07_RegressionModels/01_04_rttm/index.html

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        <h1>Historical side note, Regression to Mediocrity</h1>
        <h2>Regression to the mean</h2>
        <p>Brian Caffo, Jeff Leek, Roger Peng PhD<br/>Johns Hopkins Bloomberg School of Public Health</p>
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    <h2>A historically famous idea, Regression to the Mean</h2>
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    <ul>
<li>Why is it that the children of tall parents tend to be tall, but not as tall as their parents? </li>
<li>Why do children of short parents tend to be short, but not as short as their parents?</li>
<li>Why do parents of very short children, tend to be short, but not a short as their child? And the same with parents of very tall children?</li>
<li>Why do the best performing athletes this year tend to do a little worse the following?</li>
</ul>

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    <h2>Regression to the mean</h2>
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    <ul>
<li>These phenomena are all examples of so-called regression to the mean</li>
<li>Invented by Francis Galton in the paper &quot;Regression towvards mediocrity in hereditary stature&quot; The Journal of the Anthropological Institute of Great Britain and Ireland , Vol. 15, (1886).</li>
<li>Think of it this way, imagine if you simulated pairs of random normals

<ul>
<li>The largest first ones would be the largest by chance, and the probability that there are smaller for the second simulation is high.</li>
<li>In other words  \(P(Y < x | X = x)\) gets bigger as \(x\) heads into the very large values.</li>
<li>Similarly \(P(Y > x | X = x)\) gets bigger as \(x\) heads to very small values.</li>
</ul></li>
<li>Think of the regression line as the intrisic part.

<ul>
<li>Unless \(Cor(Y, X) = 1\) the intrinsic part isn&#39;t perfect</li>
</ul></li>
</ul>

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    <h2>Regression to the mean</h2>
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    <ul>
<li>Suppose that we normalize \(X\) (child&#39;s height) and \(Y\) (parent&#39;s height) so that they both have mean 0 and variance 1. </li>
<li>Then, recall, our regression line passes through \((0, 0)\) (the mean of the X and Y).</li>
<li>If the slope of the regression line is \(Cor(Y,X)\), regardless of which variable is the outcome (recall, both standard deviations are 1).</li>
<li>Notice if \(X\) is the outcome and you create a plot where \(X\) is the horizontal axis, the slope of the least squares line that you plot is \(1/Cor(Y, X)\). </li>
</ul>

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    <h2>Normalizing the data and setting plotting parameters</h2>
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    <pre><code class="r">library(UsingR)
data(father.son)
y &lt;- (father.son$sheight - mean(father.son$sheight)) / sd(father.son$sheight)
x &lt;- (father.son$fheight - mean(father.son$fheight)) / sd(father.son$fheight)
rho &lt;- cor(x, y)
myPlot &lt;- function(x, y) {
  plot(x, y, 
       xlab = &quot;Father&#39;s height, normalized&quot;,
       ylab = &quot;Son&#39;s height, normalized&quot;,
       xlim = c(-3, 3), ylim = c(-3, 3),
       bg = &quot;lightblue&quot;, col = &quot;black&quot;, cex = 1.1, pch = 21, 
       frame = FALSE)
}
</code></pre>

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    <h2>Plot the data, code</h2>
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    <pre><code>myPlot(x, y)
abline(0, 1) # if there were perfect correlation
abline(0, rho, lwd = 2) # father predicts son
abline(0, 1 / rho, lwd = 2) # son predicts father, son on vertical axis
abline(h = 0); abline(v = 0) # reference lines for no relathionship
</code></pre>

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    <h2>Plot the data, results</h2>
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    <div class="rimage center"><img src="fig/unnamed-chunk-2.png" title="plot of chunk unnamed-chunk-2" alt="plot of chunk unnamed-chunk-2" class="plot" /></div>

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    <h2>Discussion</h2>
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    <ul>
<li>If you had to predict a son&#39;s normalized height, it would be
\(Cor(Y, X) * X_i\) </li>
<li>If you had to predict a father&#39;s normalized height, it would be
\(Cor(Y, X) * Y_i\)</li>
<li>Multiplication by this correlation shrinks toward 0 (regression toward the mean)</li>
<li>If the correlation is 1 there is no regression to the mean (if father&#39;s height perfectly determine&#39;s child&#39;s height and vice versa)</li>
<li>Note, regression to the mean has been thought about quite a bit and generalized </li>
</ul>

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