# [R] Orthogonal polynomials used by R

Michael Dewey ||@t@ @end|ng |rom dewey@myzen@co@uk
Thu Nov 28 17:25:46 CET 2019

```Dear Ashim

As John said your two examples give the same model to within rounding
error so it is not clear what you see the problem as being. You can
always remove some of the correlation by subtracting out a large
constant from x before you use poly() on it.

Michael

On 28/11/2019 16:02, Ashim Kapoor wrote:
> On Thu, Nov 28, 2019 at 7:38 PM Fox, John <jfox using mcmaster.ca> wrote:
>
>> Dear Ashim,
>>
>> I'm afraid that much of what you say here is confused.
>>
>> First, because poly(x) and poly(x, raw=TRUE) produce the same fitted
>> values (as I previously explained), they also produce the same residuals,
>> and consequently the same CV criteria. From the point of view of CV,
>> there's therefore no reason to prefer orthogonal polynomials. And you still
>> don't explain why you want to interpret the coefficients of the polynomial.
>>
>
> The trend in the variable that I am trying to create an ARIMA model for is
> given by poly(x,4). That is why I wished to know what these polynomials
> look like.
>
> I used  :
>
> trend <- predict(lm(gdp~poly(x,4)),newdata = data.frame(
> x=94:103),interval="confidence")
>
> and I was able to (numerically) extrapolate the poly(x,4) trend, although,
> I think it would be interesting to know what polynomials I was dealing with
> in this case. Just some intuition as to if the linear / quadratic / cubic /
> fourth order polynomial trend is dominating. I don't know how I would
> interpret them, but it would be fun to know.
>
> Please allow me to show you a trick. I read this on the internet, here :-
>
> https://www.datasciencecentral.com/profiles/blogs/deep-dive-into-polynomial-regression-and-overfitting
>
> Please see the LAST comment by Scott Stelljes where he suggests using an
> orthogonal polynomial basis. He does not elaborate but leaves the reader to
> work out the details.
>
> Here is what I think of this. Take a big number say 20 and take a variable
> in which we are trying to find the order of the polynomial in the trend.
> Like this :-
>
>> summary(lm(gdp ~ poly(x,20)))
>
> Call:
> lm(formula = gdp ~ poly(x, 20))
>
> Residuals:
>       Min       1Q   Median       3Q      Max
> -1235661  -367798   -80453   240360  1450906
>
> Coefficients:
>                 Estimate Std. Error t value Pr(>|t|)
> (Intercept)    17601482      66934 262.968  < 2e-16 ***
> poly(x, 20)1  125679081     645487 194.704  < 2e-16 ***
> poly(x, 20)2   43108747     645487  66.785  < 2e-16 ***
> poly(x, 20)3    3605839     645487   5.586 3.89e-07 ***
> poly(x, 20)4   -2977277     645487  -4.612 1.69e-05 ***
> poly(x, 20)5    1085732     645487   1.682   0.0969 .
> poly(x, 20)6    1124125     645487   1.742   0.0859 .
> poly(x, 20)7    -108676     645487  -0.168   0.8668
> poly(x, 20)8    -976915     645487  -1.513   0.1345
> poly(x, 20)9   -1635444     645487  -2.534   0.0135 *
> poly(x, 20)10   -715019     645487  -1.108   0.2717
> poly(x, 20)11    347102     645487   0.538   0.5924
> poly(x, 20)12   -176728     645487  -0.274   0.7850
> poly(x, 20)13   -634151     645487  -0.982   0.3292
> poly(x, 20)14   -537725     645487  -0.833   0.4076
> poly(x, 20)15    -58674     645487  -0.091   0.9278
> poly(x, 20)16    -67030     645487  -0.104   0.9176
> poly(x, 20)17   -809443     645487  -1.254   0.2139
> poly(x, 20)18   -668879     645487  -1.036   0.3036
> poly(x, 20)19   -302384     645487  -0.468   0.6409
> poly(x, 20)20    359134     645487   0.556   0.5797
> ---
> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
> Residual standard error: 645500 on 72 degrees of freedom
> Multiple R-squared:  0.9983, Adjusted R-squared:  0.9978
> F-statistic:  2122 on 20 and 72 DF,  p-value: < 2.2e-16
>
>>
>
>
> The CV estimate of the trend is 4. I am not saying this method is perfect,
> but look above this method also suggests n=4.
>
> I CANNOT do this with raw polynomials, since they are correlated and
> JOINTLY in the presence of others they may not be significant, please see
> below.
>
>> summary(lm(gdp ~ poly(x,20,raw=T)))
>
> Call:
> lm(formula = gdp ~ poly(x, 20, raw = T))
>
> Residuals:
>       Min       1Q   Median       3Q      Max
> -1286007  -372221   -81320   248510  1589130
>
> Coefficients: (4 not defined because of singularities)
>                           Estimate Std. Error t value Pr(>|t|)
> (Intercept)             2.067e+06  2.649e+06   0.780    0.438
> poly(x, 20, raw = T)1   1.633e+06  3.556e+06   0.459    0.647
> poly(x, 20, raw = T)2  -7.601e+05  1.679e+06  -0.453    0.652
> poly(x, 20, raw = T)3   1.861e+05  3.962e+05   0.470    0.640
> poly(x, 20, raw = T)4  -2.634e+04  5.480e+04  -0.481    0.632
> poly(x, 20, raw = T)5   2.370e+03  4.854e+03   0.488    0.627
> poly(x, 20, raw = T)6  -1.434e+02  2.906e+02  -0.493    0.623
> poly(x, 20, raw = T)7   6.022e+00  1.213e+01   0.496    0.621
> poly(x, 20, raw = T)8  -1.784e-01  3.587e-01  -0.497    0.620
> poly(x, 20, raw = T)9   3.727e-03  7.503e-03   0.497    0.621
> poly(x, 20, raw = T)10 -5.373e-05  1.086e-04  -0.495    0.622
> poly(x, 20, raw = T)11  5.016e-07  1.018e-06   0.493    0.624
> poly(x, 20, raw = T)12 -2.483e-09  5.069e-09  -0.490    0.626
> poly(x, 20, raw = T)13         NA         NA      NA       NA
> poly(x, 20, raw = T)14  5.656e-14  1.167e-13   0.485    0.629
> poly(x, 20, raw = T)15         NA         NA      NA       NA
> poly(x, 20, raw = T)16 -1.933e-18  4.011e-18  -0.482    0.631
> poly(x, 20, raw = T)17         NA         NA      NA       NA
> poly(x, 20, raw = T)18  5.181e-23  1.076e-22   0.482    0.631
> poly(x, 20, raw = T)19         NA         NA      NA       NA
> poly(x, 20, raw = T)20 -7.173e-28  1.480e-27  -0.485    0.629
>
> Residual standard error: 641000 on 76 degrees of freedom
> Multiple R-squared:  0.9982, Adjusted R-squared:  0.9979
> F-statistic:  2690 on 16 and 76 DF,  p-value: < 2.2e-16
>
>>
>
> Note,however, once the orthogonal polynomials have suggested a number, 4 in
> this case, I can do this :-
>
>   summary(lm(gdp ~ poly(x,4,raw=T)))
>
> Call:
> lm(formula = gdp ~ poly(x, 4, raw = T))
>
> Residuals:
>       Min       1Q   Median       3Q      Max
> -1278673  -424315   -22357   310977  1731813
>
> Coefficients:
>                         Estimate Std. Error t value Pr(>|t|)
> (Intercept)           3.022e+06  3.676e+05   8.220 1.64e-12 ***
> poly(x, 4, raw = T)1  1.741e+05  5.357e+04   3.249  0.00164 **
> poly(x, 4, raw = T)2 -6.434e+03  2.300e+03  -2.797  0.00633 **
> poly(x, 4, raw = T)3  1.878e+02  3.667e+01   5.123 1.76e-06 ***
> poly(x, 4, raw = T)4 -8.682e-01  1.935e-01  -4.486 2.19e-05 ***
> ---
> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
> Residual standard error: 663700 on 88 degrees of freedom
> Multiple R-squared:  0.9978, Adjusted R-squared:  0.9977
> F-statistic: 1.003e+04 on 4 and 88 DF,  p-value: < 2.2e-16
>
>>
>
> Although due to correlations they may not be significant jointly, but in
> this case all 4 powers come out significant.
>
>
> Second, the model formula gdp~1+x+x^2 and other similar formulas in your
>> message don't do what you think. Like + and *, the ^ operator has special
>> meaning on the right-hand side of an R model formula. See ?Formula and
>> perhaps read something about statistical models in R. For example:
>>
>>> x <- 1:93
>>> y <- 1 + x + x^2 + x^3 + x^4 + rnorm(93)
>>> (m <- lm(y ~ x + x^2))
>>
>> Call:
>> lm(formula = y ~ x + x^2)
>>
>> Coefficients:
>> (Intercept)            x
>>    -15781393       667147
>>
>> While gpp ~ x + I(x^2) would work, a better way to fit a raw quadratic is
>> as gdp ~ poly(x, 2, raw=TRUE), as I suggested in my earlier message.
>>
>
> My bad. Yes, I have some idea of the Wilkinson-Rogers notation. I have seen
> it in books, although it slipped my mind that I had to use I( ).
>
>
>> Finally, as to what you should do, I generally try to avoid statistical
>> consulting by email. If you can find competent statistical help locally,
>> such as at a nearby university, I'd recommend talking to someone about the
>> purpose of your research and the nature of your data. If that's not
>> possible, then others have suggested where you might find help, but to get
>> useful advice you'll have to provide much more information about your
>> research.
>>
>
> My original query was about the polynomials used by R which I think is ON
> topic. My apologies that this query turned into a statistics query.
>
>
>> Best,
>>   John
>>
>>    -----------------------------
>>    John Fox, Professor Emeritus
>>    McMaster University
>>    Hamilton, Ontario, Canada
>>    Web: http::/socserv.mcmaster.ca/jfox
>>
>>> On Nov 28, 2019, at 12:46 AM, Ashim Kapoor <ashimkapoor using gmail.com>
>> wrote:
>>>
>>> Dear Peter and John,
>>>
>>> Many thanks for your prompt replies.
>>>
>>> Here is what I was trying to do.  I was trying to build a statistical
>> model of a given time series using Box Jenkins methodology. The series has
>> 93 data points. Before I analyse the ACF and PACF, I am required to
>> de-trend the series. The series seems to have an upward trend. I wanted to
>> find out what order polynomial should I fit the series
>>> without overfitting.  For this I want to use orthogonal polynomials(I
>> think someone on the internet was talking about preventing overfitting by
>> using orthogonal polynomials) . This seems to me as a poor man's cross
>> validation.
>>>
>>> So my plan is to keep increasing the degree of the orthogonal
>> polynomials till the coefficient of the last orthogonal polynomial becomes
>> insignificant.
>>>
>>> Note : If I do NOT use orthogonal polynomials, I will overfit the data
>> set and I don't think that is a good way to detect the true order of the
>> polynomial.
>>>
>>> Also now that I have detrended the series and built an ARIMA model of
>> the residuals, now I want to forecast. For this I need to use the original
>> polynomials and their coefficients.
>>>
>>> I hope I was clear and that my methodology is ok.
>>>
>>> I have another query here :-
>>>
>>> Note : If I used cross-validation to determine the order of the
>> polynomial, I don't get a clear answer.
>>>
>>> See here :-
>>> library(boot)
>>> mydf = data.frame(cbind(gdp,x))
>>> d<-(c(
>>> cv.glm(data = mydf,glm(gdp~x),K=10)\$delta[1],
>>> cv.glm(data = mydf,glm(gdp~poly(x,2)),K=10)\$delta[1],
>>> cv.glm(data = mydf,glm(gdp~poly(x,3)),K=10)\$delta[1],
>>> cv.glm(data = mydf,glm(gdp~poly(x,4)),K=10)\$delta[1],
>>> cv.glm(data = mydf,glm(gdp~poly(x,5)),K=10)\$delta[1],
>>> cv.glm(data = mydf,glm(gdp~poly(x,6)),K=10)\$delta[1]))
>>> print(d)
>>> ## [1] 2.178574e+13 7.303031e+11 5.994783e+11 4.943586e+11 4.596648e+11
>>> ## [6] 4.980159e+11
>>>
>>> # Here it chooses 5. (but 4 and 5 are kind of similar).
>>>
>>>
>>> d1 <- (c(
>>> cv.glm(data = mydf,glm(gdp~1+x),K=10)\$delta[1],
>>> cv.glm(data = mydf,glm(gdp~1+x+x^2),K=10)\$delta[1],
>>> cv.glm(data = mydf,glm(gdp~1+x+x^2+x^3),K=10)\$delta[1],
>>> cv.glm(data = mydf,glm(gdp~1+x+x^2+x^3+x^4),K=10)\$delta[1],
>>> cv.glm(data = mydf,glm(gdp~1+x+x^2+x^3+x^4+x^5),K=10)\$delta[1],
>>> cv.glm(data = mydf,glm(gdp~1+x+x^2+x^3+x^4+x^5+x^6),K=10)\$delta[1]))
>>>
>>> print(d1)
>>> ## [1] 2.149647e+13 2.253999e+13 2.182175e+13 2.177170e+13 2.198675e+13
>>> ## [6] 2.145754e+13
>>>
>>> # here it chooses 1 or 6
>>>
>>> Query : Why does it choose 1? Notice : Is this just round off noise /
>> noise due to sampling error created by Cross Validation when it creates the
>> K folds? Is this due to the ill conditioned model matrix?
>>>
>>> Best Regards,
>>> Ashim.
>>>
>>>
>>>
>>>
>>>
>>> On Wed, Nov 27, 2019 at 10:41 PM Fox, John <jfox using mcmaster.ca> wrote:
>>> Dear Ashim,
>>>
>>> Orthogonal polynomials are used because they tend to produce more
>> accurate numerical computations, not because their coefficients are
>> interpretable, so I wonder why you're interested in the coefficients.
>>>
>>> The regressors produced are orthogonal to the constant regressor and are
>> orthogonal to each other (and in fact are orthonormal), as it's simple to
>> demonstrate:
>>>
>>> ------- snip -------
>>>
>>>> x <- 1:93
>>>> y <- 1 + x + x^2 + x^3 + x^4 + rnorm(93)
>>>> (m <- lm(y ~ poly(x, 4)))
>>>
>>> Call:
>>> lm(formula = y ~ poly(x, 4))
>>>
>>> Coefficients:
>>> (Intercept)  poly(x, 4)1  poly(x, 4)2  poly(x, 4)3  poly(x, 4)4
>>>     15574516    172715069     94769949     27683528      3429259
>>>
>>>> X <- model.matrix(m)
>>>> head(X)
>>>    (Intercept) poly(x, 4)1 poly(x, 4)2 poly(x, 4)3 poly(x, 4)4
>>> 1           1  -0.1776843   0.2245083  -0.2572066  0.27935949
>>> 2           1  -0.1738216   0.2098665  -0.2236579  0.21862917
>>> 3           1  -0.1699589   0.1955464  -0.1919525  0.16390514
>>> 4           1  -0.1660962   0.1815482  -0.1620496  0.11487597
>>> 5           1  -0.1622335   0.1678717  -0.1339080  0.07123722
>>> 6           1  -0.1583708   0.1545171  -0.1074869  0.03269145
>>>
>>>> zapsmall(crossprod(X))# X'X
>>>              (Intercept) poly(x, 4)1 poly(x, 4)2 poly(x, 4)3 poly(x, 4)4
>>> (Intercept)          93           0           0           0           0
>>> poly(x, 4)1           0           1           0           0           0
>>> poly(x, 4)2           0           0           1           0           0
>>> poly(x, 4)3           0           0           0           1           0
>>> poly(x, 4)4           0           0           0           0           1
>>>
>>> ------- snip -------
>>>
>>> If for some not immediately obvious reason you're interested in the
>> regression coefficients, why not just use a "raw" polynomial:
>>>
>>> ------- snip -------
>>>
>>>> (m1 <- lm(y ~ poly(x, 4, raw=TRUE)))
>>>
>>> Call:
>>> lm(formula = y ~ poly(x, 4, raw = TRUE))
>>>
>>> Coefficients:
>>>              (Intercept)  poly(x, 4, raw = TRUE)1  poly(x, 4, raw =
>> TRUE)2  poly(x, 4, raw = TRUE)3
>>>                   1.5640                   0.8985
>>   1.0037                   1.0000
>>> poly(x, 4, raw = TRUE)4
>>>                   1.0000
>>>
>>> ------- snip -------
>>>
>>> These coefficients are simply interpretable but the model matrix is more
>> poorly conditioned:
>>>
>>> ------- snip -------
>>>
>>>> head(X1)
>>>    (Intercept) poly(x, 4, raw = TRUE)1 poly(x, 4, raw = TRUE)2 poly(x, 4,
>> raw = TRUE)3
>>> 1           1                       1                       1
>>             1
>>> 2           1                       2                       4
>>             8
>>> 3           1                       3                       9
>>            27
>>> 4           1                       4                      16
>>            64
>>> 5           1                       5                      25
>>           125
>>> 6           1                       6                      36
>>           216
>>>    poly(x, 4, raw = TRUE)4
>>> 1                       1
>>> 2                      16
>>> 3                      81
>>> 4                     256
>>> 5                     625
>>> 6                    1296
>>>> round(cor(X1[, -1]), 2)
>>>                          poly(x, 4, raw = TRUE)1 poly(x, 4, raw = TRUE)2
>> poly(x, 4, raw = TRUE)3
>>> poly(x, 4, raw = TRUE)1                    1.00                    0.97
>>                    0.92
>>> poly(x, 4, raw = TRUE)2                    0.97                    1.00
>>                    0.99
>>> poly(x, 4, raw = TRUE)3                    0.92                    0.99
>>                    1.00
>>> poly(x, 4, raw = TRUE)4                    0.87                    0.96
>>                    0.99
>>>                          poly(x, 4, raw = TRUE)4
>>> poly(x, 4, raw = TRUE)1                    0.87
>>> poly(x, 4, raw = TRUE)2                    0.96
>>> poly(x, 4, raw = TRUE)3                    0.99
>>> poly(x, 4, raw = TRUE)4                    1.00
>>>
>>> ------- snip -------
>>>
>>> The two parametrizations are equivalent, however, in that they represent
>> the same regression surface, and so, e.g., produce the same fitted values:
>>>
>>> ------- snip -------
>>>
>>>> all.equal(fitted(m), fitted(m1))
>>> [1] TRUE
>>>
>>> ------- snip -------
>>>
>>> Because one is usually not interested in the individual coefficients of
>> a polynomial there usually isn't a reason to prefer one parametrization to
>> the other on the grounds of interpretability, so why do you need to
>> interpret the regression equation?
>>>
>>> I hope this helps,
>>>   John
>>>
>>>    -----------------------------
>>>    John Fox, Professor Emeritus
>>>    McMaster University
>>>    Hamilton, Ontario, Canada
>>>    Web: http::/socserv.mcmaster.ca/jfox
>>>
>>>> On Nov 27, 2019, at 10:17 AM, Ashim Kapoor <ashimkapoor using gmail.com>
>> wrote:
>>>>
>>>> Dear Petr,
>>>>
>>>> Many thanks for the quick response.
>>>>
>>>> I also read this:-
>>>> https://en.wikipedia.org/wiki/Discrete_orthogonal_polynomials
>>>>
>>>> Also I read  in ?poly:-
>>>>      The orthogonal polynomial is summarized by the coefficients, which
>>>>      can be used to evaluate it via the three-term recursion given in
>>>>      Kennedy & Gentle (1980, pp. 343-4), and used in the ‘predict’ part
>>>>      of the code.
>>>>
>>>> I don't have access to the mentioned book.
>>>>
>>>> Out of curiosity, what is the name of the discrete orthogonal
>> polynomial
>>>> used by R ?
>>>> What discrete measure is it orthogonal with respect to ?
>>>>
>>>> Many thanks,
>>>> Ashim
>>>>
>>>>
>>>>
>>>>
>>>> On Wed, Nov 27, 2019 at 6:11 PM PIKAL Petr <petr.pikal using precheza.cz>
>> wrote:
>>>>
>>>>> You could get answer quickly by searching net.
>>>>>
>>>>>
>>>>>
>> https://stackoverflow.com/questions/39031172/how-poly-generates-orthogonal-p
>>>>> olynomials-how-to-understand-the-coefs-ret/39051154#39051154
>>>>> <
>> https://stackoverflow.com/questions/39031172/how-poly-generates-orthogonal-polynomials-how-to-understand-the-coefs-ret/39051154#39051154
>>>
>>>>>
>>>>> Cheers
>>>>> Petr
>>>>>
>>>>>> -----Original Message-----
>>>>>> From: R-help <r-help-bounces using r-project.org> On Behalf Of Ashim
>> Kapoor
>>>>>> Sent: Wednesday, November 27, 2019 12:55 PM
>>>>>> To: R Help <r-help using r-project.org>
>>>>>> Subject: [R] Orthogonal polynomials used by R
>>>>>>
>>>>>> Dear All,
>>>>>>
>>>>>> I have created a time trend by doing x<-1:93 because I have a time
>> series
>>>>>> with 93 data points. Next I did :-
>>>>>>
>>>>>> y = lm(series ~ poly(x,4))\$residuals
>>>>>>
>>>>>> to detrend series.
>>>>>>
>>>>>> I choose this 4 as the order of my polynomial using cross validation/
>>>>>> checking the absence of trend in the residuals so I think I have not
>>>>> overfit
>>>>>> this series.
>>>>>>
>>>>>> I wish to document the formula of poly(x,4). I am not able to find
>> it in
>>>>> ?poly
>>>>>>
>>>>>> Can someone please tell me what the formula for the orthogonal
>>>>>> polynomial used by R is ?
>>>>>>
>>>>>> Thank you,
>>>>>> Ashim
>>>>>>
>>>>>>       [[alternative HTML version deleted]]
>>>>>>
>>>>>> ______________________________________________
>>>>>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>>>>> 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.
>>>>>
>>>>
>>>>        [[alternative HTML version deleted]]
>>>>
>>>> ______________________________________________
>>>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>>> 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.
>>>
>>
>>
>>
>
> 	[[alternative HTML version deleted]]
>
> ______________________________________________
> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.
>

--
Michael
http://www.dewey.myzen.co.uk/home.html

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