1
$\begingroup$

Please see Chapter 3.2 from this article: http://www.turn.com.akadns.net/sites/default/files/whitepapers/TURN_Tech_WP_Data-driven_Multi-touch_Attribution_Models.pdf

Here, a method is described by which the influence of online-marketing-channels/campaigns (for example: "Display" as a channel for Display-Ads, "SEA" as a channel for Search-Engine-Ads etc.) on a binary criterion (for example: 1 = positiv User = User buys a product in online-store vs. 0 = negative User = User does does not buy a product in online-store) can be calculated. The described method is described as more easy and simple than conducting a logistic regression.

With the first equation, you can calculate for every channel the share of positive users (=users who had contact to this channel AND bought the product) on all users which had contact to this channel:

enter image description here

Example: 1000 Users had contact to the channel "Display" and 400 of them bought the product. Hence, the probability for this channel is 0.4 (=400/1000 = 400/(400 + 600))

For considering overlapping between channels, you use equation two, which includes the second-order interaction term:

enter image description here

What i am not understanding and at which point i need your help is the following: For calculation the contribute of each channel to the buy-probability (= C(xi); and this is what is important, because every marketer wants to know, which online marketing channel/campaign "works" and converts a internet-user to a buyer of the product) the following equation is used:

enter image description here

What does mean "The contribution of channel i is then computed at each positive user level" and what does mean "for a particular user"? With equation 1 and 2 we have calculated terms on a aggregated level and suddenly, in equation 3, they talk about calulation on "User level". What does this mean? And what is "N"? When i have 3 channels in total, is N = 2 (3 channels - 1 = 2 channels)? This "user level"-thing is irritating. I thougt i have to calculate for a specific channel - for example: channel "Display" - the terms according to equation 1 and 2 and just enter this terms in equation 3, and then i get C(display) = [myresult] But is it really so easy?

$\endgroup$
1
$\begingroup$

The idea of this model is that you "train" it in an aggregate manner, hence on a big chunk of data, and then you APPLY it on a user level to get your result. So first, you obtain the probabilities calculated for your training set (which can be the same as the users you're planning on applying the model on, but doesn't have to be). Then, each positive user is the same as each user who converted/generated revenue. The importance of a particular user comes is that the path/ad channels they observed could be different. N is therefore the numbers of channels the user was exposed to. After you re-distribute the revenue from each user across all your channels you sum that up to get the total value of each channel.

$\endgroup$
  • $\begingroup$ dont know, if i understood the part "rom each user across all your channels you sum that up to get the total value of each channel", therefore: let us assume we have (to make it easy to describe just) 5 Users and 2 Channels and all 5 Users have contact to channel 1, but only 3 Users have contact to Channel 2. Hence, we will get 5 times a C for Channel 1 (hence: C(Channel_1)) and 3 times a C for Channel 2 (hence: C(Channel_2)). Then, i have to sum up the 5 Cs for Channel 1 to get the Contribution of Channel 1, and sum up the 3 Cs for Channel 2 to get the Contribution of Channel 2. Correct? $\endgroup$ – flobrr Feb 3 '17 at 12:24
  • $\begingroup$ and another notice: it`s not necessary to talk about revenue, because a convertion can, but must not be linked to revenu (for example: a convertion can be opening a (fee free) bank deposit). Considerung this, a convertion is nummbered as 1, a non-convertion as 0. $\endgroup$ – flobrr Feb 3 '17 at 12:40
  • 1
    $\begingroup$ Yes you're right, you don't have to use revenue. I guess that's just what I have hence I'm used to that. In the scenario you have above, yes, you would sum them up. If instead of conversion you had a \$ amount to it, you would probably want to multiply the probabilities by the \$ and then sum up the \$s. You'll probably want to re-normalize to 1 after summing up your C's (when comparing C1 and C2). $\endgroup$ – Sharon Feb 6 '17 at 14:32
0
$\begingroup$

This is an explanation of how to implement the algorithm written by the author of the paper.

https://www.clickz.com/the-math-behind-multi-touch-attribution/50028/

$\endgroup$
  • $\begingroup$ If the link goes dead this will lose meaning. Can you edit to summarise the details? $\endgroup$ – mdewey May 2 '18 at 8:30

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.