
- If \( X_i \sim U(0,1) \), then as \( n \rightarrow \infty \), \[ \frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - 0.5) \] approaches a standard normal distribution.
- In practice, summing 12 uniforms and subtracting 6 gives a reasonably good standard normal: \[ Z = \sum_{i=1}^{12} U_i - 6 \]
Sample Code (Python):
import random
def gaussian_rand_clt():
return sum(random.uniform(0,1) for _ in range(12)) - 6
Comparison
The Box-Muller method is exact (in theory), while the CLT approach is an approximation. For most simulation purposes, both are acceptable, though Box-Muller is preferred for precision.
4. Creating a 1/7 Probability Event from a Fair Die
Question:
Given a fair die, how would you create an event with probability 1/7?
Solution:
This is a classic quant interview question that tests your understanding of randomness and simulation.
Method 1: Rejection Sampling
Roll the die twice to generate a number from 1 to 36. Use the numbers 1 to 35 to represent 35 equally likely outcomes, ignore 36, and group the 35 numbers into 7 groups of 5 outcomes each.
- Roll die once: outcome \( D_1 \in \{1,2,3,4,5,6\} \)
- Roll die again: outcome \( D_2 \in \{1,2,3,4,5,6\} \)
- Map to number: \( N = 6 \times (D_1 - 1) + D_2 \), so \( N \in \{1, ..., 36\} \)
- If \( N = 36 \), discard and repeat.
- Else, \( N \in \{1, ..., 35\} \), assign each group of 5 numbers to one of 7 outcomes.
Python Example:
import random
def die():
return random.randint(1,6)
def event_one_seventh():
while True:
n = 6 * (die() - 1) + die()
if n <= 35:
return n % 7 == 0 # True with probability 1/7
This function returns True with probability 1/7 and False otherwise.
Method 2: Waiting for a Specific Pattern
Alternatively, roll the die until you get a 1, and only then perform a secondary randomization, but this is far less efficient than the above method.
General Principle
The key is to use the die to sample from a uniform set whose size is a multiple of 7, then map those outcomes to your desired event. If your set is larger, reject the excess outcomes.
Conclusion
Quant interview questions at firms like Jump Trading, WorldQuant, and TGS Management Company are designed to probe your understanding of probability, statistics, and mathematical reasoning. Whether it’s calculating the expected waiting time for a rare pattern, understanding the properties of correlation matrices, simulating normal random variables, or constructing events with unusual probabilities, the core is deep, logical thinking. By mastering the detailed solutions and explanations above, you’ll be well-equipped for even the toughest quant interviews.
Remember to practice, understand the underlying principles, and be ready to adapt your knowledge to new and unexpected problems during your interview. Good luck!
