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POPolynomial Regression nonsense Predictions
 

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  1. POPolynomial Regression nonsense Predictions
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    <p>Suppose I want to fit a linear regression model with degree two (orthogonal) polynomial and then predict the response. Here are the codes for the first model (m1)</p> <pre><code>x=1:100 y=-2+3*x-5*x^2+rnorm(100) m1=lm(y~poly(x,2)) prd.1=predict(m1,newdata=data.frame(x=105:110)) </code></pre> <p>Now let's try the same model but instead of using $poly(x,2)$, I will use its columns like:</p> <pre><code>m2=lm(y~poly(x,2)[,1]+poly(x,2)[,2]) prd.2=predict(m2,newdata=data.frame(x=105:110)) </code></pre> <p>Let's look at the summaries of m1 and m2.</p> <pre><code>&gt; summary(m1) Call: lm(formula = y ~ poly(x, 2)) Residuals: Min 1Q Median 3Q Max -2.50347 -0.48752 -0.07085 0.53624 2.96516 Coefficients: Estimate Std. Error t value Pr(&gt;|t|) (Intercept) -1.677e+04 9.912e-02 -169168 &lt;2e-16 *** poly(x, 2)1 -1.449e+05 9.912e-01 -146195 &lt;2e-16 *** poly(x, 2)2 -3.726e+04 9.912e-01 -37588 &lt;2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.9912 on 97 degrees of freedom Multiple R-squared: 1, Adjusted R-squared: 1 F-statistic: 1.139e+10 on 2 and 97 DF, p-value: &lt; 2.2e-16 &gt; summary(m2) Call: lm(formula = y ~ poly(x, 2)[, 1] + poly(x, 2)[, 2]) Residuals: Min 1Q Median 3Q Max -2.50347 -0.48752 -0.07085 0.53624 2.96516 Coefficients: Estimate Std. Error t value Pr(&gt;|t|) (Intercept) -1.677e+04 9.912e-02 -169168 &lt;2e-16 *** poly(x, 2)[, 1] -1.449e+05 9.912e-01 -146195 &lt;2e-16 *** poly(x, 2)[, 2] -3.726e+04 9.912e-01 -37588 &lt;2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.9912 on 97 degrees of freedom Multiple R-squared: 1, Adjusted R-squared: 1 F-statistic: 1.139e+10 on 2 and 97 DF, p-value: &lt; 2.2e-16 </code></pre> <p>So m1 and m2 are basically the same. Now let's look at the predictions prd.1 and prd.2</p> <pre><code>&gt; prd.1 1 2 3 4 5 6 -54811.60 -55863.58 -56925.56 -57997.54 -59079.52 -60171.50 &gt; prd.2 1 2 3 4 5 6 49505.92 39256.72 16812.28 -17827.42 -64662.35 -123692.53 </code></pre> <p>Q1: Why prd.2 is significantly different from prd.1? </p> <p>Q2: How can I obtain prd.1 using the model m2?</p>
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    1. CONot an answer, but sufficiently high values of R-squared (0.99 something) always freak me out...
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    2. COThat's not a problem at all. We can change $y$ with something like $y=-2+3*x-5*x^2+x^5+rnorm(100,15)$ and R-squared reduces to 95%, but the problem remains for the predictions.
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    3. COThe first model's results look like an ill-conditioned matrix somewhere. The predictions simply follow from the nonsense coefficients estimated by the first model.
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