Quantitative research interviews at top-tier hedge funds and prop trading firms like Citadel, Jane Street, and WorldQuant are renowned for their rigor, mathematical depth, and emphasis on problem-solving skills. These interviews test not only your understanding of advanced probability, statistics, and finance, but also your ability to think clearly and communicate your reasoning. In this guide, we will explore some of the most frequently asked quant research interview questions, provide detailed solutions, and help you build the intuition needed to succeed in such highly competitive environments.

Quant Research Interview Questions – Citadel, Jane Street, WorldQuant

1. Given \( \log X \sim N(0,1) \). Compute the expectation of \( X \).

Understanding the Problem

This question tests your understanding of distributions, particularly the log-normal distribution, and your ability to compute expectations.

Step 1: Recognize the Distribution

If \( \log X \sim N(0,1) \), that means \( X \) is log-normally distributed with parameters \( \mu = 0 \) and \( \sigma^2 = 1 \). The probability density function of a log-normal variable is:

$$ f_X(x) = \frac{1}{x \sqrt{2\pi}} \exp\left( -\frac{(\log x)^2}{2} \right), \quad x > 0 $$

Step 2: The Expected Value Formula

For a log-normal variable \( X \sim \text{LogNormal}(\mu, \sigma^2) \), the expectation is:

$$ \mathbb{E}[X] = e^{\mu + \frac{1}{2}\sigma^2} $$

Plugging in \( \mu = 0 \), \( \sigma^2 = 1 \):

$$ \mathbb{E}[X] = e^{0 + \frac{1}{2} \cdot 1} = e^{0.5} $$

Step 3: Boxed Final Answer

Therefore, the expectation of \( X \) is:

$$ \boxed{e^{1/2}} $$

Step 4: Detailed Solution (Derivation Method)

To derive it directly from the integral:

$$ \mathbb{E}[X] = \int_{0}^{\infty} x f_X(x) dx = \int_{0}^{\infty} x \frac{1}{x \sqrt{2\pi}} \exp\left( -\frac{(\log x)^2}{2} \right) dx \\ = \int_{0}^{\infty} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{(\log x)^2}{2} \right) dx $$

Let \( y = \log x \), then when \( x = 0 \), \( y \to -\infty \) and when \( x \to \infty \), \( y \to \infty \), and \( dx = e^y dy \):

$$ \mathbb{E}[X] = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{y^2}{2} + y \right) dy $$

$$ = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{y^2 - 2y}{2} \right) dy \\ = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{(y - 1)^2 - 1}{2} \right) dy \\ = e^{1/2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{(y-1)^2}{2} \right) dy \\ = e^{1/2} $$

2. Probability Game: Dice Throwing – Who Has a Greater Chance?

The Problem Statement

There is a game:

• You: Throw 1 die 4 times, trying to get at least one 6.
• Opponent: Throws 2 dice 24 times, trying to get at least one double 6 (both dice show 6 in a single throw).

Who has a higher probability of winning?

Step 1: Probability for You (At Least One 6 in 4 Throws)

The probability of not getting a 6 in one throw of a die is \( \frac{5}{6} \).

The probability of not getting a 6 in all 4 throws:

$$ P(\text{No 6 in 4 throws}) = \left(\frac{5}{6}\right)^4 $$

Therefore, the probability of getting at least one 6:

$$ P(\text{At least one 6}) = 1 - \left(\frac{5}{6}\right)^4 \approx 1 - \left(\frac{625}{1296}\right) \approx 1 - 0.482 = 0.518 $$

Step 2: Probability for Opponent (At Least One Double 6 in 24 Throws of 2 Dice)

The probability of getting double 6 in a single throw of 2 dice:

There are \( 6 \times 6 = 36 \) possible outcomes, only 1 is (6,6):

$$ P(\text{Double 6 in one throw}) = \frac{1}{36} $$

The probability of not getting a double 6 in one throw is \( 1 - \frac{1}{36} = \frac{35}{36} \).

Probability of not getting any double 6 in 24 throws:

$$ P(\text{No double 6 in 24 throws}) = \left(\frac{35}{36}\right)^{24} $$

Therefore, the probability of getting at least one double 6:

$$ P(\text{At least one double 6}) = 1 - \left(\frac{35}{36}\right)^{24} $$

Calculate \( \left(\frac{35}{36}\right)^{24} \):

\( \ln\left(\frac{35}{36}\right) \approx -0.0286 \)
\( 24 \times -0.0286 = -0.686 \)
\( e^{-0.686} \approx 0.503 \)

So,

$$ P(\text{At least one double 6}) \approx 1 - 0.503 = 0.497 $$

Step 3: Who Has the Higher Probability?

You have a slightly higher probability (0.518 vs 0.497) of success.

Step 4: Generalizable Python Code

# Calculate probability of at least one 6 in 4 throws
prob_no_6 = (5/6)**4
prob_at_least_one_6 = 1 - prob_no_6

# Probability of at least one double 6 in 24 throws
prob_no_double6 = (35/36)**24
prob_at_least_one_double6 = 1 - prob_no_double6

print("At least one 6 in 4 throws:", prob_at_least_one_6)
print("At least one double 6 in 24 throws:", prob_at_least_one_double6)

3. Correlation Bounds: Given \( \text{Corr}(X, Y) = 0.9 \), \( \text{Corr}(Y, Z) = 0.8 \), what are the minimum and maximum possible values for \( \text{Corr}(X, Z) \)?

Understanding the Problem

This is a classic question on the constraints of correlation coefficients among three variables. The answer involves the positive semi-definiteness of the correlation matrix.

Step 1: The Correlation Matrix

The correlation matrix must be positive semi-definite, i.e.,

$$ \begin{pmatrix} 1 & \rho_{XY} & \rho_{XZ} \\ \rho_{XY} & 1 & \rho_{YZ} \\ \rho_{XZ} & \rho_{YZ} & 1 \\ \end{pmatrix} $$

Here, \( \rho_{XY} = 0.9 \), \( \rho_{YZ} = 0.8 \), \( \rho_{XZ} = ? \)

Step 2: Determinant Must be Non-Negative

For positive semi-definiteness:

$$ \det(C) = 1 + 2\rho_{XY} \rho_{YZ} \rho_{XZ} - \rho_{XY}^2 - \rho_{YZ}^2 - \rho_{XZ}^2 \ge 0 $$

Plug in values:

$$ 1 + 2 \times 0.9 \times 0.8 \times \rho_{XZ} - (0.9^2 + 0.8^2 + \rho_{XZ}^2) \ge 0 \\ 1 + 1.44 \rho_{XZ} - (0.81 + 0.64 + \rho_{XZ}^2) \ge 0 \\ 1 + 1.44 \rho_{XZ} - 1.45 - \rho_{XZ}^2 \ge 0 \\ 1.44 \rho_{XZ} - \rho_{XZ}^2 - 0.45 \ge 0 $$

Rewrite:

$$ -\rho_{XZ}^2 + 1.44\rho_{XZ} - 0.45 \ge 0 $$

Multiply both sides by -1 (reverse sign):

$$ \rho_{XZ}^2 - 1.44\rho_{XZ} + 0.45 \le 0 $$

Step 3: Solve the Quadratic Inequality

Solve \( \rho_{XZ}^2 - 1.44\rho_{XZ} + 0.45 = 0 \)

Quadratic formula:

$$ \rho_{XZ} = \frac{1.44 \pm \sqrt{1.44^2 - 4 \times 1 \times 0.45}}{2} $$

Calculate:

\( 1.44^2 = 2.0736 \)
\( 4 \times 0.45 = 1.8 \)
\( \sqrt{2.0736 - 1.8} = \sqrt{0.2736} \approx 0.523 \)

So,

$$ \rho_{XZ,1} = \frac{1.44 + 0.523}{2} = \frac{1.963}{2} \approx 0.9815 $$

$$ \rho_{XZ,2} = \frac{1.44 - 0.523}{2} = \frac{0.917}{2} \approx 0.4585 $$

Step 4: Final Answer

Therefore,

$$ \boxed{0.4585 \leq \text{Corr}(X, Z) \leq 0.9815} $$

Step 5: Python Code for General Case

import numpy as np

def corr_bounds(rxy, ryz):
    a = 1
    b = - (rxy + ryz)
    c = rxy * ryz - 1
    # Actually, use the explicit formula as derived above:
    disc = (1.44**2 - 4 * 1 * 0.45)
    root = np.sqrt(disc)
    return ( (1.44 - root)/2, (1.44 + root)/2 )

print(corr_bounds(0.9, 0.8))
# Output should match: (0.458, 0.981)

4. Joint Gaussian: Best Guess for \( x \) Given \( z = x + y \)

Problem Statement

Suppose you draw a pair \( (x, y) \) from a joint Gaussian distribution with zero covariance (i.e., \( x \) and \( y \) are independent). You know the standard deviations of \( x \) and \( y \) (\( \sigma_x, \sigma_y \)), and you are told that \( z = x + y \). What is your best guess (conditional expectation) for \( x \)?

Step 1: Notation and Setup

Let:

• \( x \sim N(\mu_x, \sigma_x^2) \)
• \( y \sim N(\mu_y, \sigma_y^2) \)
• \( x, y \) independent (\( \text{Cov}(x, y) = 0 \))
• \( z = x + y \)

We are asked: What is \( \mathbb{E}[x | z] \)?

Step 2: Conditional Expectation for Multivariate Normal

In general, for jointly normal variables:

$$ \mathbb{E}[x | z] = \mu_x + \text{Cov}(x, z) \ /Var(z) \cdot (z - \mathbb{E}[z]) $$

Let’s compute all the terms step by step.

Step 3: Compute Means and Variances

• Mean of z: \( \mathbb{E}[z] = \mathbb{E}[x] + \mathbb{E}[y] = \mu_x + \mu_y \)
• Variance of z: \( \text{Var}(z) = \text{Var}(x) + \text{Var}(y) + 2\,\text{Cov}(x,y) = \sigma_x^2 + \sigma_y^2 \) (since covariance is zero)
• Covariance between x and z:
  • Since \( z = x + y \), \( \text{Cov}(x, z) = \text{Cov}(x, x+y) = \text{Cov}(x, x) + \text{Cov}(x, y) = \sigma_x^2 + 0 = \sigma_x^2 \)

Step 4: Plug Into the Conditional Expectation Formula

So:

$$ \mathbb{E}[x \mid z] = \mu_x + \frac{\sigma_x^2}{\sigma_x^2 + \sigma_y^2} (z - (\mu_x + \mu_y)) $$

Step 5: Special Case (Zero Means)

If the means are zero (\( \mu_x = \mu_y = 0 \)), then:

$$ \boxed{ \mathbb{E}[x \mid z] = \frac{\sigma_x^2}{\sigma_x^2 + \sigma_y^2} z } $$

This is the best linear unbiased estimator (BLUE) for \( x \) given \( z \).

Step 6: Intuition

The formula makes sense: your best guess for \( x \) is a weighted fraction of \( z \). If \( x \) is much more variable than \( y \) (i.e., \( \sigma_x^2 \gg \sigma_y^2 \)), your guess for \( x \) is close to \( z \). If \( y \) is much more variable, your best guess for \( x \) is close to 0.

Step 7: Python Code to Compute the Estimate

def best_guess_x(z, sigma_x, sigma_y, mu_x=0, mu_y=0):
    return mu_x + (sigma_x**2 / (sigma_x**2 + sigma_y**2)) * (z - mu_x - mu_y)

# Example: sigma_x = 2, sigma_y = 3, z = 10
print(best_guess_x(10, 2, 3))
# Output: 2^2 / (2^2 + 3^2) * 10 = 4/13 * 10 ≈ 3.08

Summary Table of Citadel Quant Research Interview Questions

Final Thoughts and Interview Tips

The quantitative research interview at firms like Citadel, Jane Street, and WorldQuant is built to test your mathematical and logical skills under pressure. Here’s how to prepare:

• Master Probability & Statistics: Most questions are grounded in these areas. Know your distributions, expectations, and properties.
• Brush Up on Linear Algebra: Correlation matrices, eigenvalues, and positive semi-definiteness often come up.
• Practice Coding: Be ready to write code to simulate or verify your answers, especially for probability puzzles.
• Communicate Clearly: Interviewers value clear, step-by-step reasoning as much as correct answers.
• Think Out Loud: Share your thought process so the interviewer can follow your logic and give hints if needed.

With thorough preparation and an analytical mindset, you can excel in quant interviews at the world’s most competitive finance firms.

Good luck on your journey to becoming a top quant researcher!

Related Articles

• Quant Research Interview Questions - Jane Street
• Quant Interview Questions - Akuna Capital
• Quant Research Interview Test - Jane Street
• Quant Interview Question - Jane Street
• Quant Interview Questions - WorldQuant