Sortostat/Optimal sorting cutoffs
From OpenWetWare
Contents |
Problem
What is the optimal cut-off percentile for choosing a chamber to be sorted if you have N sorts (trials) remaining until you must take the sort to preserve a constant dilution rate?
Analytical Solution
Definition of variables
Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph E[X_N] =
expected value of the optimal percentage that can be returned from N trials
Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph S_i =
random variable representing the percentile returned from the ith trial
- all trials are assumed to be independent therefore Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph S_i = S
, for all i Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph C_i =
the cut-off percentile for the ith trial.
General
Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph E[X_N] = P(S>C_1) E[S|S>C_1] +
Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph (1-P(S>C_1))(P(S>C_2)E[S|S>C_2]) +
Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph (1-P(S>C_1))(1-P(S>C_2))(P(S>C_3)E[S|S>C_2]) +
Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph ...
Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph (1-P(S>C_1))(1-P(S>C_2))...(1-P(S>C_{N-1})E[S_N]
Simplified
Since Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph (1-P(S>C_1))
can be factored out of every term after the first above, the solution can be simplified and solved recursively:
Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph E[X_N] = P(S>C_N) E[S|S>C_N] + (1-P(S>C_N))E[X_{N-1}]
base case:
Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): \emph E[X_1] = \int_0^\infty P(S)*S dS
- e.g., if you have only 1 trial then you expect to get the mean of the distribution for S.
Simulation Solution
Since our probability skills were pretty sad, we (Alex Mallet) simulated it to confirm our analytical results. MATLAB file can be found here.
