2
$\begingroup$

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

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

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

What happens when you write $TSS-ESS=RSS$? What algebra can you use from this re-writing to show $$ R^2=1-\frac{||\hat{\epsilon}||^2}{(\hat{y}-y)^2}?$$

Improve this answer
$\endgroup$
5
  • $\begingroup$ okay, should I recognize a binomial expansion from $TSS-ESS=RSS$? such as $\sum\limits_{i=1}^{n}(y_i-\bar y)^2-\sum\limits_{i=1}^{n}(\hat y_i-\bar y)^2 =(\sum\limits_{i=1}^{n}(y_i-\bar y)-\sum\limits_{i=1}^{n}(\hat y_i-\bar y))(\sum\limits_{i=1}^{n}(y_i-\bar y)+\sum\limits_{i=1}^{n}(\hat y_i-\bar y))$? $\endgroup$
    – Revolucion for Monica
    Commented Oct 19, 2015 at 17:36
  • $\begingroup$ Ja, I do $\frac{TSS-ESS}{TSS}=\frac{RSS}{TSS}$, I get $1-\frac{ESS}{TSS}=\frac{RSS}{TSS}$ ... $\endgroup$
    – Revolucion for Monica
    Commented Oct 19, 2015 at 17:47
  • $\begingroup$ I'm missing the addition step... The definition I had for $\rho_{y,\epsilon}$ was $$\frac{\sum (y_i -\bar y)(\epsilon_i-\bar\epsilon)}{n * var(x_i)var(\epsilon_i)}$$ $\endgroup$
    – Revolucion for Monica
    Commented Oct 19, 2015 at 17:59
  • $\begingroup$ I have a better understanding of where you are stuck now. When I first composed my answer, I thought you were stuck earlier on in the equation starting with $R^2$. I've got to run now, but if I have time I'll work out the steps. In the meantime, I'm pretty sure this question is answered elsewhere on this site. Perhaps you could try the archives. $\endgroup$
    – Sycorax ♦
    Commented Oct 19, 2015 at 18:07
  • $\begingroup$ Sure, I've already looked if it wouldn't it duplicated but I will try again and try also by myself, thanks for your help nonetheless. $\endgroup$
    – Revolucion for Monica
    Commented Oct 19, 2015 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.