# Help needed to Interpret ln(y) = a +b (Standardized X)

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:

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

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.

• 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... Jul 2 '19 at 8:12
• Dear @Noah , Also , my Y is a continuous value ... Jul 2 '19 at 8:26
• The term standardized would lead me to guess $Z = (X - \bar X)\ /\ \text{SD}(X)$ Jul 2 '19 at 8:40
• Dear @NickCox, you are correct... Jul 2 '19 at 8:41
• Dears , I have updated my post for more clarity... Jul 2 '19 at 8:45