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if we have two vectors $Y$ and $X$. Lets first standardise $Y$ by its mean so we get ($y$):

$$ y = Y/\overline{Y}$$ We can get the slope of the least squares regression of $y$ and $X$ as:

$$ \beta = cov(X,y)/var(X)$$

Now say we wanted to standardise $X$ as well to make it ($x$). We could standardise $X$ then do the regression again. But instead we can just multiply the gradient like this: $\beta_s = \bar{X}\beta$.

My question is, if we have only an unstandardised slope where both $Y$ and $X$ are not standardised ( $\beta_u $ ) i.e.

$$ \beta_u = cov(X,Y)/var(X)$$

can we get $\beta_s$ when we have both means and variances aswell?

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I think you almost answered the questions yourself, but let me go through the derivation anyway.

Your question is, given $\beta_u$, $\overline{X}$ and $\overline{Y}$ can you calculate: $$\beta_s = \frac{cov(x,y)}{var(x)}$$

Substituting the unstandardized vectors into the expression gives,

$$\beta_s = \frac{cov(x,y)}{var(x)} = \frac{cov(X/\overline{X},Y/\overline{Y})}{var(X/\overline{X})}$$

Applying the multiplicative rule for the covariance and variance it becomes,

$$\beta_s = \frac{cov(X,Y)/(\overline{X}\overline{Y})}{var(X)/{\overline{X}}^2} = \frac{cov(X,Y)}{var(X)}\frac{\overline{X}}{\overline{Y}}$$

The relation is hence,

$$\beta_s = \beta_u\frac{\overline{X}}{\overline{Y}}$$

The basic properties for the covariance and variance can be found at:

https://en.wikipedia.org/wiki/Covariance#Properties https://en.wikipedia.org/wiki/Variance#Basic_properties

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