1
$\begingroup$

We run regression specifications of the form,

lm(outcome_log_s ~ x1_s + x1_x^2)

In which the outcome variable is logged and standardized. We want to be able to convey what, for example, a coefficient of ".1" on one of our covariates means is in terms of the original y-scale. How would we go about doing this given standard deviation information from mean centered and log mean centered versions of the same variable

Improve this question
$\endgroup$
4
  • $\begingroup$ I'm not sure I understand the question. Could formulating it as having $Y$ as the outcome, $Z = \log(Y) - \overline{\log(Y)}$ and the model $Z = \sum_{i=1}^{p} \beta_{i} X_{i} + \sum_{i=1}^{p} \beta_{i+p} X_{i}^2$? Or do we have only linear dependence on $X$? $\endgroup$
    – tmrlvi
    Commented Jul 25, 2019 at 23:04
  • $\begingroup$ And when centering the outcome, is it done with the sample mean of true population mean? What happens at different observations? The centering is different? $\endgroup$
    – tmrlvi
    Commented Jul 25, 2019 at 23:08
  • $\begingroup$ Centering is done on the sample mean. Not sure what you mean by "what happens at different observations." The outcome is centered by taking (Y_i - Y_mean) for all observations. $\endgroup$
    – Parseltongue
    Commented Jul 25, 2019 at 23:27
  • $\begingroup$ The point is then to compare two (or more) observations from the same sample? $\endgroup$
    – tmrlvi
    Commented Jul 25, 2019 at 23:36

1 Answer 1

Reset to default
1
+50
$\begingroup$

TL;DR

Additive changes in log imply multiplicative effect. The scaling of the output also plays a role as it changes the scale of the multiplicative effect. That is, if change of 1 in the units of $X$ corresponds to a change of 0.69 in the units of scaled log $Y$, then this implies $e^{0.69 \cdot \text{sd}(\log(Y))}$.

As standard deviation is not affected by centering, we can use sd(data$outcome_log_m) as the value of $\text{sd}(\log(Y))$. Thus, we get a multiplicative affect of $ e^{0.69 \cdot 0.8339012 } \approx 1.70$.

In terms of interpertation, this means 70% increase in the value measured.

Explanation

Assume we transform the data as $$ Z_{i} = \frac{\log(Y_{i}) - \overline{\log(Y)}}{\text{sd}(\log(Y))}$$ and model it as $$ Z_i = \beta^{\top} X_i + \eta_{i}$$ where $\eta_{i}$ models the noise.

Assume that for a specific point $i$, $X_{i}$ changes to $X_{i,\varepsilon} = X_{i} + \varepsilon$, without affecting the value of $\eta$. This implies $$Z_{i,\varepsilon} - Z_{i} = \beta^{\top} \varepsilon.$$ On the other hand, assuming the mean and standard deviation are close to the true mean and standard deviation (so the $\varepsilon$ won't change them a lot), $$Z_{i,\varepsilon} - Z_{i} = \frac{\log(Y_{i,\varepsilon}) - \overline{\log(Y)}}{\text{sd}(\log(Y))} - \frac{\log(Y_{i}) - \overline{\log(Y)}}{\text{sd}(\log(Y))} = \frac{1}{\text{sd}(\log(Y))} \log\left(\frac{Y_{i,\varepsilon}}{Y_{i}}\right).$$ Together, they imply, $$ \log\left(\frac{Y_{i,\varepsilon}}{Y_{i}}\right) = \beta^{\top}\varepsilon \cdot \text{sd}\left(\log(Y)\right)$$ which can be written as $$ Y_{i,\varepsilon} = e^{\beta^{\top}\varepsilon \cdot \text{sd}\left(\log(Y)\right)} Y_{i}$$

Precise calculation would take into account the different sources of error - from the mean calculation, standard deviation calculation and estimation error to create a confedence interval.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Sorry -- I appear to still be struggling. Can you explain a bit more how you get the line after the sentence: "which in turn implies...?" How did you go from Z_ie - Z_i to log(Y_ie / Yi), and why do you multiply the right side by the sd(log(y))? Pardon the ignorance. $\endgroup$
    – Parseltongue
    Commented Jul 26, 2019 at 0:47
  • $\begingroup$ See updates. I assumed that the sample size is large enough so that the affect of $Y_{i,\varepsilon}$ in the mean and sd is negligible. If it is not the case, we can make a more carefull analysis. Another way to analyze it, is to consider the difference between two observations in the same sample. This type of analysis require assumptions on the error $\eta$. $\endgroup$
    – tmrlvi
    Commented Jul 26, 2019 at 1:08
  • $\begingroup$ Hey tmrlvi. Quick question, do these calculations change if you include X and X^2 in your model (e.g. a polynomial term). It seems that, in this case, the 𝜀 amount that you add to X will be affected as you also have to increase by 𝜀^2 $\endgroup$
    – Parseltongue
    Commented Jul 29, 2019 at 21:06
  • 1
    $\begingroup$ In fact, if we our model is $\beta_{1} X + \beta_{2} X^2$, then an $\epsilon$ change in $X$ would lead to a $(\left(\beta_{1}+2\beta_{2}X\right)\varepsilon+\beta_{2}\varepsilon^{2}) \cdot \text{sd}(\log(Y))$ change in $\log(Y)$. That is - the value of $X$ also plays a role. $\endgroup$
    – tmrlvi
    Commented Jul 29, 2019 at 21:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.