# How to interpret standardised regression coefficients in negative binomial regression

I need some advice on interpreting regression coefficients as this is very new to me.

I am running a negative binomial regression on over-dispersed count-data $Y$. In this example, $Y$ is hospital admissions in the previous year.

Many of my variables are on different scales so they were all standardised (zero-meaned with unit standard deviation). I am using a log-link.

I think I understand what the coefficient means on unstandardised variables.

For instance: if $\beta_{i}=0.169$ then for that particular variable $x_{i}$, an $\text{exp}(0.169)=1.1841$-fold increase in $Y$ occurs for every unit change in $x_{i}$.

However, I am confused with standardised variables using log-links.

I have read that for standardised variables (if only $x$ has been standardised), the model gives the expected change in $Y$ when $x_{i}$ increases one standard deviation. So if I calculate the standard-deviation for $x_{i}$ beforehand ($\sigma(x_{i})=a$), and $\beta_{i}=0.169$, this means we can expect $Y$ to increase by 1.1841 when $x_{i}$ increases by $a$.

Is that correct?

• You understanding is correct. Note that this is one of the disadvantages of standardised coefficients because they do not generalise readily to anyone else's data-set as their standard deviations may be different from yours. – mdewey Apr 13 '17 at 14:06
• How do you calculate a standard deviation beforehand? You either assume you know it or you need o collect the data to estimate it. – Michael R. Chernick Apr 13 '17 at 14:08
• @MichaelChernick I already have the data that I am fitting the model to so I assume I can just calculate the standard deviation from there? – user2037067 Apr 13 '17 at 14:10
• @mdewey The goal of my work is to determine the relative importance of several features in a multivariate model. If I only standardise the independent variables (I can't standardise $Y$ because that will lead to non-count, negative data), can I assume the magnitude of $\beta$ provides a value of the relative effect on $Y$? – user2037067 Apr 13 '17 at 14:13
• As I said you don't calculate it you estimate it. – Michael R. Chernick Apr 13 '17 at 14:15