Questions / Quant research, Research decisions
Allocate runs where the uncertainty is larger
Two independent experiment arms have known outcome standard deviations σ_A=6 and σ_B=3. With equal cost per task and 300 tasks in total, minimize Var(mean_A−mean_B)=36/n_A+9/n_B subject to n_A+n_B=300. How many tasks should go to A?
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Answer: 200 tasks
- Differentiating 36/n_A + 9/(300−n_A) gives n_A/(300−n_A)=6/3=2. Thus A receives 200 tasks and B receives 100; allocating by variance instead of standard deviation would be too extreme.
Interview shortcut. Neyman allocation: n ∝ σ, so n_A/n_B = 6/3.
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