> -----Original Message-----
> From: stephen sefick [mailto:ssefick at gmail.com]
> Sent: April-04-11 2:49 PM
> To: Steven McKinney
> Subject: Re: [R] Linear Model with curve fitting parameter?
>
> Steven:
>
> I am really sorry for my confusion. I hope this now makes sense.
>
> b0 == y intercept == y-intercept == (intercept) fit by lm
>
> a <- 1:10
> b <- 1:10
>
> summary(lm(a~b))
> #to show what I was calling b0
>
> So...
>
> ################################################
> manning
>
> Q = K*A*(R^b2)*(S^b3)
>
> log(Q) = log(K)+log(A)+(b2*log(R))+(b3*log(S))
Okay, using this notation, this appears to be the original
model you queried about. So for this model, as I showed
before,
Let Z = log(Q) - log(A)
E(Z) = b0 + b2*log(R) + b3*log(S)
= log(K) + b2*log(R) + b3*log(S)
Fitting the model lm(Z ~ log(R) + log(S))
will yield parameter estimates b_hat_0, b_hat_2, b_hat_3
where
b_hat_0 (the fitted model intercept) is an estimate of b0 (which is log(K)),
b_hat_2 is an estimate of b2,
b_hat_3 is an estimate of b3.
So in answer to your previous question, b0 is an
estimate of log(K), not ( log(Qintercept)+log(K) )
so an estimate for K is exp(b_hat_0)
>
> ################################################
> dingman
> Q = K*(A^b1)*(R^b2)*(S^b3*log(S))
>
> log(Q) = log(K)+(b1*log(A))+(b2*log(R))+(b3*(log(S))^2)
The dingman model notation is ambiguous. Is the last
term S^(b3*log(S)) or (S^b3)*log(S) ?
Previous email showed
> dingman
> log(Q)=log(b0)+log(K)+a*log(A)+r*log(R)+s*(log(S))^2
which implies (if I ignore the log(b0) term)
Q = K*(A^a)*(R^r)*(exp(log(S)*log(S))^s)
= K*(A^a)*(R^r)*(S^(log(S)*s))
This is linearizable as
log(Q) = log(K) + a*log(A) + r*log(R) + s*(log(S))^2
= b0 + b1*log(A) + b2*log(R) + b3*(log(S)^2)
Fitting lm(log(Q) ~ log(A) + log(R) + I(log(S)^2) ... )
will yield estimates b_hat_0, b_hat_1, b_hat_2 and b_hat_3
where b_hat_0 is an estimate of b0 = log(K) so an estimate of K is exp(b_hat_0),
b_hat_1 is an estimate of b1 = a,
b_hat_2 is an estimate of b2 = r,
b_hat_3 is an estimate of b3 = s
>
> ################################################
>
> Bjerklie
>
> Q = K*(A^b1)*(R^b2)*(S^b3)
>
> log(Q) = log(K)+(b1*log(A))+(b2*log(R))*(b3*log(S))
Fitting lm(log(Q) ~ log(A) + log(R) + log(S) ... )
will yield estimates b_hat_0, b_hat_1, b_hat_2 and b_hat_3
where b_hat_0 is an estimate of b0 = log(K) so an estimate of K is exp(b_hat_0),
b_hat_1 is an estimate of b1 = a,
b_hat_2 is an estimate of b2 = r,
b_hat_3 is an estimate of b3 = s
Best
Steve McKinney
>
> ################################################
>
>
>
>
>
> On Mon, Apr 4, 2011 at 2:58 PM, Steven McKinney <smckinney at
bccrc.ca> wrote:
> >
> >> -----Original Message-----
> >> From: stephen sefick [mailto:ssefick at gmail.com]
> >> Sent: April-03-11 5:35 PM
> >> To: Steven McKinney
> >> Cc: R help
> >> Subject: Re: [R] Linear Model with curve fitting parameter?
> >>
> >> Steven:
> >>
> >> You are exactly right sorry I was confused.
> >>
> >>
> >> #######################################################
> >> so log(y-intercept)+log(K) is a constant called b0 (is this
right?)
> >
> > Doesn't look right to me based on the information you've
provided.
> > I don't see anything labeled "y" in your previous
emails, so I'm
> > not clear on what y is and how it relates to the original model
> > you described
> >
> > > >> I have a model Q=K*A*(R^r)*(S^s)
> > > >>
> > > >> A, R, and S are data I have and K is a curve fitting
parameter.
> >
> > If the model is
> >
> > Q=K*A*(R^r)*(S^s)
> >
> > then
> >
> > log(Q) = log(K) + log(A) + r*log(R) + s*log(S)
> >
> > Rearranging yields
> >
> > log(Q) - log(A) = log(K) + r*log(R) + s*log(S)
> >
> > Let Z = log(Q) - log(A) = log(Q/A)
> >
> > so
> >
> > Z = log(K) + r*log(R) + s*log(S)
> >
> > and a linear model fit of
> >
> > Z ~ log(R) + log(S)
> >
> > will yield parameter estimates for the linear equation
> >
> > E(Z) = B0 + B1*log(R) + B2*log(S)
> >
> > (E(Z) = expected value of Z)
> >
> > so B0 estimate is an estimate of log(K)
> > B1 estimate is an estimate of r
> > B2 estimate is an estimate of s
> >
> > and these are the only parameters you described in the original model.
> >
> >
> >>
> >> lm(log(Q)~log(A)+log(R)+log(S)-1)
> >>
> >> is fitting the model
> >>
> >> log(Q)=a*log(A)+r*log(R)+s*log(S) (no beta 0)
> >>
> >> and
> >>
> >> lm(log(Q)~log(A)+log(R)+log(S))
> >>
> >>
> >> is fitting the model
> >>
> >> log(Q)=b0+a*log(A)+r*log(R)+s*log(S)
> >
> > K has disappeared from these equations so these model fits do
> > not correspond to the model originally described. Now a b0
> > appears, and is used in models below. I think changing notation
> > is also adding confusion. What are "y" and
"intercept" you
> > discuss above, in relation to your original notation?
> >
> >>
> >> ######################################################
> >>
> >> These are the models I am trying to fit and if I have reasoned
> >> correctly above then I should be able to fit the below models
> >> similarly.
> >
> > You will be able to fit models appropriately once you have a
> > clearly defined system of notation that allows you to map between
> > the proposed data model, the parameters in that model, and the
> > corresponding regression equations.
> >
> > Once you have consistent notation, you will be able to see
> > if you can express your model as a linear regression, or
> > if not, what kind of non-linear regression you will need to
> > do to get estimates for the parameters in your model.
> >
> > Best
> >
> > Steve McKinney
> >
> >>
> >> manning
> >> log(Q)=log(b0)+log(K)+log(A)+r*log(R)+s*log(S)
> >>
> >> dingman
> >> log(Q)=log(b0)+log(K)+a*log(A)+r*log(R)+s*(log(S))^2
> >>
> >> bjerklie
> >> log(Q)=log(b0)+log(K)+a*log(A)+r*log(R)+s*log(S)
> >>
> >> #######################################################
> >>
> >> Thank you for all of your help!
> >>
> >> Stephen
> >>
> >> On Fri, Apr 1, 2011 at 2:44 PM, Steven McKinney <smckinney at
bccrc.ca> wrote:
> >> >
> >> >> -----Original Message-----
> >> >> From: stephen sefick [mailto:ssefick at gmail.com]
> >> >> Sent: April-01-11 5:44 AM
> >> >> To: Steven McKinney
> >> >> Cc: R help
> >> >> Subject: Re: [R] Linear Model with curve fitting
parameter?
> >> >>
> >> >> Setting Z=Q-A would be the incorrect dimensions. I could
Z=Q/A.
> >> >
> >> > I suspect this is confusion about what Q is. I was presuming
that
> >> > the Q in this following formula was log(Q) with Q from the
original data.
> >> >
> >> >> >> I have taken the log of the data that I have and
this is the model
> >> >> >> formula without the K part
> >> >> >>
> >> >> >> lm(Q~offset(A)+R+S, data=x)
> >> >
> >> > If the model is
> >> >
> >> > Q=K*A*(R^r)*(S^s)
> >> >
> >> > then
> >> >
> >> > log(Q) = log(K) + log(A) + r*log(R) + s*log(S)
> >> >
> >> > Rearranging yields
> >> >
> >> > log(Q) - log(A) = log(K) + r*log(R) + s*log(S)
> >> >
> >> > so what I labeled 'Z' below is
> >> >
> >> > Z = log(Q) - log(A) = log(Q/A)
> >> >
> >> > so
> >> >
> >> > Z = log(K) + r*log(R) + s*log(S)
> >> >
> >> > and a linear model fit of
> >> >
> >> > Z ~ log(R) + log(S)
> >> >
> >> > will yield parameter estimates for the linear equation
> >> >
> >> > E(Z) = B0 + B1*log(R) + B2*log(S)
> >> >
> >> > (E(Z) = expected value of Z)
> >> >
> >> > so B0 estimate is an estimate of log(K)
> >> > B1 estimate is an estimate of r
> >> > B2 estimate is an estimate of s
> >> >
> >> > More details and careful notation will eventually lead
> >> > to a reasonable description and analysis strategy.
> >> >
> >> >
> >> > Best
> >> >
> >> > Steve McKinney
> >> >
> >> >
> >> >
> >> >> Is fitting a nls model the same as fitting an ols? These
data are
> >> >> hydraulic data from ~47 sites. To access predictive
ability I am
> >> >> removing one site fitting a new model and then accessing
the fit with
> >> >> a myriad of model assessment criteria. I should get the
same answer
> >> >> with ols vs nls? Thank you for all of your help.
> >> >>
> >> >> Stephen
> >> >>
> >> >> On Thu, Mar 31, 2011 at 8:34 PM, Steven McKinney
<smckinney at bccrc.ca> wrote:
> >> >> >
> >> >> >> -----Original Message-----
> >> >> >> From: r-help-bounces at r-project.org
[mailto:r-help-bounces at r-project.org] On Behalf Of stephen
> >> >> sefick
> >> >> >> Sent: March-31-11 3:38 PM
> >> >> >> To: R help
> >> >> >> Subject: [R] Linear Model with curve fitting
parameter?
> >> >> >>
> >> >> >> I have a model Q=K*A*(R^r)*(S^s)
> >> >> >>
> >> >> >> A, R, and S are data I have and K is a curve
fitting parameter. I
> >> >> >> have linearized as
> >> >> >>
> >> >> >> log(Q)=log(K)+log(A)+r*log(R)+s*log(S)
> >> >> >>
> >> >> >> I have taken the log of the data that I have and
this is the model
> >> >> >> formula without the K part
> >> >> >>
> >> >> >> lm(Q~offset(A)+R+S, data=x)
> >> >> >>
> >> >> >> What is the formula that I should use?
> >> >> >
> >> >> > Let Z = Q - A for your logged data.
> >> >> >
> >> >> > Fitting lm(Z ~ R + S, data = x) should yield
> >> >> > intercept parameter estimate = estimate for log(K)
> >> >> > R coefficient parameter estimate = estimate for r
> >> >> > S coefficient parameter estimate = estimate for s
> >> >> >
> >> >> >
> >> >> >
> >> >> > Steven McKinney
> >> >> >
> >> >> > Statistician
> >> >> > Molecular Oncology and Breast Cancer Program
> >> >> > British Columbia Cancer Research Centre
> >> >> >
> >> >> >
> >> >> >
> >> >> >>
> >> >> >> Thanks for all of your help. I can provide a
subset of data if necessary.
> >> >> >>
> >> >> >>
> >> >> >>
> >> >> >> --
> >> >> >> Stephen Sefick
> >> >> >> ____________________________________
> >> >> >> | Auburn University
|
> >> >> >> | Biological Sciences
|
> >> >> >> | 331 Funchess Hall
|
> >> >> >> | Auburn, Alabama
|
> >> >> >> | 36849
|
> >> >> >> |___________________________________|
> >> >> >> | sas0025 at auburn.edu
|
> >> >> >> | http://www.auburn.edu/~sas0025
|
> >> >> >> |___________________________________|
> >> >> >>
> >> >> >> Let's not spend our time and resources
thinking about things that are
> >> >> >> so little or so large that all they really do
for us is puff us up and
> >> >> >> make us feel like gods. We are mammals, and
have not exhausted the
> >> >> >> annoying little problems of being mammals.
> >> >> >>
> >> >> >> -K. Mullis
> >> >> >>
> >> >> >> "A big computer, a complex algorithm and a
long time does not equal science."
> >> >> >>
> >> >> >> -Robert Gentleman
> >> >> >> ______________________________________________
> >> >> >> R-help at r-project.org mailing list
> >> >> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> >> >> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> >> >> >> and provide commented, minimal, self-contained,
reproducible code.
> >> >> >
> >> >>
> >> >>
> >> >>
> >> >> --
> >> >> Stephen Sefick
> >> >> ____________________________________
> >> >> | Auburn University
|
> >> >> | Biological Sciences
|
> >> >> | 331 Funchess Hall
|
> >> >> | Auburn, Alabama
|
> >> >> | 36849
|
> >> >> |___________________________________|
> >> >> | sas0025 at auburn.edu
|
> >> >> | http://www.auburn.edu/~sas0025 |
> >> >> |___________________________________|
> >> >>
> >> >> Let's not spend our time and resources thinking about
things that are
> >> >> so little or so large that all they really do for us is
puff us up and
> >> >> make us feel like gods. We are mammals, and have not
exhausted the
> >> >> annoying little problems of being mammals.
> >> >>
> >> >> -K. Mullis
> >> >>
> >> >> "A big computer, a complex algorithm and a long time
does not equal science."
> >> >>
> >> >> -Robert Gentleman
> >> >
> >>
> >>
> >>
> >> --
> >> Stephen Sefick
> >> ____________________________________
> >> | Auburn University |
> >> | Biological Sciences |
> >> | 331 Funchess Hall |
> >> | Auburn, Alabama |
> >> | 36849
|
> >> |___________________________________|
> >> | sas0025 at auburn.edu |
> >> | http://www.auburn.edu/~sas0025 |
> >> |___________________________________|
> >>
> >> Let's not spend our time and resources thinking about things
that are
> >> so little or so large that all they really do for us is puff us up
and
> >> make us feel like gods. We are mammals, and have not exhausted
the
> >> annoying little problems of being mammals.
> >>
> >> -K. Mullis
> >>
> >> "A big computer, a complex algorithm and a long time does not
equal science."
> >>
> >> -Robert Gentleman
> >
>
>
>
> --
> Stephen Sefick
> ____________________________________
> | Auburn University |
> | Biological Sciences |
> | 331 Funchess Hall |
> | Auburn, Alabama |
> | 36849 |
> |___________________________________|
> | sas0025 at auburn.edu |
> | http://www.auburn.edu/~sas0025 |
> |___________________________________|
>
> Let's not spend our time and resources thinking about things that are
> so little or so large that all they really do for us is puff us up and
> make us feel like gods. We are mammals, and have not exhausted the
> annoying little problems of being mammals.
>
> -K. Mullis
>
> "A big computer, a complex algorithm and a long time does not equal
science."
>
> -Robert Gentleman