# Logistic regression with an log transformed variable, how to determine economic significance

I am using a logistic regression model with continuous independent variables and two log transformed size variables (total assets and total deposits).

My question is how to interpret the results and measure the economic impact? The normal steps in interpretation I take is as follows: I firstly run stata test on summary statistics (sum), then a logistic regression (logit) and than the marginal effects for the economic significance (mfx).

These results can be explained by sentences as "a one standard deviation increase in $X$ (an independent variable) increases the probability of participation by $X\%$ I will multiply the $dy/dx$ of the marginal effects by the SD to get the % impact.

So far so good, but how do I interpreted the results with the two ln(size) variables? To clarify my problem I will show you a part of the results:

sum stat:

obs       mean       stdD        min         max
lnTCD|      2755   -1.469624    2.992985  -15.01948   4.827369
lnTA |      2767   -1.131045    2.796627  -16.51857   6.731258

logit res:

Coeff       stERR       Z      P>Z             95%
lnTCD|   1.088141   .1424768     0.65   0.519     .8418451    1.406494
lnTA |   .7880959    .105272    -1.78   0.075     .6065653    1.023954

Margin eff:

dy/dx       stERR      Z      P>Z           95%          X
lnTCD|    .0104326      .01619    0.64   0.519  -.021308  .042174  -1.09399
lnTA |   -.0294112       .0165   -1.78   0.075  -.061749  .002926   -.7886


In a normal situation I would multiply lnTA of the marginal effects (-0.0294112) by the Standard deviation of the summary statistic (2.796627) which results in -8.22%

Although this holds for the other non log variables, intuitively this does not sound correct for these (ln) variables.

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possible duplicate of Interpretation of log transformed predictors in logistic regression –  Macro Jul 29 '12 at 15:13
@Macro, I suppose I thought about that question differently than this one. If I remember correctly, there I was addressing the OP's belief that they needed to transform their X variables to achieve normality (although since it was merged the idiosyncrasies of that question have been lost). I'm OK w/ merging these, but I'm also OK w/ not doing so; I think there are subtle differences in emphasis. Eg, my answer doesn't make much sense in the context of the merged question (perhaps I'll delete it). Here, I think Michael's answer that logit(Y) is linear in ln(X), not X, gets at the right nuance. –  gung Jul 29 '12 at 15:29
@gung, I think onestop's answer to the linked question gets directly at the question posed here - how do I interpreted the results with the two ln(size) variables?. To me it seems like an exact duplicate but if you disagree then that's fine - I'm not trying to pressure you into close voting - my first impulse was "Hmm why didn't gung close?" since you are one of the few that sometimes does close vote, which is why I posted the now-deleted comment. –  Macro Jul 29 '12 at 15:33
@Macro, no problem (trust me, I'm difficult to offend). I think your position is reasonable, & I don't necessarily disagree. I just think it's a little bit ambiguous & I tend to give the benefit of the doubt by default. However, the real reason I didn't vote to close here is that I wasn't reminded of the other question because I thought about that question as about whether covariates were required to be normally distributed in LR. –  gung Jul 29 '12 at 15:53
@macro, thank your possible duplicate suggestion. However, I still do not know how to interpret the results. Is it possible to explain it in plain simple terms? Regarding the explanation of the other question. Does this mean that I am dealing with an ln(y)=b+bln(x) or with y=b+bln(x)? –  Peter Teunissen Jul 29 '12 at 16:36

Please disregard my answer if it repeats what it has been suggested to you in linked posts...

With log variables, your interpretation of the mfx $dy/dx$ becomes:

$\frac{dy}{d\log x}$ which is equivalent to $\frac{dy}{dx/x}$ so the interpretation implies that 1% variation in $X$ causes a $\frac{dy}{d\log x}$ variation on $y$

In your case the std. deviation(2.79) is still an average variation of the log scale so:

$\frac{dy}{d\log x}\Delta(\log x) = -0.029*2.79= -0.08$

a 2.79% variation (or standard deviation)of $X$ causes the probability to decrease in $8\%$ .

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