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I am reading the Hidden Markov Model note http://cs229.stanford.edu/section/cs229-hmm.pdf from Stanford and have problem finding the meaning of the notation described below.

What does the semicolon (;) means in below (Page 2 in above link)?

$$P( z_t, z_{t-1},\ldots, z_1; A )$$

$A$ is the transition matrix. How should it be read? For example, I know $P( x | y )$ is "probability of $x$ GIVEN $y$"; $P( x=s_1 | y=o_1 )$ is "probability of $x$ being $s_1$ GIVEN that $y$ is $o_1$".

Thanks in advance for any help for this possibly trivial, but surprisingly hard to find answer.

Further question on same notation issue.

In Wikipedia article on EM algorithm, They have

$$L( \theta;X ) = p( X,Z | \theta )$$

In the above, X is the set of observed data, Z is the set of observed data and $\theta$ is a vector of parameters. Based on my current understanding of the notation, I will read the above as "the likelihood of theta given X is equal to the joint probability of X and Z given theta". In this context, X contains observed data, which to me appears to be random variables rather than fixed parameters. Why is semicolon ';' used here?


I have just noticed a few other threads asking the same question, when these threads shown up as related threads on right pane. My apology that I couldn't find them earlier. The following two threads have elaborated answers, in agreement with the accepted answer of this thread.

Meaning of probability notations $P(z;d,w)$ and $P(z|d,w)$

What is the meaning of the semicolon in $f(x;\Theta)$?


marked as duplicate by whuber Sep 2 '13 at 15:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


It seems that $P(z_{t},\ldots,z_1 ; A)$ should be read as "the probability of $z_{t},\ldots,z_1$, for a given value of parameters $A$".

There is a difference in notation because $A$ is not a random variable but a fixed parameter. If $A$ were a random variable, you'd use $P(X|A)$ instead of $P(X; A)$.

  • $\begingroup$ So, it is "given" similar to " | ", however for parameters instead of random variables. Thanks a lot for the clarification, much appreciated. $\endgroup$ – hong Aug 23 '13 at 14:46
  • $\begingroup$ I hope you don't mind I extend the question in same thread, since it is related issue. I added into original post above. $\endgroup$ – hong Sep 2 '13 at 9:04
  • $\begingroup$ Yes, this is a bit tricky. The point to realize is that in this case, $p(X,Z|\theta)$ really is a function $\theta$ and not of $(X,Z)$. $\endgroup$ – Stijn Sep 2 '13 at 9:08
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    $\begingroup$ Thanks again for the reply. I have seen another article using $L(\theta|X)$; I guess the use of the notation is confusing to many. I came across another thread here discussing this notation. $\endgroup$ – hong Sep 2 '13 at 14:23

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