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My course notes (3rd-year module in Bayesian Statistics, unpublished) have a paragraph,

Gaussian data with known variance

Suppose we have $\textbf{x}=\{x_1,\dots,x_n\}$ iid given $\theta$ and $\sigma^2$ as N$(x_i\,|\,\theta,\sigma^2)$, and assume $\theta$ unknown but $\sigma^2$ known. Given these data you are interested in the unknown expected value $\theta$. We can write down the likelihood function as $$f(\textbf{x}\,|\,\theta)\propto\prod_{i=1}^n\sqrt{\frac{1}{2\pi\sigma^2}}\exp\left(-\frac{1}{2\sigma^2}(x_i-\theta)^2\right).$$

Would I be right in thinking that calling $f$ "the likelihood function" is actually an abuse of notation/terminology (albeit a conventional one), and that it would be more accurate to say the following?

We can write down the likelihood function as $$\mathcal{L}(\theta\,|\,\textbf{x})\propto f(\textbf{x}\,|\,\theta)=\prod_{i=1}^n\sqrt{\frac{1}{2\pi\sigma^2}}\exp\left(-\frac{1}{2\sigma^2}(x_i-\theta)^2\right),$$ where $f(\textbf{x}\,|\,\theta)$ is the joint pdf of the data $\textbf{x}$ conditional on $\theta$.

ETA: I asked a related question a while ago, and accepted an answer. I've spent most of today staring at the image included in that answer. enter image description here

mjc
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  • Still, the density of the data could be proportional to the rhs if, e.g., the data is produced in an ordered way, with $x_1<x_2<\cdots$. – Xi'an Dec 07 '22 at 17:02
  • I would agree calling $f$ a likelihood is an abuse of terminology. It's also bizarre, most obviously it's a density function, but then the "propto" symbol is there, but they have included the normalizing constant. Much easier to write $f(x|\theta) \propto \exp(c (x_i - \theta) ^2) $ – AdamO Dec 07 '22 at 17:06
  • @AdamO Thanks, that's a relief to have confirmed. The proportionality you mention is actually added as a further step in the notes, which I omitted from my quote, but the notes do as quoted use the $\propto$ sign for the first step, between $f$ and the product with the proportionality constant. – mjc Dec 07 '22 at 17:10
  • @Xi'an I'm afraid I don't understand. How does the ordering of the data affect the equality of the joint pdf to the product of the individual pdfs? – mjc Dec 07 '22 at 17:13
  • The density of the ordered sample is $n!$ times the density of the unordered one. – Xi'an Dec 07 '22 at 18:20

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