Based on the simulation below, it appears that standardizing all variables in a data set affects OLS results in the following ways:

  1. Coefficient estimates change
  2. Standard errors change
  3. P-values remain the same except the p-value for the intercept coefficient.

Are these results general? Do they also apply to other models, not only OLS?


#Standardization simulation
remove(list = ls())

n = 50
t <- rnorm(n, mean = 2.2, sd = 5)
x1 <- rnorm(n, mean = 1.5, sd = 5)
x2 <- rnorm(n, mean = 3.3, sd = 6)
x3 <- rnorm(n, mean = 2, sd = 7)

betas <- matrix(runif(4, min = -5, max = 5))

inputs <- as.matrix(cbind(t, x1, x2, x3))

y <- (inputs %*% betas) + rnorm(n, mean = 0, sd = 20)

data <- data.frame(cbind(y, inputs))

standardize <- function(variable){
    demeaned <- variable - mean(na.omit(variable))
    sd <- sqrt(var(na.omit(variable)))

stan.data <- data.frame(apply(data, 2, FUN = standardize))

summary(lm(y ~ t + x1 + x2 + x3, data = data))

summary(lm(y ~ t + x1 + x2 + x3, data = stan.data))

2 Answers 2


This method is called beta coefficients (at least by Wooldridge). We compute the Z-scores of all variables, and then run the regression on the transformed data. Now what happens?

The slopes are different: before we had a slope of say $\hat \beta_1$, but now we have a slope of $\hat \sigma_1 /\hat \sigma_y \cdot \hat \beta_1$, where $\hat \sigma_i$ is the sample standard deviation.

There is no intercept; or rather it will be estimated as very (very) close to 0. Hence the change in p-value of the intercept you mentioned.

The interpretation of $\hat \beta_i$ is now: what happens with y, when we increase the standard deviation of $x_i$ by 1. In a sense, we can say something about the most "important" variable (but I dislike that word).

One final note: just as you you stated, the p-values are not going to be different and so you will always have the same significance as before.

  • $\begingroup$ The intercept is not close to zero at all based on OPs simulation and standardized data. It's further aways from zero than it was before.. $\endgroup$
    – Talik3233
    Commented Jul 5, 2019 at 10:56
  • $\begingroup$ @Talik3233 then OP is doing it wrong $\endgroup$
    – Repmat
    Commented Jul 26, 2022 at 19:44

Yes, they are general for OLS.

Let me take the example of a simple regression. We know the slope coefficient is $$ \hat\beta_1=\frac{\sum_i(x_i-\bar{x})(y_i-\bar{y})}{\sum_i(x_i-\bar{x})^2} $$ or $$ \hat\beta_1=\frac{\widehat{Cov}(y_i,x_i)}{\widehat{Var}(x_i)} $$ and the intercept estimate is $$ \hat\beta_0=\bar y-\hat\beta_1\bar x. $$ Standardizing yields $\bar x=\bar y=0$ and $\widehat{Var}(x_i)=\widehat{Var}(y_i)=1$. Hence, $$ \hat\beta_1=\widehat{Cov}(y_i,x_i)=\widehat{Corr}(y_i,x_i) $$ and $$ \hat\beta_0=0. $$ The standard errors are the square roots of the diagonal elements of $s^2(X'X)^{-1}$, e.g., $s.e.(\hat\beta_1)=s/(\sum_i(x_i-\bar{x})^2)^{1/2}$.

In the scaled case, this standard error then becomes $$\tilde s/(n-1)^{1/2},$$ where $\tilde s$ is the square root of the error variance estimate of the scaled regression. We get $n-1$ in the denominator because $\widehat{Var}(x_i)=1/(n-1)\sum_i(x_i-\bar{x})^2=1$, so $\sum_i(x_i-\bar{x})^2=n-1$. Hence, the standard errors are different.

The $t$-ratios are nevertheless the same. For the unscaled case, write \begin{align*} t&=\frac{\hat{\beta_1}}{s.e.(\hat\beta_1)}\\ &=\frac{\frac{\widehat{Cov}(y_i,x_i)}{\widehat{Var}(x_i)}}{s/(\sum_i(x_i-\bar{x})^2)^{1/2}}\\ &=\frac{\frac{\widehat{Cov}(y_i,x_i)}{\widehat{Var}(x_i)}}{s/((n-1)^{1/2}sd(x))}\\ &=(n-1)^{1/2}\frac{\widehat{Cov}(y_i,x_i)sd(x)}{s\widehat{Var}(x_i)}\\ &=(n-1)^{1/2}\frac{\widehat{Corr}(y_i,x_i)\widehat{Var}(x_i)sd(y)}{s\widehat{Var}(x_i)}\\ &=(n-1)^{1/2}\frac{\widehat{Corr}(y_i,x_i)sd(y)}{s}, \end{align*} where the 5th equality follows from rearranging $$ \widehat{Corr}(y_i,x_i)=\frac{\widehat{Cov}(x_i,y_i)}{sd(x)sd(y)} $$ For the unscaled case, write \begin{align*} \tilde t&=\frac{\widehat{Corr}(y_i,x_i)}{\frac{\tilde s}{(n-1)^{1/2}}}\\ \end{align*} An application of the Frisch Waugh Lovell theorem will demonstrate that scaling the regressor does nothing to the residuals. Scaling $y$, however, gives residuals $\tilde u_i$ which are related to the unscaled ones $\hat u_i$ via $\tilde u_i=\hat u_i/sd(y)$. Hence, $\tilde s=s/sd(y)$. Thus, \begin{align*} \tilde t&=\frac{(n-1)^{1/2}\widehat{Corr}(y_i,x_i)}{\frac{s}{sd(y)}}\\ &=\frac{(n-1)^{1/2}\widehat{Corr}(y_i,x_i)sd(y)}{s}=t\\ \end{align*} If the test statistics are the same, so will of course the p-values.

That the standard error on the intercept is different follows from the fact that the scaled intercept is estimated to be zero very precisely.

P.S.: scale achieves the goals of your function, too.

  • $\begingroup$ Does this mean that for all independent variables the standard errors are the same when using standardized variables since x_1 is cancelled out in the formula? $\endgroup$
    – Irazall
    Commented Aug 11, 2020 at 22:35

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