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 regression, when $Y$ is also standardized, in addition to standardizing on $X$ the slope simply becomes the sample correlation coefficient between $X$ and $Y$, $S_{X,Y}$.  To see this, note the following:
In simple regression, there is no constant term in the model, so this is always zero (i.e. $\hat{\beta}_0=0$).  To determine the form of the slop, we have to set up the notation and then do the algebra.
In the standardized model, since there is no intercept term, the model matrix is simply a vector of your $X$-values (i.e. there is no column of 1's which represents the intercept term in non-standarized regression).  In other words, $\boldsymbol{X}$ here is a 1-D vector of your standardized $X$-values:
\begin{eqnarray*}
\boldsymbol{X} & = & \frac{1}{S_{X}}\begin{bmatrix}X_{1}-X\\
X_{2}-\bar{X}\\
\vdots\\
X_{n}-\bar{X}
\end{bmatrix}
\end{eqnarray*}
Where $S_x$ is the sample standard deviation of the $X$-values. It is easily shown that:
\begin{eqnarray}
\boldsymbol{X^{\prime}X} & = & \boldsymbol{r_{XX}}                  
\end{eqnarray}
where $\boldsymbol{r_{XX}}$ is the correlation matrix.  But since there is always perfect correlation between $X$ and itself is 1, $\boldsymbol{r_{XX}}=1$.
Now, the least squares normal equations are written as:
\begin{eqnarray*}
\boldsymbol{X^{\prime}X}\boldsymbol{b} & = & \boldsymbol{X^{\prime}Y}
\end{eqnarray*}
or using the result above, that $\boldsymbol{X^{\prime}X}=\boldsymbol{r_{XX}}=1$, this reduces to:
\begin{eqnarray*}
\boldsymbol{b} & = & \boldsymbol{X^{\prime}Y}
\end{eqnarray*}
So all that's left to do is find the form of $\boldsymbol{X^{\prime}Y}$ to obtain the estimate for $\hat{\beta_1}$.
\begin{eqnarray*}
\boldsymbol{X^{\prime}Y} & = & \frac{1}{S_{X}}\begin{bmatrix}X_{1}-\bar{X} & X_{2}-\bar{X} & \cdots & X_{n}-\bar{X}\end{bmatrix}\frac{1}{S_{Y}}\begin{bmatrix}Y_{1}-Y\\
Y_{2}-\bar{Y}\\
\vdots\\
Y_{n}-\bar{Y}
\end{bmatrix}\\
 & = & \frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)\left(Y_{i}-\bar{Y}\right)}{(n-1)S_{X}S_{Y}}\\
 & = & \boldsymbol{r_{XY}}
\end{eqnarray*}
So we see when both $X$ and $Y$ are standardized, the $\boldsymbol{X^{\prime}Y}=\boldsymbol{r_{XY}}$, which is just the correlation coefficient in simple regression.  So, in this case, $\hat{\beta}_1$ is the correlation coefficient between the unstandardized $X$ and $Y$.
We can verify these results in R:
#Generate some data
x<-rnorm(10, 5, 3)
y<-rnorm(10, 18, 2)

#Standardize x and y
x.standardized<-scale(x)
y.standardized<-scale(y)

#Compute estimated regression coefficients with both x and y standardized
betas.hats<-lm(y.standardized~x.standardized)$coeff
round(betas.hats, 5)

  (Intercept) x.standardized 
   0.00000        0.60224 

#Verify that beta_1 = correlation(x,y)
round(cor(x,y), 5)
  [1] 0.60224

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?
