This one took me a minute to even set up correctly.
Model each day's output as a signal plus independent noise, where the correlation of 0.95 implies a signal-to-noise ratio. Then derive how averaging n days reduces noise variance by a factor of n, and solve for the n that yields a correlation of 0.99 between two independent n-day averages.
Pro tip: Explicitly state the assumptions (e.g., equal variance, independence of noise across days) and note that the result is an approximation; this shows rigor and awareness of real-world complexities.
Assume each day's output vector is the sum of a fixed true signal and independent noise with equal variance. The correlation between two days is then signal variance divided by total variance.
Given correlation ρ = 0.95, express the signal variance as ρ times the total variance, and the noise variance as (1-ρ) times the total variance.
Averaging n independent days reduces the noise variance by a factor of n while the signal variance remains unchanged. Thus, the correlation between two independent n-day averages becomes ρ_n = signal variance / (signal variance + noise variance / n).
Set ρ_n = 0.99 and solve for n. Using ρ = 0.95, the equation is 0.95 / (0.95 + 0.05/n) = 0.99, which simplifies to n = 19.
State that approximately 19 days are needed. Mention that this assumes independence and equal variances, and that in practice the required n may differ due to correlations or non-stationarity.
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