Capital One·Data Scientist·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

May 2026

Summary

Capital One data scientist interview that leaned pretty heavily into probability and statistics theory. The MGF question was the kind of thing I hadn't touched since grad school and it showed.

Questions Asked (1)

You're given the moment-generating function M_X(t) = e^(t^2). What distribution does X follow, and what is E[X^2]? Walk through the second derivative of the MGF evaluated at zero and reconcile that with the variance formula.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I recognized the MGF form eventually but took an embarrassingly long pause before connecting it to a normal distribution with mean 0 and variance 2.

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AI HintsAI Generated

Suggested Approach

Recognize that the MGF e^(t^2) matches the MGF of a normal distribution with mean 0 and variance 2. Then compute the second derivative of the MGF at t=0 to find E[X^2], and relate it to the variance formula Var(X) = E[X^2] - (E[X])^2, noting that E[X] = 0 from the first derivative.

Pro tip: Show that you can derive the distribution from the MGF by comparing it to known forms, and explicitly state that the variance is the second central moment, which here equals E[X^2] because the mean is zero.

1. Identify the distribution from the MGF

Compare M_X(t) = e^(t^2) to the standard MGF of a normal distribution: M(t) = e^(μt + (σ^2 t^2)/2). Match parameters to get μ = 0 and σ^2 = 2.

2. Compute the first derivative to find E[X]

Differentiate M_X(t) = e^(t^2) to get M'(t) = 2t e^(t^2). Evaluate at t=0: M'(0) = 0, so E[X] = 0.

3. Compute the second derivative to find E[X^2]

Differentiate again: M''(t) = (2 + 4t^2) e^(t^2). Evaluate at t=0: M''(0) = 2, so E[X^2] = 2.

4. Reconcile with the variance formula

Use Var(X) = E[X^2] - (E[X])^2. Since E[X] = 0, Var(X) = E[X^2] = 2. This matches the variance from the normal distribution identified in step 1.

Key Points to Mention

The MGF of a normal distribution N(μ, σ^2) is e^(μt + (σ^2 t^2)/2).
Matching e^(t^2) to this form gives μ = 0 and σ^2 = 2, so X ~ N(0, 2).
The first derivative of the MGF evaluated at zero gives the mean: M'(0) = E[X].
The second derivative of the MGF evaluated at zero gives the second moment: M''(0) = E[X^2].
Variance formula: Var(X) = E[X^2] - (E[X])^2.
Since E[X] = 0, Var(X) = E[X^2] = 2, consistent with the normal distribution parameters.

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