1
$\begingroup$

I'm looking for a model to predict CTR (click-through-rate)

I have the following data: For each ad I know the number of impressions, clicks and some other attributes (which are mainly dummy variables).

The CTR per ad is calculated as follows: #clicks / #impressions.

I have two questions regarding predicting CTR:

  1. I am wondering which model should be used to predict the CTR. I tried a linear regression, but the R-squared is very low (around 10%-15%). A logistic regression is not an option as my dependent variable is not a 0/1 variable.

  2. When I run a linear regression with clicks as dependent variable and impressions, etc. as explanatory variables, my R-squared suddenly is around 85-95%. How is it possible that this differs so much from taking CTR as dependent variable?

EDIT: I followed the approach from kjetil, which works perfectly.

$\endgroup$
  • $\begingroup$ Have you considered zero inflated model? Presumably, there'd be a lot of zeroes from people who don't click at all. $\endgroup$ – Huy Pham Jan 8 at 5:03
  • $\begingroup$ @HuyPham I don't have user-level data, I only have aggregated data per ad, so for one ad I for instance know there have been 10000 impressions and 4 clicks. There are no ads with zero clicks in my dataset. Or should I maybe un-group the dataset such that for instance I get 9996 rows with zeros and 4 with ones (clicks)? $\endgroup$ – M09 Jan 8 at 8:48
  • $\begingroup$ No the answer by Kiejtil works better. I just threw an idea out there. $\endgroup$ – Huy Pham Jan 8 at 22:20
0
$\begingroup$

You should try logistic regression. Let $x=\text{number of clicks}$, $n=\text{number of impressions}$. Then $\text{CTR}=x/n$, and in modeling that proportion directly you loose information. A logistic regression (possibly quasibinomial) gets at least the variance structure correct. In R you could do something like:

mod <- glm( cbind(x,n-x) ~ size + etc..., data=your_data_frame, family=quasibinomial())

A similar post with answer and example is Count explanatory variable, proportion dependent variable

EDIT

Good that you have tried some of my suggestions. Here some answers to your further edits:

  1. Look at this part of the output *Dispersion parameter for quasibinomial family * do there seem to be a substantial reduction?

  2. Look at the Deviance Residuals: from the output. Maybe most of the difference is in the extremes, but you could extract all of the residuals by resid(your_glm_object, type="deviance") and then plot against each other the residuals for each model, or their histograms.

  3. There is a version of AIC for quasi-models called QAIC (similar for BIC I suppose). A small paper about this in R (by Ben Bolker) is here. QAIC is implemented in some R packages, listed there.

  4. Getting predictions from this models: Use something like predict(your_glm_object, type="response", newdata=your_data_frame_with_new_data) For details see ?predict.glm.

$\endgroup$
  • 1
    $\begingroup$ Thank you for your answer! The quasibinomial regression does not return a AIC value, which makes it hard to determine what the optimal model is (which variables may be excluded). How can I solve this? Also, how should I interpret the coefficient estimates, suppose size_small has a coefficient of -2.04, is it correct to say that the number of clicks decreases by approximately 2 if the ad is small compared to large? I think this is a wrong interpretation as it isn't a linear regression, but I don't know how to read it othterwise. $\endgroup$ – M09 Jan 7 at 15:04
  • $\begingroup$ You could post some output from (one of) the analysis ... and see the links in my answer, which has some explication of interpretation. Logistic regression models a probability, so you can compare the predicted probabilities from the model, search this site for "interpreting logistic regression" $\endgroup$ – kjetil b halvorsen Jan 7 at 16:16

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.