lscov vs \
in [Matlab]
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From: Peter Perkins on 29 May 2010 18:13 On 5/25/2010 5:04 AM, Michael wrote: > Although I cannot help you on your actual problem, I want to throw in > that the usage of lscov to me only seems necessary if you have a priori > covariance information available. If you just want to solve an > overdetermined equation system, "\" should be totally sufficient. That is the origin of LSCOV, yes (thus the name). But if you look at the help since about R14, you'll find that LSCOV can be called with a (known) cov matrix, or a (known) weight vector, or with neither. More importantly, it returns not only regression coefficients, but also std errs, MSE, and an estimate of the cov matrix of the coef estimates. So if all you want is the least squares estimates themselves, backslash is quite easy to type. If you want all the rest, LSCOV is where you should look. They return the same coefs (as Bruno explained).
From: John D'Errico on 30 May 2010 05:57
"Flyx" <flyxylf(a)hotmail.com> wrote in message <htg581$5fl$1(a)adenine.netfront.net>... > "Michael " <michael.schmittNOSPAM(a)bv.tum.de> ha scritto nel messaggio > news:htg3q5$j8k$1(a)fred.mathworks.com... > > Although I cannot help you on your actual problem, I want to throw in that > > the usage of lscov to me only seems necessary if you have a priori > > covariance information available. If you just want to solve an > > overdetermined equation system, "\" should be totally sufficient. > > Yes, I supposed. But lscov accepts even only the two parameters A and b > given a result that completely matches to the "linsolve" one. Why does it > differ from "\" ? It should be theoretically the same. > Even though they should be theoretically the same, they will never be exactly the same in practice. Floating point arithmetic assures this fact, that any two algorithms for the same result may give slightly different results in the least significant bits of the result. Next, despite the fact that you claim m >> n, so that the system is strongly overdetermined, this tells us absolutely NOTHING. I can easily give you a matrix that is well-scaled, with 1000000 rows and only 2 columns, yet it is singular. And all bets are off on singular matrices. A = ones(1e6,2); b = rand(1e6,1); A\b Warning: Rank deficient, rank = 1, tol = 2.2204e-07. ans = 0.50032124992531 0 lscov(A,b) Warning: A is rank deficient to within machine precision. > In lscov at 197 ans = 0.500321249925306 0 Well, at least the two solutions are close, but as you see, they differ in the LSBs. And this is with a trivial matrix. It could be worse, if your problem is poorly scaled, or poorly conditioned. So tell us, what is the condition number of your matrix? Don't just tell us the relative sizes, which is no information at all. John |