# Why is $R^2=\rho^2$

Considering $y_i=\beta_1+\beta_2x_i+\epsilon_i$

$\bar y_i=\hat\beta_1+\hat\beta_2\bar x_i+\bar\epsilon_i$

a linear equation of least square used when it seems that there is a like between two data, $\epsilon_i$ is the noise, usually $\epsilon$~$N(0,\sigma)$

$\beta_i$ are constant we find by the least square regression method.

$\hat\beta_i$ are estimated values of those $\beta_i$

Having $$||y-\bar y||^2=||\hat y-\bar y||^2+ ||\epsilon||^2$$

$$\underbrace{\sum\limits_{i=1}^{n}(y_i-\bar y)^2}_{TSS}=\underbrace{\sum\limits_{i=1}^{n}(\hat y_i-\bar y)^2}_{ESS} +\underbrace{\sum\limits_{i=1}^{n}(\hat\epsilon_i^2)}_{RSS}$$

I have to show that $R^2=\frac{ESS}{TSS}=\frac{(\hat y-\bar y)^2}{(y_-\bar y)^2}=1-\frac{||\hat \epsilon||^2}{(\hat y-\bar y)^2}=\underbrace{1-\frac{RSS}{TSS}=\rho_{xy}^2}_{\text{The step I don't get}}$

with

$$\rho = \rho_{xy} =\frac{\sum ^n _{i=1}(x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum ^n _{i=1}(x_i - \bar{x})^2} \sqrt{\sum ^n _{i=1}(y_i - \bar{y})^2}}$$

I don't even know where to start... Any hint appreciate.

• were does $x$ in the formula for $\rho$ come from?
– Max
Oct 20, 2015 at 17:31
• Without telling us where $x$ comes from, this might be hard to answer...
– 5xum
Oct 20, 2015 at 17:33
• just please $\hat{\epsilon}$ differs from $\epsilon$ ? Oct 20, 2015 at 18:42

It's just a calculation with using a lot of definitions from the linear regression. Let's start with the $R^2$ (I use $\sum_{i=1}^n = \sum$):

$$R^2 = \frac{ESS}{TSS} = \frac{\sum (\hat{y}_i - \bar{y})^2}{\sum(y_i - \bar{y})^2}$$

With the sample variance

\begin{align*} &\hat{\sigma}^2_y = \frac{1}{n-1} \sum (y_i - \bar{y})^2 \\ \Leftrightarrow\ &\hat{\sigma}^2_y (n-1) = \sum (y_i - \bar{y})^2 \qquad\qquad (1) \end{align*}

you get

$$R^2 = \frac{\sum (\hat{y}_i - \bar{y})^2}{\sum(y_i - \bar{y})^2} = \frac{\sum (\hat{y}_i - \bar{y})^2}{\hat{\sigma}^2_y (n-1)} \qquad\qquad (2)$$

Keep this equation in mind! Now we take a look at the estimator for $\beta_2$ in our linear regression. With some calculations you can see that the minimum quadrat estimator $\hat{\beta}_2$ is:

$$\hat{\beta}_2 = \frac{\sum(x_i - \bar{x})y_i}{\sum(x_i - \bar{x})^2}$$

With $(1)$ (just for the sample x) we get:

$$\hat{\beta}_2 = \frac{\sum(x_i - \bar{x})y_i}{\sum(x_i - \bar{x})^2} \overset{(1)}{=} \frac{\sum(x_i - \bar{x})y_i}{\hat{\sigma}^2_x(n-1)} \qquad\qquad (3)$$

Now we take a closer look at the pearson correlation coefficient $\rho_{xy}$:

\begin{align*} \rho_{xy} &= \frac{\sum(x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum(x_i - \bar{x})^2}\sqrt{\sum(y_i - \bar{y})^2}} = \\ \\ &= \frac{\sum(x_i - \bar{x})(y_i - \bar{y})}{\frac{n-1}{n-1}\sqrt{\sum(x_i - \bar{x})^2}\sqrt{\sum(y_i - \bar{y})^2}} = \\ \\ &= \frac{\sum(x_i - \bar{x})(y_i - \bar{y})}{(n-1)\sqrt{\frac{1}{n-1}\sum(x_i - \bar{x})^2}\sqrt{\frac{1}{n-1}\sum(y_i - \bar{y})^2}} = \\ \\ &= \frac{\sum(x_i - \bar{x})(y_i - \bar{y})}{(n-1)\hat{\sigma}_x\hat{\sigma}_y} \overset{on\ your\ own}{=} \frac{\sum(x_i - \bar{x})y_i}{(n-1)\hat{\sigma}_x\hat{\sigma}_y} \end{align*}

With this you get:

$$\rho_{xy} (n-1)\hat{\sigma}_x\hat{\sigma}_y = \sum(x_i - \bar{x})y_i \qquad\qquad (4)$$

Now we can put $(4)$ in $(3)$ and get: $$\hat{\beta}_2 = \frac{\rho_{xy} (n-1)\hat{\sigma}_x\hat{\sigma}_y}{\hat{\sigma}^2_x (n-1)} = \frac{\rho_{xy} \hat{\sigma}_y}{\hat{\sigma}_x} \qquad\qquad (5)$$

Now we use $\hat{y}_i = \hat{\beta}_1 + \hat{\beta}_2x_i\ \ (i)$ and $\hat{\beta}_1 = \bar{y}-\hat{\beta}_2\bar{x} \ \ (ii)$ and can finally compute by using equation $(2)$:

\begin{align*} R^2 &= \frac{\sum (\hat{y}_i - \bar{y})^2}{\hat{\sigma}^2_y (n-1)} \overset{(i)}{=} \frac{\sum (\hat{\beta}_1 + \hat{\beta}_2x_i - \bar{y})^2}{\hat{\sigma}^2_y (n-1)} \overset{(ii)}{=} \frac{\sum (\bar{y}-\hat{\beta}_2\bar{x} + \hat{\beta}_2x_i - \bar{y})^2}{\hat{\sigma}^2_y (n-1)} = \\ \\ &= \frac{\sum (\hat{\beta}_2 (x_i-\bar{x}))^2}{\hat{\sigma}^2_y (n-1)} = \frac{\hat{\beta}_2^2\overbrace{\sum (x_i-\bar{x})^2}^{=(n-1)\hat{\sigma}_x^2\ (1)}}{\hat{\sigma}^2_y (n-1)} = \hat{\beta}_2^2\frac{(n-1)\hat{\sigma}_x^2}{\hat{\sigma}^2_y (n-1)} = \\ \\ &\overset{(5)}{=}\left(\frac{\rho_{xy} \hat{\sigma}_y}{\hat{\sigma}_x}\right)^2\frac{\hat{\sigma}_x^2}{\hat{\sigma}_y^2} = \rho_{xy}^2 \left(\frac{\hat{\sigma}_y}{\hat{\sigma}_x}\right)^2\frac{\hat{\sigma}_x^2}{\hat{\sigma}_y^2} = \rho_{xy}^2 \frac{\hat{\sigma}_y^2}{\hat{\sigma}_x^2}\frac{\hat{\sigma}_x^2}{\hat{\sigma}_y^2} = \rho_{xy}^2 \end{align*}

Hope you get it with that. :)

• I know that according to the policy of this site (and the Habeas corpus and the Magna Carta... ;) ) I shouldn't say that but this is awsooooooooooooooooooooome! I think I'm your best fan until the end of your life! Because I have to ask a question to comment and because I want to get on: how did you had the intuition of the proof? Do you advise me to try it again on my own? Oct 21, 2015 at 8:59
• It was a proposition in my statistic course, but our professor didn't proof this, he just says the clue is to use the sample variance. I picked out this one as an exercise for the test and put in equation (1) everywhere it's possible. After very much calculations and desperate hours I get it. In my opinion it's more important to understand the interpretation behind this indicators. You can try it again with the tip to use the sample variance. But don't investigate to much time in it, there are more important things to deal with in the linear regression (resp. in the linear model). Oct 21, 2015 at 10:05
• The teacher just gaved me the answer today! The proof may interest you, it's a bit shorter! thanks again for the time you spent on this one! Nov 5, 2015 at 12:16

Here is the proof my teacher gaved me today, shorter than the very good one from Daniel.

\begin{align*} R^2&=\frac{ESS}{TSS}=\frac{\sum(\hat y -\bar y)^2}{\sum (y_i-\bar y)^2}\\ &=\frac{\sum (\hat\beta_1+\hat \beta_2x_i-\bar y)^2}{\sum(y_i-\bar y)^2}\\ &=\frac{\sum (\bar y- \hat \beta_2\bar x+\hat \beta_2x_i-\bar y)^2}{\sum(y_i-\bar y)^2}\\ &=\frac{\hat\beta_2\sum(x_i-\bar x)^2}{\sum(y_i-\bar y)^2}\\ &=\frac{cov(x,y)^2*var(x)}{var(x)^2*var(y)}\\ &=\frac{cov(x,y)^2}{var(x)*var(y)}\\ &=\rho_{xy}^2 \end{align*}