Bank of America·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Quant researcher screen with a stats question that sounds basic until you actually think through the details. Short round, probably a phone screen.

Questions Asked (1)

You have 100 iid draws from a N(0, sigma^2) distribution. How do you estimate sigma, and is your estimator unbiased?

Technical Trade-offs

Author's notes

I went straight to the sample variance and then had to backtrack when they asked about bias.

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AI HintsAI Generated

Suggested Approach

Start by stating the maximum likelihood estimator for sigma, which is the square root of the average of squared deviations from the mean. Then discuss its bias, noting that the MLE is biased but can be corrected by using n-1 in the denominator. Finally, mention the unbiased estimator and its properties.

Pro tip: Demonstrate awareness that in practice, for large samples the bias is negligible, but for small samples the correction matters. Also, mention that if the mean is known to be zero, the estimator changes and is unbiased.

1. Identify the parameter and estimator

State that we want to estimate the standard deviation sigma from 100 iid N(0, sigma^2) draws. The natural estimator is the maximum likelihood estimator (MLE): sigma_hat = sqrt( (1/n) * sum_{i=1}^n (X_i - X_bar)^2 ).

2. Discuss bias of the MLE

Explain that the MLE is biased because it uses n in the denominator, underestimating sigma on average. The bias arises from Jensen's inequality and the fact that the sample variance with n is biased for sigma^2.

3. Provide the unbiased estimator

Introduce the unbiased estimator: s = sqrt( (1/(n-1)) * sum_{i=1}^n (X_i - X_bar)^2 ). Note that while s^2 is unbiased for sigma^2, s itself is not unbiased for sigma; however, it is a common correction and reduces bias.

4. Clarify the nuance about unbiasedness

Point out that strictly speaking, no estimator of sigma based on the sample variance is exactly unbiased for sigma due to the square root. The unbiased estimator for sigma^2 is s^2, but for sigma, one might use c4(n) correction or acknowledge that s is approximately unbiased for large n.

5. Consider known mean case

If the mean is known to be zero (as given), the MLE for sigma is sqrt( (1/n) * sum X_i^2 ). This estimator is also biased, but the unbiased version uses n-1: sqrt( (1/(n-1)) * sum X_i^2 ).

Key Points to Mention

Maximum likelihood estimator for sigma: sqrt( (1/n) * sum (X_i - X_bar)^2 )
Bias of MLE: underestimates sigma due to Jensen's inequality
Unbiased estimator for variance: s^2 = (1/(n-1)) * sum (X_i - X_bar)^2
Unbiasedness for sigma: s is not exactly unbiased, but approximately so for large n
Known mean case: if mean=0, use sum X_i^2 / n for MLE, and sum X_i^2 / (n-1) for unbiased variance
Large sample properties: bias diminishes as n increases

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