lecture 10.1.key

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Transforms Revisited

Transforms are used to

Change the mean function so that it islinear.

Adjust for non-constant variance problem Fix non-Normal residuals Although you won't always solve all three

problems (or any problem for that matter.)

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Youve already studied

log transforms and square-root transforms

Now

were going to consider a more generalclass of transforms and discuss strategiesfor finding the best transform

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Strategy

First, transform Y. If that doesn't work, transform the

predictors, but not Y.

If that improves things but not perfectly, seeif you can now transform Y.

There are also approaches that considertransforming ALL variables simultaneously.

Keep in mind

Don't remove outliers, influential points,etc. until the transforming is done. These points might not really be so outlying

once the transform is done.

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Keep in Mind

Simple is better than complicated If you are expected to interpret the

parameters, then transformations mightmake this impossible.

Transform Y

E(Y|X)6=0+ 1x1+ . . .+ pxp

Basic idea: What if

but instead:

E(Y|X) = g(0+ 1x1+ . . . + pxp)

so we need to discover g()

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E(Y|X) = g(0+ 1x1+ . . . + pxp)

if we knew g(), we could invert it:

g1(E(Y|X)) = g1(g(0+ 1x1+ . . . + pxp))

Ynew = 0 + 1x1 + . . .+ pxp

Transform Y: 2 approaches

Inverse Response Plots Box-Cox Method

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Inverse Response Plots

a technique for guessing g()

If the predictors have an elliptically symmetricdistribution (so joint Normal is one example of this), then

plot y-hat against y.

The shape of the resulting curve gives you an idea as to the

shape of g inverse.

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> m1=lm(ozone~temperature+pressure,data=ozonetext)> plot(m1)

A plot of the predictors show that their joint distributionis roughly elliptical.

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> library(alr3)

> invResPlot(m2)

lambda RSS1 0.3658881 1989.771

2 -1.0000000 3412.9123 0.0000000 2082.3774 1.0000000 2196.992

Suggests that the best transform is Ynew = Y0.365881

(lambda=0 refers to the log transform)

Note log transform isnt to different from optimal

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> ozone.t1=transform(ozonetext,ozone.t = ozone^(.37) )> m2=lm(ozone.t~temperature+pressure,data=ozone.t1)> plot(m2) transformed

original

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transform

original

transformed

original

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On the whole, the transformationimproved the validity of the model.

But interpretation may now be quitedifficult.

Still, improved validity means we bettertrust p-values and confidence intervals andprediction intervals.

> summary(m2)

Call:lm(formula = ozone.t ~ temperature + pressure, data = ozone.t1)

Coefficients: Estimate Std. Error t value Pr(>|t|)(Intercept) -0.4004629 0.1774149 -2.257 0.0256 *temperature 0.0423812 0.0027663 15.321

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Another approach:

(Useful when the distribution of the variable to betransformed is not Normal.)

Box-Cox

Choose a transform of Y, (Y)

such that distribution of Y is closer to Normalwhere

(Y) = gm(Y)1(Y 1)/

(Y) =gm(Y)log(Y) for = 0

(gmis the geometric mean)

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(Y) = gm(Y)1(Y 1)/

gm(Y) is the geometric mean of y =

ni=1Y1/ni

To find lambda....

maximum likelihood estimation of lambda.

> library(MASS)

> boxcox(m1)

or

> library(alr3)

>summary(powerTransform(y~x1+x2,data=))

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which confirmsour previous

transformationusing lambda = .37

> boxcox(m1)

> summary(powerTransform(m1))

bcPower Transformation to Normality

Est.Power Std.Err. Wald Lower Bound Wald Upper Bound

Y1 0.2343 0.0866 0.0646 0.4041

Likelihood ratio tests about transformation parameters

LRT df pval

LR test, lambda = (0) 7.568201 1 5.940706e-03

LR test, lambda = (1) 66.558671 1 3.330669e-16

In fact, optimal transform is .23, which is smaller thanprevious .37. However, .37 is within the confidence interval

of 0.0646 to 0.4041

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Likelihood ratio tests about transformation parameters

LRT df pval

LR test, lambda = (0) 7.568201 1 5.940706e-03

LR test, lambda = (1) 66.558671 1 3.330669e-16

Null: lambda=0Alt: lambda 0

Small p-value, so we reject.Thus, it is best to notdo a

log transform.

Null: no transform (lambda=1)Alt: do a transform

Reject. We need a transform.

Transform Predictors

You can use BoxCox to transformpredictors when Y is NOT transformed

Then, if necessary, use inverse responseplot to transform Y

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In this approach, we find a transformationthat makes the joint distribution of all the

predictorsmultivariate Normal.

(or as close to it as we can get)

once thats done, we try to find atransform for Y.

Then we see if it helps.

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o these predictors look like they come from a Normaldistribution?

(probably not)

>library(alr3)

> summary(powerTransform(ozone~temperature+height,data=o2.mini))

box.cox Transformations to Multinormality

Est.Power Std.Err. Wald(Power=0) Wald(Power=1)

temperature 1.1383 0.3246 3.5070 0.426height 18.9126 4.5176 4.1864 3.965

LRT df p.valueLR test, all lambda equal 0 25.50600 2 2.893633e-06

LR test, all lambda equal 1 17.30179 2 1.749703e-04

Best lambda could be within two Std. Errors of Estimated.

For temp, use a lambda between 0.5 to 1.7, roundinggenerously.

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> summary(powerTransform(cbind(o2.mini$temperature, o2.mini

$height,data=o2.mini)~1)

box.cox Transformations to Multinormality

Est.Power Std.Err. Wald(Power=0) Wald(Power=1)

temperature 1.1383 0.3246 3.5070 0.426height 18.9126 4.5176 4.1864 3.965


LR test, all lambda equal 1 17.30179 2 1.749703e-04

Temp: try square-root transform or no transform

Height: Transform to a high power, which is very unusual

and probably not helpful. But let's try the 20th poweranyways.

> o2.minit=transform(o2.mini,temp.t = sqrt(temperature),height.t =height^20)> plot(o2.minit)

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> o2.minit=transform(o2.mini,temp.t = sqrt(temperature),height.t = height^20)> plot(o2.minit)

residuals: no transform

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not much better, so look at transforming

Y

transformedpredictors > m.t1 = lm(ozone~temp.t+height.t,data=o2.minit)

> plot(m.t1)

> invResPlot(m.t1)

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once again,

Y Y

1/3

looks best.> o2.minit2 = transform(o2.minit,ozone.t =ozone^(1/3))> m.t2 = lm(ozone.t~temp.t

+height.t,data=o2.minit2)

> plot(m.t2)

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A third approach is to use boxcox totransform the predictors and the response

simultaneously

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Use BoxCox to transform ALL at once.>

summary(powerTransform(with(o2.mini,cbind(ozone,height,temperature))

)box.cox Transformations to Multinormality

Est.Power Std.Err. Wald(Power=0) Wald(Power=1)ozone 0.2503 0.0888 2.8178 -8.4416

height 18.8959 4.4542 4.2422 4.0177

temperature 1.1590 0.2661 4.3550 0.5976


LR test, all lambda equal 1 83.53574 3 0.000000e+00

This is consistent with the 1/3 power of ozone, a 20th power forheight, and no change (raise to the 1 power) for temp.

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2 (p+ 1)/n= 2 3/141 = 0.04 = "big" leverage

lecture 10.1.key

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