How to interpret k=1 when summing over k? Is there something wrong with this equation?

Question

I am having a hard time understanding the following equation to do with document clustering.

Is there something wrong with it?

$$\ln \left(\prod_{n=1}^N \sum_{k=1}^K p\left(x_n \mid z_n, k=1\right) p\left(z_n, k=1\right)\right)$$

In pseudo code I initially want to read this as

The natural log of
  The product of (for each n)
    The sum of (for each k)
      The joint probability of
        The (data variable given the latent variable) and the first k
        Multiplied by the joint probability of the latent variable and k being 1

But then why would we enumerate over K?

For example if k=K and K !=1

Is there a more concise way to write the equation?

[Update] GPT advises

$$\ln \left(\prod_{n=1}^N \sum_{k=1}^K p\left(x_n \mid z_n, k\right) p\left(z_n \mid k\right)\right)$$

[Update]

The full question is

The answer is

[Update]

From my course notes I understand that $z_i=k$ and $k \in \{1...K\}$ and it represents the class label for the ith document.

Broele · Accepted Answer · 2023-06-26T21:06:09.280

Summery:

The correct equation (in this way of notion) would be $$\mathrm{ln}\left(\prod_{n=1}^Np(x_n)\right)=\mathrm{ln}\left(\prod_{n=1}^N\sum_{k=1}^Kp(x_n|z_n=k)p(z_n=k)\right)$$

This is a direct application of the law of total probability, which (roughtl) states that given an event $A$ and a finite partition $B_k$ of the sample space, the probability can be partitioned as follows: $$P(A) = \sum_k P(A\cap B_k)$$

Why split it this way?

In mixture models, we typically do not model the total density (here $p(x_n$)), but for each part/cluster individually. So we define the distribution for each cluster individually, here given by: $p(x_n|z_n=k)$.

Typically, this is easier to model and the total probability is given by the sum shown above. In case of your question we can define $p(x_n|z_n=k)$ for each cluster:
Each cluster $k$ has it's own probability $\mu_{k,w}$ that a given word $w$ occurs at a given position in the document. Under the assumption that all words are independent, a given document that consists of the words sequence $w_1,w_2,...w_M$ gets the probability $$p(x_n|z_n=k)=\mu_{k,w_1}\cdot\mu_{k,w_2}\cdot\ldots\cdot\mu_{k,w_M}$$, which can be compressed into $$\prod_{w}\mu_{k.w}^{c(x_n,w)}$$ if $c(x_n,w)$ just counts how often $w$ appears in $x_n$.

About the question

With this in mind, the question that you posted asks for $p(x|\mu)$, which deals with a single sample $x$ and omits the index $n$. Given $\mu$ here means, for a given probability vector $\mu$. Since we have one $\mu_k$ for each cluster $k$, this notion also omits the index $k$. Hence, we can see $p(x|\mu)$ as a short version of $p(x_n|\mu_k)$. An even better version would be $p(x_n|z_n=k)$ since $\mu_k$ is only relevant if document $x_n$ belongs to cluster $k$.

In this light the answer to the question would be: $$p(x|\mu) = \prod_{w}\mu_{w}^{c(x,w)}$$ The answer you scanned and posted is including this into the computation of the total likelihood, while the question only asked for this piece of it.

The notion

From a probability theory point of view, this is a strong shortening of the correct notion. You typically need to distinguish between random variables and observed values / values they could take. This would lead to the equation

$$\mathrm{ln}\left(\prod_{n=1}^Np(X_n=x_n)\right)=\mathrm{ln}\left(\prod_{n=1}^N\sum_{k=1}^Kp(X_n=x_n|Z_n=k)p(Z_n=k)\right)$$ In machine learning literature, where the observations and the random variables are often names the same (except upper/lower case), it is common to shorten the notion to the one we have seen in the summery above.

The versions from the question

Original

$p(z_n,k=1)$ will be either p(z_n) (for $k=1$) or 0 (for $k \not= 1$). Note that $k=1$ is not a probabilistic event that might either occure with a certain probability, but a binary statement, that is either true (i.e. $p(K=1)=1$) or false (i.e. $p(k=1)=0$)

GPT

It shows that you should not trust GPT without checks. First GPT produces also the machine-learning-community-type shortcut equation (which is totally fine), but this only works if implicitly you could define a longer probability-theory-style equation. But $p(z_n|k)$ only makes sense if there would be an implicit random variable corresponding to $k$. We do not have this, so $p(z_n|k)$ just looks like a ML-style equation, but has not meaning (of course you are free to define a meaning for this notion. It would just differ from the common notions)

score 0 · Answer 2 · answered Jun 21 '23 at 20:03

To answer your question regarding why we enumerate over K, it's because the equation considers the possibility of assigning a document to any of the K clusters. The sum over k accounts for all possible cluster assignments, allowing for the calculation of the joint probability of a document belonging to any cluster.