James McDermott wrote: > Would the unique quadratic defined by the three points be the same > curve as the curve predicted by a quadratic B-spline (fit to all of > the data) through those same three points?
Yes, if you restrict attention to an interval between knots. You'll need to re-evaluate it for each such interval (but since quadratic splines are continuous, you can reuse evaluations at the knots, and you just need one new point in each interval). From a practical point of view, you need to make sure that COBS really is giving you a quadratic spline and really is reporting all of the knots correctly. Watch out for coincident knots (zero length intervals); you don't care about the derivative on those, but they might cause overflows in some calculations. Duncan Murdoch > > Jim > > On 7/19/05, Duncan Murdoch <[EMAIL PROTECTED]> wrote: > >>On 7/19/2005 3:34 PM, James McDermott wrote: >> >>>I wish it were that simple (perhaps it is and I am just not seeing >>>it). The output from cobs( ) includes the B-spline coefficients and >>>the knots. These coefficients are not the same as the a, b, and c >>>coefficients in a quadratic polynomial. Rather, they are the >>>coefficients of the quadratic B-spline representation of the fitted >>>curve. I need to evaluate a linear combination of basis functions and >>>it is not clear to me how to accomplish this easily. I was hoping to >>>find an alternative way of getting the derivatives. >> >>I don't know COBS, but doesn't predict just evaluate the B-spline? The >>point of what I posted is that the particular basis doesn't matter if >>you can evaluate the quadratic at 3 points. >> >>Duncan Murdoch >> >> >>>Jim McDermott >>> >>>On 7/19/05, Duncan Murdoch <[EMAIL PROTECTED]> wrote: >>> >>>>On 7/19/2005 2:53 PM, James McDermott wrote: >>>> >>>>>Hello, >>>>> >>>>>I have been trying to take the derivative of a quadratic B-spline >>>>>obtained by using the COBS library. What I would like to do is >>>>>similar to what one can do by using >>>>> >>>>>fit<-smooth.spline(cdf) >>>>>xx<-seq(-10,10,.1) >>>>>predict(fit, xx, deriv = 1) >>>>> >>>>>The goal is to fit the spline to data that is approximating a >>>>>cumulative distribution function (e.g. in my example, cdf is a >>>>>2-column matrix with x values in column 1 and the estimate of the cdf >>>>>evaluated at x in column 2) and then take the first derivative over a >>>>>range of values to get density estimates. >>>>> >>>>>The reason I don't want to use smooth.spline is that there is no way >>>>>to impose constraints (e.g. >=0, <=1, and monotonicity) as there is >>>>>with COBS. However, since COBS doesn't have the 'deriv =' option, the >>>>>only way I can think of doing it with COBS is to evaluate the >>>>>derivatives numerically. >>>> >>>>Numerical estimates of the derivatives of a quadratic should be easy to >>>>obtain accurately. For example, if the quadratic ax^2 + bx + c is >>>>defined on [-1, 1], then the derivative 2ax + b, has 2a = f(1) - f(0) + >>>>f(-1), and b = (f(1) - f(-1))/2. >>>> >>>>You should be able to generalize this to the case where the spline is >>>>quadratic between knots k1 and k2 pretty easily. >>>> >>>>Duncan Murdoch >>>> >> >> ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
