On 21/08/13 11:23, Ye Lin wrote:
> T
> hanks for your insights Rolf! The model I want to fit is y=x/a+x with
> no intercept, so I transformed it to 1/y=1+a/x as they are the same.
For crying out loud, they are ***NOT*** the same. The equations y =
x/(a+x) and
1/y = 1 + a/x are indeed algebraically identical, but if an "error" or
"noise" term is added
to each then then the nature of the error term is vastly different. It
is the error or
noise term that is of central concern in a statistical context.
cheers,
Rolf
> but i will look up nls() and see how to fit the model without
> transformation.
>
>
> On Tue, Aug 20, 2013 at 2:45 PM, Rolf Turner <rolf.tur...@xtra.co.nz
> <mailto:rolf.tur...@xtra.co.nz>> wrote:
>
>
> (1) It is not acceptable to use "wanna" in written English. You
> should say
> "I want to fit a model ....".
>
> (2) The model you have fitted is *not* equivalent to the model you
> first state.
>
> If you write "y ~ x/(a+x)" you are tacitly implying that
>
> y = x/(a+x) + E
>
> where the "errors" E are i.i.d. with mean 0.
>
> If this is the case then it will *not* be the case that
>
> 1/y = 1 + a/x + E
>
> with the E values being i.i.d. with mean 0.
>
> If the model "y ~ x/(a+x)" is really what you want to fit, then
> you should
> be using non-linear methods, e.g. by applying the function nls().
>
> cheers,
>
> Rolf Turner
>
>
>
> On 21/08/13 09:39, Ye Lin wrote:
>
> Hey All,
>
> I wanna to fit a model y~x/(a+x) to my data, here is the code
> I use now:
>
> lm((1/y-1)~I(1/x)+0, data=b)
>
> and it will return the coefficient which is value of a
>
> however, if I use the code above, I am not able to draw a
> curve the
> presents this equation. How can I do this?
>
>
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