text: Q:
  
  Unstandardized $\beta_0$ and $\beta_1$ When Both $X$ and $Y$ are Standardized
  
  In a univariate linear regression model, I understand that the unstandardized slope for the standardized predictor $z=(x - \mu)/\sigma$ is equal to $\beta_1 \sigma_y / \sigma_x$, but I am wondering what the slope is when $y$ is also standardized?  What about the constant term when $y$ is standardized?
  Below are some results of regressions:
  
  regression of standardized $y$ on standardized $x$: $\beta_{0}=0$ and $\beta_{1}=0.5585$
  regression of unstandardized $y$ on unstandardized $x$: $\beta_{0}=0.338$ and $\beta_{1}=1.556$.
  mean and sigma of $x$ are $\bar{x}=0.206$ and $\sigma_x=1.047$
  mean and sigma of $y$ are $\bar{y}=0.661$ and $\sigma_y=2.934$
  
  Use of the typical
  $
  \beta_{unstd}=\beta_1 \left( \frac{\sigma_y}{\sigma_x} \right)
  $
  $
  1.565=0.5585 \left( \frac{2.934}{1.047} \right)
  $
  which is close, but not equal to 1.556.  However, I am not sure this typical relationship holds if $y$ is also standardized(?).
  
  A:
  
  In simple linear regression:
  
  The slope is $\beta_1 = r_{xy}\frac{\sigma_y}{\sigma_x}$
  
  The constant term is $\beta_0 = \bar{y} -\bar{x}\beta_1$
  
  You can standardize the data, or shift and scale the data in any other way, but this correlation coefficient, $r_{xy}$, will remain the same.
  
  In a univariate linear regression model, I understand that the unstandardized slope for the standardized predictor $z=(x - \mu)/\sigma$ is equal to $\beta_1 \sigma_y / \sigma_x$,  but I am wondering what the slope is when $y$ is also standardized?
  
  However, I am not sure this typical relationship holds if $y$ is also standardized(?).
  
  This typical relationship $\beta_{unstd} = \beta_{std} \sigma_y / \sigma_x$ is especially the case when $y$ is also standardized, and not when only $x$ is standardized.
  For standardized $x$ and $y$ you get $\sigma_x=\sigma_y =1$ and $\beta_{std} = r_{xy}$. For unstandardized $x$ and $y$ you get $\beta_{unstd} = r_{xy}\frac{\sigma_y}{\sigma_x}$. You can similarly work out what happens when only $x$ or only $y$ is standardized.
  Intuition: the changes in $\sigma_x$ and $\sigma_y$ when you are standardizing, they are like stretching or squeezing the graph. If $\sigma_y$ goes from $1$ to $2.934$ then it is like stretching the graph in $y$ direction which makes the slope $2.934$ times larger and that is why $\sigma_y$ is in the numerator of $\frac{\sigma_y}{\sigma_x}$. For a stretch in $x$ direction the slope becomes smaller and changes with the rate (inverse) of the stretch and that is why $\sigma_x$ is in the denominator of $\frac{\sigma_y}{\sigma_x}$.
  
  What about the constant term when $y$ is standardized?
  
  You can just fill in the formula above:
  $\bar{y} - \bar{x}\beta_1=0.661 - 0.206 \times 1.556  \approx 0.340$
  This is not exactly the same as $0.338$, just like your $1.556$ and $1.565$ did not match. Possibly this is due to some computation error and the results of the regressions that you provided. This computation error I can actually not imagine since linear regression, a simple computation, should not give you an error larger than a round of error. Such error is not enough to explain the difference, but maybe you did the computation by hand?). Or, maybe you made a typing error and switched a 5 and a 6? How did you get these regression results?
  
  