Choosing a cut using MIttas and Angelsis

[Sai Ram Nellutla]

Mittas and Angelsis find the cut by collecting:

• The mean mu of all the data below the cut;
• The mean mu0,mu1 of all the data below,above the cut.
• Then return the cut that most divides the data.

Is mu=mu0??

[Jason Hooks]

Possibly a typo.  I think mu is the mean of all the data.  mu0 would be the mean of the first part of the data, and mu1 would be the mean of the second part.

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[Tim Menzies]

jason is right

before the cut, we have n=n0+n1 items with mean mu

after the cut we have 

==> n0 items on LHS with mean mu0  

==> n1 items on RHS with mean mu1

expected value of LHS delta is

     lhs = n0/(n0+n1) * (mu - mu0)^2 

expected value of RHS delta   is

     rhs = n1/(n0 + n1) * (mu - mu1)^2

so find the split that generates max values for lhs+rhs

(note that dividing by (n0+n1) is optional. its a constant that effects both sides equally so it can be dropped. but i like typing n0/(n0+n1) cause that reminds me that we are talking about the probability of falling into the LHS since that is n0/(n0+n1))

t

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<hubris>

  vita=   http:// goo.gl/8eNhY  

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  stats=   http:// goo.gl/vggy1 

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</hubris>