# How to convert standardized coefficients to unstandardized coefficients?

My goal is to use the coefficients derived by previous research on the subject to predict actual outcomes given a set of independent variables. However, the research paper lists the Beta coefficients and t-value, only. I would like to know if it is possible to convert the standardized coefficients to unstandardized ones.

Would it be useful to convert my unstandardized independent variables to standardized ones to calculate the predicted value? How would I return to an unstandardized predicted value (if that is even possible..)

Number of bus route (buslines) | 0.275(Beta) | 5.70*** (t-value)

I am also given this regarding the independent variables:

Number of bus route (buslines) | 12.56(avg) | 9.02(Std) | 1(min) | 53(max)

• How have the coefficients been standardized? In general the $\beta$'s have a unit which is the unit of $Y$ divided by the unit of $X$, what is their unit in the paper? Jun 21, 2012 at 14:03
• I am not sure I understand your question. Here is a sample row of an independent variable after regression analysis from the paper. Transit supply characteristics: Number of bus route (buslines) | 0.275(Beta) | 5.70*** (t-value)
– user12127
Jun 21, 2012 at 14:11
• The coefficient itself is not standardized as gui11aume mentioned. But the t statistic it the estimated coefficient divided by its estimated standard deviation. Given t and the degrees of freedom you could calculate the p-value and the estimated standard deviation because Beta=t-value x estimated standard deviation. But I am not sure whether or not this is what you are looking for. The beta estimate is not standardized. The t statistic is the standardized form of the beat estimate. So you already have the standardized coefficient. Jun 21, 2012 at 14:33

It sounds like the paper uses a multiple regression model in the form

$$Y = \beta_0 + \sum_i \beta_i \xi_i + \varepsilon$$

where the $\xi_i$ are standardized versions of the independent variables; viz.,

$$\xi_i = \frac{x_i - m_i}{s_i}$$

withe $m_i$ the mean (such as 12.56 in the example) and $s_i$ the standard deviation (such as 9.02 in the example) of the values of the $i^\text{th}$ variable $x_i$ ('buslines' in the example). $\beta_0$ is the intercept (if present). Plugging this expression into the fitted model, with its "betas" written as $\hat{\beta_i}$ (0.275 in the example), and doing some algebra gives the estimates

$$\hat{Y} = \hat{\beta_0} + \sum_i \hat{\beta_i} \frac{x_i - m_i}{s_i}=\left(\hat{\beta_0}-\left(\sum_i\frac{\hat{\beta_i m_i}}{s_i}\right)\right)+\sum_i\left(\frac{\hat{\beta_i}}{s_i}\right)x_i.$$

This shows that the coefficients of the $x_i$ in the model (apart from the constant term) are obtained by dividing the betas by the standard deviations of the independent variables and that the intercept is adjusted by subtracting a suitable linear combination of the betas.

This gives you two ways to predict a new value from a vector $(x_1, \ldots, x_p)$ of independent values:

1. Using the means $m_i$ and standard deviations $s_i$ as reported in the paper (not recomputed from any new data!), calculate $(\xi_1,\ldots, \xi_p) = ((x_1-m_1)/s_1, \ldots, (x_p-m_p)/s_p)$ and plug those into the regression formula as given by the betas or, equivalently,

2. Plug $(x_1, \ldots, x_p)$ into the algebraically equivalent formula derived above.

If the paper is using a Generalized Linear Model, you may need to follow this calculation by applying the inverse "link" function to $\hat{Y}$. For example, with logistic regression it would be necessary to apply the logistic function $1/(1 + \exp(-\hat{Y}))$ to obtain the predicted probability ($\hat{Y}$ is the predicted log odds).

• Perfect, thank you! Got some help from a colleague. One more question though: My new value (Y-hat) is very low. The author uses a logarithmically transformed dependent variable in his regression. Does that mean I should exp(Y-hat) to expand back up to the untransformed unit of measurement.
– user12127
Jun 22, 2012 at 0:08
• Also, there is no Y-intercept included in the paper, and testing the exp(Y-hat) method seems to indicate that there should be a value for Y-intercept that represents some of the variance not explained by the model, in order to raise the predicted outcome to a reasonable level.
– user12127
Jun 22, 2012 at 0:18
• Then it is not the coefficients that are stadnardized. It is the variables. Jun 22, 2012 at 1:09
• Michael M, yes, $\exp(\hat{y})$ is probably what you want and yes, you need to find out what the intercept is. You might have to fudge it by guessing the intercept and varying it until your model appears to reproduce any graphics and tables in the paper sufficiently accurately.
– whuber
Jun 22, 2012 at 3:21
• If you are looking to do what the title asks, look here: www3.nd.edu/~rwilliam/stats1/x92.pdf if the y is also standarized. Also see stats.stackexchange.com/questions/235057/… Sep 15, 2016 at 4:47

$$B = p \times \frac{sy}{sx}$$

• $x$ is the independent variable
• $y$ is the dependent variable
• $s$ is the standard deviation
• $p$ is the path coefficient
• $B$ is the regression coefficient.
• I am not sure what a path coefficient is. It looks like perhaps B is a regression coefficient which would not be dimensionless. It would be in y units per 1 x unit. However p=B sx/sy where sx is the estimated standard deviation in x divided by the estimated standard deviation in y and p is dimensionless. It represents an estimated correlation between x and y. Lance if this is what you intended please make the changes by editing your post. Jan 5, 2017 at 23:37