Sortostat/Optimal sorting cutoffs: Difference between revisions
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==Problem== | ==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 | 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== | ==Analytical Solution== | ||
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==Simulation Solution== | ==Simulation Solution== | ||
Since our probability skills were pretty sad, we simulated it to confirm our analytical results. MATLAB file can be found here. | Since our probability skills were pretty sad, we ([[Alex Mallet]]) simulated it to confirm our analytical results. MATLAB file can be found here. | ||
===Results=== | ===Results=== |
Latest revision as of 10:26, 11 January 2006
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
[math]\displaystyle{ \emph E[X_N] = }[/math] expected value of the optimal percentage that can be returned from N trials
[math]\displaystyle{ \emph S_i = }[/math] random variable representing the percentile returned from the ith trial
- all trials are assumed to be independent therefore [math]\displaystyle{ \emph S_i = S }[/math], for all i
[math]\displaystyle{ \emph C_i = }[/math] the cut-off percentile for the ith trial.
General
[math]\displaystyle{ \emph E[X_N] = P(S\gt C_1) E[S|S\gt C_1] + }[/math]
[math]\displaystyle{ \emph (1-P(S\gt C_1))(P(S\gt C_2)E[S|S\gt C_2]) + }[/math]
[math]\displaystyle{ \emph (1-P(S\gt C_1))(1-P(S\gt C_2))(P(S\gt C_3)E[S|S\gt C_2]) + }[/math]
[math]\displaystyle{ \emph ... }[/math]
[math]\displaystyle{ \emph (1-P(S\gt C_1))(1-P(S\gt C_2))...(1-P(S\gt C_{N-1})E[S_N] }[/math]
Simplified
Since [math]\displaystyle{ \emph (1-P(S\gt C_1)) }[/math] can be factored out of every term after the first above, the solution can be simplified and solved recursively:
[math]\displaystyle{ \emph E[X_N] = P(S\gt C_N) E[S|S\gt C_N] + (1-P(S\gt C_N))E[X_{N-1}] }[/math]
base case:
[math]\displaystyle{ \emph E[X_1] = \int_0^\infty P(S)*S dS }[/math]
- 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.