Percentages transformation in regression I am performing an econometric analysis, but I am quite of a beginner, so I am wondering of how to include the percentage values in the regression. Should I leave the interest rates as they are (ex. 3.25% or 0.0325) or should I transform them in percentage points (1+3.25/100)*100? At the end, should I log transform them (I believe I should not, because I do not need the percentage change of the interest rate, but the nominal change of the interest rate itself). 
My second dilemma is about the indices. Should I log transform the indices? When I include indices, how do I seasonally adjust them? Is, for example, taking indices with base value 2010=100 considered as seasonal adjustment?
 A: Several ideas to keep in mind for linear regression and many linear models more generally:
In some sense, it doesn't matter whether you represent a percent as 3.25 or .0325 or 1.0325:


*

*Linear transformations of your data in some sense don't matter:


*

*Eg. the coefficient of a variable measured in feet will simply be 12 times larger than the coefficient on the same variable measured in inches.

*You can see this formally with a derivation like here: Is there ever an advantage to using raw data in supervised learning over normalized/standardized data?


*Multiplying by 100, dividing by 100, these are all linear transformations. If you leave interest rates as 3.25 or write them as .0325 doesn't matter.

*If you subtract a constant from any of your variables, this is ALSO a linear transformation of your data as long as a constant is included in your regression. This will change the estimated coefficient on your constant as well! But it doesn't matter in the sense that you still get the same fit.


For small $r$, we have $\log(1+r) \approx r$
What's the first order Taylor expansion of $\log$ around 1 (i.e. we draw a tangent line to $\log(x)$ at $x=1$)?
$$ \log(x) \approx x - 1 \quad \quad \text{ for $x$ near $1$} $$
Hence $\log(1.0325) \approx .0325$. For returns, interest rates etc... near 0, the log return (i.e. $\log(1+r)$ is almost the same thing). For numbers far from 1 though, this obviously isn't true. 
Difference in logarithms is close to the percent change
For $x$ and $y$ close together, $\frac{x}{y}$ will be near 1, hence if we use the taylor expansion around 1 of the logarithm:
$$ \begin{align*} \log(x) - \log(y) &\approx  \frac{x}{y} - 1 \\
&\approx \frac{x - y}{y}
\end{align*}
$$ 
