$L(\theta;x)=f(x;\theta)$ vs. $L(\theta;x)\propto f(x|\theta)$ My second-year notes in Statistical Inference and Modelling (unpublished) have a definition,

The likelihood function of $X$, given the data $x$, is $L∶\Theta\rightarrow\mathbb{R}$ defined by $L(\theta;x)=f(x;\theta)$.

My third-year notes in Bayesian Statistics (unpublished) have a statement,

the likelihood is proportional to the joint distribution, i.e. $L(\theta;x)\propto f(x|\theta)$.

This makes me wonder several things.

*

*What's the significance of the order of parameters separated by a semicolon, e.g. $(\theta;x)$ vs. $(x;\theta)$?

*Do $f(x;\theta)$ and $f(x|\theta)$ refer to the same function?

*$f(x|\theta)$ has a 'given' bar like a conditional but the third-year notes appear to refer to it as a 'joint'. Which is it? and why the apparent mis-match?

*Is the likelihood $L$ equal to the other function $f$ or proportional? Is it proportional in general and sometimes equal as a special case, with a proportionality constant of $1$?

Edit: The paragraph from the third-year notes.


*

*& 2. $;$ means $|$, i.e., conditional dependence: $p(x|\theta) = p(x;\theta)$ and $f(x|\theta) = f(x;\theta)$.


*Also, because $\theta$ is not a random variable per se, it's the model's $f$ parameters, it is possible that they refer to $f(x|\theta)=f(x;\theta)=f_\theta(x)$ as joint, if $x\in R^d$ and $d>1$.


*The likelihood function is defined as $L(\theta)=f(x|\theta)$. It can be proportional if you omit the constants.
