a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9) b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1) cor(a,b)= -1 a'=qbinom(a, 1, 0.5) b'=qbinom(b, 1, 0.5) why cor(a',b') becomes -0.5 ? -- View this message in context: http://r.789695.n4.nabble.com/qbinom-tp4651460.html Sent from the R help mailing list archive at Nabble.com.
qbinom
4 messages · Peter Langfelder, jaybell, David Winsemius
On Fri, Nov 30, 2012 at 9:47 AM, jaybell <stephenn9 at yahoo.com.tw> wrote:
a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9) b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1) cor(a,b)= -1 a'=qbinom(a, 1, 0.5) b'=qbinom(b, 1, 0.5) why cor(a',b') becomes -0.5 ?
On my computer the correlation is -0.8. It is not 1 because you did a non-linear transformation of a and b. Plot a vs. b, then plot a' vs b' and you will see why the correlation is not -1: a vs b is a straight line, a' vs b' is not a straight line. Peter
sorry, I repost the question again a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9) b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1) cor(a,b)= -1 pa=qbinom(a, 1, 0.5) pb=qbinom(b, 1, 0.5) cor(pa,pb)=-0.8 but when pa=qbinom(a,10,0.5) pb=qbinom(b,10,0.5) cor(pa,pb) becomes -1 again -- View this message in context: http://r.789695.n4.nabble.com/qbinom-tp4651460p4651496.html Sent from the R help mailing list archive at Nabble.com.
On Nov 30, 2012, at 1:05 PM, jaybell wrote:
sorry, I repost the question again
Why? This was already answered.
a=c(0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9) b=c(0.9, 0.8, 0.7, 0.6, 0.5, 0.4, 0.3, 0.2, 0.1) cor(a,b)= -1 pa=qbinom(a, 1, 0.5) pb=qbinom(b, 1, 0.5) cor(pa,pb)=-0.8 but when pa=qbinom(a,10,0.5) pb=qbinom(b,10,0.5) cor(pa,pb) becomes -1 again
What part of "plot the values" was difficult to understand?
-- View this message in context: http://r.789695.n4.nabble.com/qbinom-tp4651460p4651496.html Sent from the R help mailing list archive at Nabble.com.
David Winsemius, MD Alameda, CA, USA