This is the warm-up question and I knew it, but I still fumbled the bias correction explanation a bit.
Start by proposing the maximum likelihood estimator (MLE) for the standard deviation, which is the square root of the average of squared observations. Then check its bias by computing the expected value, noting that it underestimates the true standard deviation. Finally, explain the bias correction using the gamma function or an unbiased estimator based on the chi distribution.
Pro tip: Mention that while the MLE is biased, it is consistent and often preferred for large samples; however, for small samples, the unbiased estimator is more accurate. Also, note that the bias correction factor depends only on the sample size and can be computed using the gamma function.
Define the estimator as the square root of the average of squared observations: σ_hat = sqrt((1/n) * Σ X_i^2). This is the maximum likelihood estimator for σ when the mean is known to be zero.
Compute the expected value E[σ_hat]. Since n * σ_hat^2 / σ^2 follows a chi-squared distribution with n degrees of freedom, σ_hat is proportional to the square root of a chi-squared variable. Use the known expectation of the chi distribution to show E[σ_hat] = σ * sqrt(2/n) * Γ((n+1)/2) / Γ(n/2), which is less than σ.
Express the bias as E[σ_hat] - σ = σ * (c_n - 1), where c_n = sqrt(2/n) * Γ((n+1)/2) / Γ(n/2). Note that c_n < 1 for all finite n, so the MLE underestimates σ.
Define the unbiased estimator as σ_unbiased = σ_hat / c_n. This correction factor depends only on n and can be computed using the gamma function. For large n, c_n ≈ 1 - 1/(4n), so the correction is small.
Mention that the unbiased estimator is not the MLE and may have larger variance. Also, note that for large n, the bias is negligible, and the MLE is consistent. Optionally, mention that if the mean were unknown, the denominator would be n-1 for the variance, but here the mean is known.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, clarify that the estimators are for volatility (sigma) of the underlying Brownian motion, and that calibration involves scaling by known constants to make them unbiased. Then, compare their mean-squared errors analytically or via simulation, noting that each uses different information from the path. Finally, state a preference based on the MSE results, typically favoring the estimator with the lowest MSE.
Pro tip: Mention that the range-based estimator R is highly efficient for volatility but can be sensitive to jumps; however, under pure Brownian motion, it often outperforms |C|. Also, note that the combination R^2 + C^2 might have higher variance due to squaring, so it's rarely the best under MSE.
Explain that for Brownian motion observed at n=100 points, the volatility sigma is to be estimated. Calibrate each estimator by multiplying by a constant so that its expectation equals sigma. For |C|, the scaling is sqrt(pi/(2n))? Actually, for Brownian motion, C ~ N(0, n sigma^2), so E|C| = sigma sqrt(2n/pi). Thus calibrated |C| = |C| * sqrt(pi/(2n)). For R, the expected range of Brownian motion over n steps is sigma * sqrt(n) * E[range of standard BM], so calibrate by dividing by sqrt(n)*E[range]. For R - |C|/2, calibrate similarly. For R^2 + C^2, calibrate by taking square root and scaling.
Since calibration ensures unbiasedness, compare variances. For |C|, variance can be computed from the half-normal distribution. For R, the distribution is known (Feller) but complex; simulation is easier. For R - |C|/2, note that it's a linear combination of range and absolute terminal value, which are correlated. For R^2 + C^2, it's a quadratic form; its square root is biased, so calibration may involve more than a constant.
MSE = variance (since unbiased). Use known results or simulate many paths (e.g., 100,000) to estimate MSE for each estimator. Compare numerically. Typically, the range-based estimator R has the lowest MSE because it uses more information about the path's extremes. The estimator R - |C|/2 might improve upon R alone if it reduces variance, but often it's worse. R^2 + C^2 is likely poor due to high variance from squaring.
Based on MSE, prefer the estimator with the smallest MSE. Under pure Brownian motion, the calibrated range R is often the most efficient. However, if the problem expects a specific answer, note that R - |C|/2 might be designed to be uncorrelated with C? Actually, it's a known result that R and C are independent for Brownian motion? No, they are not independent. But there is a known unbiased estimator: (R + |C|)/2? Not sure. Better to rely on simulation.
Mention that in practice, financial returns may have jumps or stochastic volatility, so the best estimator under Brownian motion may not be robust. Also, note that the number of observations (100) is fixed, so asymptotic results may not apply perfectly.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.