0
$\begingroup$

I am analysing server data and I have a scenario where I need to get the % by which Y is changed because of a unit change in X:

EDIT: I am doing a Linear Regression in Python (and its other forms like Lasso - ultimate aim is to find feature importances)

My Y is a continuous variable. My Xs are all standardized (meaning : x-xmean/xstd.dev)

Case 1:

ln(y) = a + b (Standardized X)

When X is Increased by 1 standard deviation, then Y increases by b *100 % or [ exp(b) -1 ] *100 %

So when X increases by 1 unit , does Y increases by b*100/std.deviation of X % or [ exp(b) -1 ] *100 / std.dev(X) % ?

or should I un-standardize the coeff and take it as:

% change in Y for 1 standard deviation change in X is [ exp{ b1 / std.dev(X) } -1 ] *100 ?

Case 2:

ln(y) = a + b (Standardized X)
Here X is a % , Eg: % of memory used at the moment, or % of cpu time spent on a job , etc.

How should I interpret % change in Y in this case?

Data in my target (Y) is as shown in the pic below:

enter image description here

$\endgroup$
3
  • $\begingroup$ Would you please post the raw data before taking logs or standardizing? $\endgroup$ Jul 1 '19 at 14:21
  • $\begingroup$ @JamesPhillips, I have added a pic of the raw data for Y , andI have many X cols , like 1300 cols or so ... $\endgroup$ Jul 2 '19 at 4:42
  • $\begingroup$ Dears, I have edited my question with more clarity on my understanding ... $\endgroup$ Jul 2 '19 at 6:30
1
$\begingroup$

For a one standard deviation increase in $X$, $\ln y$ is expected to increase by $b$ units. That's the only interpretation you can get from this model.

To use the % change interpretation, you need to model $\ln(E[y]) = a + b Z$ (where $Z = X/\sigma$). You've modeled $E[\ln y] = a + b Z$. The first model is a generalized linear model with a log link. The second model is a linear model with a log-transformed outcome.

In the first model, if you take $\exp$ of both sides, you get $$E[y] = \exp(a + bZ)=\exp(a)\exp(bZ)=\alpha \ \exp(bZ)$$ To see how $E[y]$ changes when we increase $Z$ by 1 (i.e., increase $X$ by one stndard deviation), we can simply plug, going from $Z = 0$ to $Z=1$. $$E[y|Z=0]=\alpha \ \exp(b \times 0) = \alpha$$ $$E[y|Z=1]=\alpha \ \exp(b \times1) = \alpha \ \exp(b)$$ So, for a one standard deviation increase in $X$, $E[y]$ increase by a factor of $\exp(b)$. In the second model, if you take $\exp$ of both sides, you get $$\exp(E[\ln y]) = \exp(a + bZ)$$ The left side is not reducible, so we can't go further down this path. The only way to interpret this model is by interpreting the linear change in $E[\ln y]$, as I did in the beginning of this post.

This distinction has been discussed here, here, and here on CV and here.

Another note is that you shouldn't standardize a predictor that is already in interpretable units like percentage points. It only muddies the interpretation.

$\endgroup$
7
  • $\begingroup$ Dear @Noah , I am using Python - Linear Regression and was wanting to use a Log-Linear model .... I will check your links and get back to you... $\endgroup$ Jul 2 '19 at 8:12
  • $\begingroup$ Dear @Noah , Also , my Y is a continuous value ... $\endgroup$ Jul 2 '19 at 8:26
  • $\begingroup$ The term standardized would lead me to guess $Z = (X - \bar X)\ /\ \text{SD}(X)$ $\endgroup$
    – Nick Cox
    Jul 2 '19 at 8:40
  • $\begingroup$ Dear @NickCox, you are correct... $\endgroup$ Jul 2 '19 at 8:41
  • $\begingroup$ Dears , I have updated my post for more clarity... $\endgroup$ Jul 2 '19 at 8:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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