I mean it is written in a book "Statistics for Management and Economics", that coefficient of determination is coefficient of correlation squared. Well, am I the only one to whom this is surprising fact as he expected something more clear or natural?? I mean, if someone can present me the proof why, or why some other, more natural things do not work, like I don't know, absolute value of the coefficient of correlation or something similar to Chebyshev theorem ($1-\text{coefficient of correlation}$)?
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1The title of the question is very confusing, could you please edit it so it is more clear? – Yuriy S Jan 17 '19 at 11:35
1 Answers
Suppose you have $n$ paired observations $(x_i,y_i)$ on $(x,y)$ and you want to predict $y$ on the basis of $x$.
You consider the prediction model $$y=\phi(x)+e$$ where $\phi$ is the part of $y$ explained by $x$ through $\phi$ and $e$ is the unexplained part.
Suppose you see from the scatter plot of $x$ and $y$ that $\phi$ is more or less linear.
So you choose $$\phi(x)=a+bx$$
Then the least square linear predictor of $y$ obtained on the basis of $x$ is $$\hat y=\hat a+\hat b x$$, where $$\hat a=\bar y-\hat b\bar x\quad,\quad \hat b=\frac{\operatorname{cov}(x,y)}{\operatorname{var}(x)}$$
It can be shown that
\begin{align} \operatorname{var}(\hat y)&=\operatorname{var}(\hat a+\hat b x) \\&=\hat b^2\operatorname{var}(x) \\&=r^2 \operatorname{var}(y) \end{align}
, where $r$ is the correlation coefficient between $x$ and $y$.
A measure of efficacy of the predictor $\phi$ is given by the proportion of variation in $y$ explained by $\phi$, i.e., $$\frac{\operatorname{var}(\hat y)}{\operatorname{var}(y)}=r^2$$, which is termed as coefficient of determination.
Of course, the coefficient of determination is numerically equal to the square of the correlation coefficient, but that is hardly a definition or a motivation for the former.
When $r^2=0$ the linear prediction of $y$ obtained on the basis of $x$ is worst, and when $r^2=1$ the prediction is perfect as $\phi$ explains the variability in $y$ completely.
For more details, the following threads might be helpful:
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