IMC·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026

Summary

IMC quant engineer interview, one probability question under a tight time constraint. Pretty clean setup but the execution under pressure is where it gets you.

Questions Asked (1)

U and V are independent uniform [0,1] random variables. Let M be the max and m be the min of the two. What is P(M^2 > m)? Set up the integral, reduce it, and give a closed-form answer. You have about 3 minutes.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

The symmetry angle is the key move here.

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AI HintsAI Generated

Suggested Approach

First, express the event M^2 > m in terms of U and V, then use the joint density to set up a double integral over the region where the condition holds. Evaluate the integral by splitting into cases or using symmetry to simplify, and finally compute the closed-form probability.

Pro tip: Leverage the symmetry between U and V by assuming U > V (so M=U, m=V) and then multiplying by 2, which simplifies the region and integral. Also, sketch the unit square and shade the region to visualize the bounds.

1. Define the event in terms of U and V

Since M = max(U,V) and m = min(U,V), the condition M^2 > m means that the larger value squared exceeds the smaller value. Consider the two cases: U > V and V > U.

2. Set up the double integral

Use the joint density f(u,v)=1 on [0,1]^2. For U > V, the condition becomes U^2 > V, so the region is {0 ≤ v ≤ u ≤ 1, v < u^2}. The integral is ∫_{u=0}^1 ∫_{v=0}^{min(u, u^2)} dv du. By symmetry, multiply by 2 for the other case.

3. Simplify the integral bounds

For 0 ≤ u ≤ 1, min(u, u^2) = u^2 because u^2 ≤ u. So the inner integral is ∫_0^{u^2} dv = u^2. Thus the integral for U > V is ∫_0^1 u^2 du = 1/3.

4. Account for symmetry and compute final probability

By symmetry, the total probability is 2 * (1/3) = 2/3. Verify that the regions for U > V and V > U are disjoint and cover the event, so no overlap.

Key Points to Mention

Independence and uniform distribution imply joint density is 1 on the unit square.
The condition M^2 > m can be split into two symmetric cases: U > V and V > U.
For U > V, the inequality becomes U^2 > V, which defines a region under the curve v = u^2.
The inner integral simplifies because u^2 ≤ u for u in [0,1].
The integral evaluates to 1/3 for one case, and symmetry gives 2/3 total.
The final answer is a closed-form probability: 2/3.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.