# Given a predictor $x$. under what circumstance would you have high $R^2$ but low $\beta$

Assume I have a time series $$y$$ and a predictor $$x$$. Let's say they are both centered at zero.

$$R^2 = 1 - \frac{ \sum (y_i - x_i)^2 }{\sum y_i^2}$$

Now I run a new regression $$y \sim \beta x$$, in an attempt to "rescale" my predictor. $$\beta = \frac{\operatorname{cov}(x,y)}{\operatorname{var}(x)} = \frac{\sum x_i y_i}{\sum x_i^2}$$

I want to find a relationship between $$\beta$$ and $$R^2$$ (note this $$R^2$$ is what I computed above $$R^2 = 1 - \frac{ \sum (y_i - x_i)^2 }{\sum y_i^2}$$, not $$R^2$$ of my new regression that produced $$\beta$$, $$x$$ is already a predictor )

My hypothesis is that $$R^2$$ is maximized when $$\beta = 1$$. Because that means my predictor $$x$$ is "well-scaled". But I have trouble proving it... if that's not true, why would $$R^2$$ not maximize when my predictor is "best scaled"?

$$\sum x_i y_i = \beta \sum x_i^2,$$

plug into \begin{align} R^2 & = 1 - \frac{ \sum (y_i - x_i)^2 }{\sum y_i^2} \\[6pt] & = 1 - \frac{ \sum x_i^2 - 2\sum x_i y_i + \sum y_i^2 }{\sum y_i^2} \\[6pt] & = \frac{ 2 \sum x_i y_i - \sum x_i^2 }{\sum y^2_i} = (2\beta-1) \frac{ \sum x_i^2 }{ \sum y_i^2}. \end{align}

Nothing suggests that $$R^2$$ is maximized when $$\beta = 1$$ but as I mentioned, not sure why $$R^2$$ not maximize when my predictor is "best scaled"?

Let's consider a regression relationship between, say, driving time $$y$$ and distance travelled $$x$$:

$$y = x\beta + e$$

$$x$$ is denominated in kilometers, just for fun. Now, let's rescale $$x$$, so that it is denominated in centimeters. What is the new $$\beta$$, label it $$\beta'$$? As all the $$x$$-values are multiplied by $$100,000$$, clearly $$\beta' = \beta / 100,000$$.

We can continue this rescaling of $$x$$ in any direction, making $$\beta$$ as large or small as we want. The underlying model, of course, has not changed, just the units of measurement of $$x$$ - implying that $$R^2$$ hasn't changed either. So... there is no relationship between $$\beta$$ and $$R^2$$.

• I tried to stress in my question. The $R^2$ in question is not the $R^2$ from the new regression that produced $\beta$. It's from directly compute $R^2 = 1 - \frac{ \sum (y_i - x_i)^2 }{\sum y_i^2}$. $x$ itself is already a predictor Commented Jul 30 at 23:38
• Sure, but you've never specified what the true value of $\beta$ is. If you are saying that the true value of $\beta = 1$, a) you should put that in the question, and b) of course that will be the value that maximizes $R^2$. My point is that, by rescaling $x$, any value of $\beta$ can be the value that maximizes $R^2$, not just $1$. Commented Jul 30 at 23:43
• I am trying to prove that will really maximize $R^2$ though.. I am not sure why the "of course" part is obvious Commented Jul 30 at 23:46
• To clarify a point - in any given sample, it's extremely unlikely that the optimal $\hat{\beta} = 1$, but that's not what you're asking, right? You're asking what maximizes $R^2$ for the population? Commented Jul 31 at 0:52
• Suggest clarifying what you denote by β since you and the OP seem to be using it to indicate the UNstandardized regression coefficient. What then is b? And what then is the standardized coefficient? Commented Jul 31 at 9:19