I have data containing several variables. I ran a regression model. Prior to running the model I have normalized the dependent variable Y and the independent variables X1 and X2.

After receiving the output I want to interpret the results. For example, if the coefficient of X1 is 0.15, I know that it means that for addition of one standard deviation of X1, there is an increase of 0.15 standard deviations in Y, but this is not clear.

I want to go back to the original units of Y and X for interpretation. How do I do that ? Can I simply take the "normalized coefficients", multiply by the standard deviation of Y and add the mean of Y?

Something about it doesn't make me comfortable.

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    $\begingroup$ Would you please post a link to the raw data? I would like to surface fit "Y = f(X1, X2)"and see what I find with an equation search. $\endgroup$ – James Phillips Jan 23 '19 at 16:02

Firstly, why did you normalized Y? It will make your output harder to interpret, and it is often not necessary to standardize the dependent variable.

I presume you have centered and scaled you X's, you can backtransform them to interpret,

        # run your model with the X's standardized

        mean <- mean(x1)
        sd <- sd(x1)


#you can also plot

curve(b0 + b1*(x1-mean)/sd, add=TRUE) 

I also recommend to use Y at its original scale, since standardization does not change the distribution shape.

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  • $\begingroup$ Can I say that the change in Y is actually the "standardized coefficient" multiplied by the standard deviation of Y ? I think it makes sense $\endgroup$ – user3275222 Jan 23 '19 at 16:08

Standardising both the dependent and independent variables can be useful for presentation and coefficient interpretation, normally in simple linear regression, whenever the Pearson correlation coefficient $r_{xy}$ is of interest. The OLS estimation $\hat{\beta} = r_{xy} \times \frac{s_y}{s_x} $. Clearly, if a variable $z$ has been standardised to ${z\prime}$ then it's standard deviation must equal unity, since $$ \sqrt{\mathbb{V}[{z\prime}]} = \sqrt{\mathbb{V}\left[\frac{z-\bar{z}}{\sqrt{\mathbb{V}[z]}} \right]} = \sqrt{\frac{1}{\mathbb{V}[z]}{\mathbb{V}[z-\bar{z}]}} = 1 $$ Therefore, when both $y$ and $x$ have been standardised $s_y=s_x=1$ and the OLS estimator for $\hat\beta$ is the correlation coefficient $r_{xy}$. Note also that the OLS estimator of $\alpha$ is zero for standardised variables since $\bar{x}=\bar{y}=0$.

In your multiple regression context, if your $X_i$ are independent then the $\beta_i$ will be equal to the correlation coefficients if you had performed separate simple regressions.

If the $X_i$ are not independent, each $\beta_k$, a partial correlation coefficient, shows the relationship between $Y$ and $X_k$ with the other $X_i$ fixed. This can have a different value to the simple correlation coefficient between $Y$ and $X_k$ in a simple linear regression ignoring the other $X_i$.

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