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$y_i=\beta_0+\beta_1x_i+\varepsilon $ is a simple linear model.

We know $R^2=1-\frac{SSE/(n-k-1)}{SST/(n-1)} $ where

$SSE=Y'(I-H)Y=\sum(y_i-\hat{y})^2$

and $SST=Y'(I-P_1)Y=\sum(y_i-\bar{y})^2$

(for $H=X(X'X)^{-1}X'$ and $P_1=\frac{1}{n}\mathbf{11'}$ and $\mathbf{1}=(1,\cdots,1)_{1\times n}'$, $Y=(y_1,y_2,\cdots,y_n)'$,$X=\begin{bmatrix} 1 & x_{11}\\ 1& x_{21}\\ \vdots&\vdots \\ 1&x_{n1 } \end{bmatrix}$)

Define $r=\frac{\sum (y_i-\bar{y})(x_i-\bar{x})}{\sqrt{\sum (y_i-\bar{y})^2\sum (x_i-\bar{x} )^2}}$.

Show $R^2=r^2$

Thanks in advance.

Ian
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    Please show us your thoughts about tackling this problem and share with us the progress we have made, so we can focus on what will help you understand this material. – whuber Sep 16 '13 at 22:14
  • This does not answer your question 100%, but it should give you enough hints to get started on it: http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient#Pearson.27s_correlation_and_least_squares_regression_analysis –  Sep 16 '13 at 22:28
  • So your formula for $r$ is the general formula for a correlation, but are you sure that what you want is $\bar x$ in there (this is probably where you trouble lies)? –  Sep 17 '13 at 00:41
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    The question seems to be using the *adjusted* $R^2$, which won't equal the square of $r$. – Silverfish Jan 12 '15 at 22:47
  • [This question](http://stats.stackexchange.com/questions/99669/the-equivalence-of-sample-correlation-and-r-statistic-for-simple-linear-regressi) also asks why $R^2=r^2$. – Silverfish Jan 13 '15 at 16:32

1 Answers1

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This answer will be written in a several regressors setting. You can directly apply it to the your simple case of only one regressor.

Let $N=I_n-\frac{1}{n}\mathbf{1}\mathbf{1}'$

$R^2=\frac{SSE}{SST}=\frac{(Xb)'NXb}{Y'NY}$

The square of the sample correlation is $\displaystyle r^2(\hat{Y},Y)=\frac{\left(\sum(\hat{y_i}-\bar{\hat{y}})(y_i-\bar{y})\right)^2}{ SST\cdot SSE}=\frac{(b'X'NY)'b'X'NY}{(Xb)'NXb \cdot Y'NY}$

Note: $N$ is symmetric and idempotent and $Ne=e$, and $X'e=0$.

$b'X'NY=b'X'NXb+b'X'Ne=b'X'NXb$.

So $r^2(\hat{Y},Y)=\frac{(b'X'NXb)'b'X'NXb}{(Xb)'NXb \cdot Y'NY}=\frac{(Xb)'NXb}{Y'NY}=R^2$

An old man in the sea.
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  • This is not the formula for $r$ invoked in the question: how exactly are your formula and the OP's formula related? – whuber Nov 30 '14 at 02:23
  • @whuber As soon as I get some free time, I'll answer your question. – An old man in the sea. Dec 01 '14 at 15:08
  • @whuber is it clearer now? And by the way, why is this being marked as a duplicate when the mod's don't provide a link of the question from which this one would be considered a duplicate of? – An old man in the sea. Jan 14 '15 at 15:16
  • (1) I think there's still a gap: your formula for $r^2$ still differs from that used in the question. (2) The link to the duplicate is given at the top in a highlighted box beginning "This question already has an answer here:". (3) No moderators were involved in closing this thread: five high-reputation community members closed it. – whuber Jan 14 '15 at 15:45