In the article "A survey of transfer learning": https://ieeexplore.ieee.org/abstract/document/5288526 Pan and Yang give a good mathematical definition of transfer learning which seems to be widely adopted, given that the article has been cited more than 5000 times.

There is one thing though that confuses me, and that is the definition of a domain and the purpose of the marginal probability distribution. Here is an excerpt from the article:

"In this survey, a domain 𝓓 consists of two components: a feature space 𝓧 and a marginal probability distribution P(X), where $X=\{x_1,…,x_n\}$βˆˆπ“§. For example, if our learning task is document classification, and each term is taken as a binary feature, then 𝓧 is the space of all term vectors, $x_i$ is the ith term vector corresponding to some documents, and X is a particular learning sample. In general, if two domains are different, then they may have different feature spaces or different marginal probability distributions."

Later in the article they say that when two domains are not equal: 𝓓$_S$ β‰  𝓓$_T$, corresponding to source and target respectively, then that can mean either (or both) of two things:

(1) 𝓧$_S$ β‰  𝓧$_T$


(2) P$_S$(X) β‰  P$_T$(X).

Their example is document classification, where the case of (1) could mean that the two domains are in different languages. In the case of (2) that could mean that the source domain and target domain documents focus on different topics.

What is confusing me here is, how does P(X) represent that? What is the purpose of this P? They call it a marginal probability distribution. Marginal with respect to what, are we summing over some other variables as one does when working with marginal distributions?

I can relate to this f.x. when doing image classification, when one domain is a set of images of animals, while another domain could be a set of peoples faces. But how does the P(X) encode that information? Does P give the probability of an instance x of a random variable X to be included in the domain?

Thanks in advance for all answers :)


Frimann wrote, "Does P give the probability of an instance x of a random variable X to be included in the domain?"

The answer is yes. More specifically, for every possible classifier input $x\in\mathscr{X}$, the "marginal probability distribution" $P(X=x)$ is the probability that $x$ would be randomly sampled from a distribution over $\mathscr{X}$. They are calling it the marginal probability distribution because you may have to sum over other random variables to compute $P(X=x)$.

If the source "domain" is 1024x768 RGB pictures of animals and the target "domain" is 1024x768 RGB pictures of cats, then the space of inputs $\mathscr{X}=\{0,1,2,\ldots, 255\}^{1024\times768\times3}$ is the same for both domains, but the marginal distribution functions are different. The marginal probability function $P_S$ (respectively $P_T$) maps $\mathscr{X}$ into [0,1] and is defined as the probability $P_S(x)$ (respectively $P_T(x)$) that a particular image $x\in\mathscr X$ would be randomly chosen from the set of possible 1024x768 animal images (respectively cat images).

  • $\begingroup$ So my understanding was somewhat correct. If I understand correctly then the domain definition not only describes what's included in the domain, but also what's the probability of getting a certain sample. $\endgroup$ – Frimann Bjornsson Feb 20 '19 at 16:55
  • $\begingroup$ One thing I'd like to ask regarding the application of this definition to my resource. I'm working with endgame tablebases in chess, which give you the value (win, draw, loss) for a given board state. I'm doing transfer learning from the dataset of 5 piece board states, to the dataset of 6 piece board states. So one domain is the 5 piece dataset, and the other is the 6 piece dataset. Am I right to say that 𝒳 is the space of all possible board states in chess, and P$_5$(X) is non zero for all 5 piece states, but zero for all others? And similar for P$_6$(X) $\endgroup$ – Frimann Bjornsson Feb 20 '19 at 17:01
  • $\begingroup$ @FrimannBjornsson Wow, what a great application. I would be very interested in your research. With regard to your question, you could let $\mathscr X$ be the set of all chess positions and use $P_5$ and $P_6$ as the marginals as you described above. That seems like a good idea. There are many alternatives. Another possibility is that you let $\mathscr X_5$ be the set of 5 piece positions and let $\mathscr X_6$ be the set of all 6 piece positions. Maybe we should discuss this by email. My email address is "hundal" followed by "hh" at yahoo.com. $\endgroup$ – irchans Feb 20 '19 at 18:22
  • $\begingroup$ @FrimannBjornsson Frimann wrote, "the domain..., but also what's the probability of getting a certain sample." Yes, in the paper you cited, that is correct. $\endgroup$ – irchans Feb 20 '19 at 18:24

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