I have to compute some standard errors using the delta
method and so have to use the command "numericDeriv"
to get the desired gradient. Befor using it on my
complicated function, I've done a try with a simple
exemple :
x <- 1:5
numericDeriv(quote(x^2),"x")
and i get :
[1] 1 8 27 64 125 216
attr(,"gradient")
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] Inf 0 0 NaN 0 0
[2,] 0 0 0 NaN 0 0
[3,] 0 Inf 0 NaN 0 0
[4,] 0 0 0 NaN 0 0
[5,] 0 0 Inf NaN 0 0
[6,] 0 0 0 NaN 0 0
I don't understand the result. I thought I will get :
[1] 1 8 27 64 125 216
attr(,"gradient")
[,1]
[1,] 1
[2,] 4
[3,] 6
[4,] 8
[5,] 10
[6,] 12
The derivative of x^2 is still 2x, isn't it ?
Thanks for help
On Wed, 16 Nov 2005, Florent Bresson wrote:> I have to compute some standard errors using the delta > method and so have to use the command "numericDeriv" > to get the desired gradient. Befor using it on my > complicated function, I've done a try with a simple > exemple : > > x <- 1:5 > numericDeriv(quote(x^2),"x") > > and i get : > > [1] 1 8 27 64 125 216 > attr(,"gradient") > [,1] [,2] [,3] [,4] [,5] [,6] > [1,] Inf 0 0 NaN 0 0 > [2,] 0 0 0 NaN 0 0 > [3,] 0 Inf 0 NaN 0 0 > [4,] 0 0 0 NaN 0 0 > [5,] 0 0 Inf NaN 0 0 > [6,] 0 0 0 NaN 0 0 > > I don't understand the result. I thought I will get : > > [1] 1 8 27 64 125 216 > attr(,"gradient") > [,1] > [1,] 1 > [2,] 4 > [3,] 6 > [4,] 8 > [5,] 10 > [6,] 12 > > The derivative of x^2 is still 2x, isn't it ?and (1:5)^2 is still [1] 1 4 9 16 25 ! Try> x <- as.numeric(1:5) > numericDeriv(quote(x^2),"x")since the author of numericDeriv has forgotten some coercions. -- Brian D. Ripley, ripley at stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UK Fax: +44 1865 272595
On 16-Nov-05 Florent Bresson wrote:> I have to compute some standard errors using the delta > method and so have to use the command "numericDeriv" > to get the desired gradient. Befor using it on my > complicated function, I've done a try with a simple > exemple : > > x <- 1:5 > numericDeriv(quote(x^2),"x") > > and i get : > > [1] 1 8 27 64 125 216 > attr(,"gradient") > [,1] [,2] [,3] [,4] [,5] [,6] > [1,] Inf 0 0 NaN 0 0 > [2,] 0 0 0 NaN 0 0 > [3,] 0 Inf 0 NaN 0 0 > [4,] 0 0 0 NaN 0 0 > [5,] 0 0 Inf NaN 0 0 > [6,] 0 0 0 NaN 0 0 > > I don't understand the result. I thought I will get : > > [1] 1 8 27 64 125 216 > attr(,"gradient") > [,1] > [1,] 1 > [2,] 4 > [3,] 6 > [4,] 8 > [5,] 10 > [6,] 12 > > The derivative of x^2 is still 2x, isn't it ?The trap you've fallen into is that "x <- 1:5" makes x of integer type, and (believe it or not) you cannot differentiate when the support of a function is the integers. Wrong topology (though I'm not sure that this is quite how R thinks about it). So give x a bit of elbow-room ("numeric" type has "continous" -- well, nearly -- topology):> x <- as.numeric(1:5) > numericDeriv(quote(x^2),"x")[1] 1 4 9 16 25 attr(,"gradient") [,1] [,2] [,3] [,4] [,5] [1,] 2 0 0 0 0 [2,] 0 4 0 0 0 [3,] 0 0 6 0 0 [4,] 0 0 0 8 0 [5,] 0 0 0 0 10 Cheers, Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <Ted.Harding at nessie.mcc.ac.uk> Fax-to-email: +44 (0)870 094 0861 Date: 16-Nov-05 Time: 13:11:49 ------------------------------ XFMail ------------------------------