Seeking clarity regarding kernels With regards to Bayesian statistics, I understand the kernel of a probability density function (pdf) or probability mass function (pmf) to be the form of the pdf or pmf in which any factors that are not functions of any of the variables in the domain are omitted.  So to test my understanding, for a Normal distribution we would have:
$N(\mu,\sigma^2)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ $\rightarrow$ and its kernel would be $\rightarrow$ $e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
and for a Binomial distribution, we would have:
$Bin(n,p) = \left(n \atop x\right)p^x(1-p)^{n-x}$ $\rightarrow$ and its kernel would be $\rightarrow$ $\frac{p^x(1-p)^{n-x}}{x!(n-x)!}$
First, is my definition correct and second, is my understanding correct?
 A: Let's stay on topic with Bayesian statistics. Call the binomial success parameter $\theta.$
If you put a prior probability distribution on $\theta,$ it might be $\mathsf{Beta}(2,3)$ and the relevant beta density might be written as
$$p(\theta) = \frac{\Gamma(2+3)}{\Gamma(2)\Gamma(3)}\theta^{2-1}(1-\theta)^{3-1} = 12\theta^{2-1}(1-\theta)^{3-1}.$$ for $0<\theta<1.$
This is a function in $\theta$ with kernel $p(\theta) \propto \theta(1-\theta)^2.$
Then if you observe $x=41$ successes in $n=60$ trials, the likelihood function $f(x|\theta)$ is the binomial PDF considered as a function of $\theta$ proportional to $$\theta^{x}(1-\theta)^{n-x} = \theta^{41}(1-\theta)^{19},$$
where $0 <\theta<1.$ (Even in frequentist statistics it is customary to omit
irrelevant constants of integration (or summation) from the likelihood function.)
So neither the prior nor the likelihood has gamma functions or factorials. Then, because the beta prior and the binomial likelihood are 'conjugate' (mathematically compatible)
you can use Bayes' Theorem to multiply the prior by the likelihood to get
an expression which you can recognize as the kernel
$$\theta^{2+41 -1}(1-\theta)^{3+19=1} = \theta^{43-1}(1-\theta)^{22-1}$$ of the posterior distribution
$\mathsf{Beta}(2 + 41, 3 + 19) \equiv \mathsf{Beta}(43, 22)$
Note: Notice that the parameters of a beta distribution are 'out of phase by 1' with the exponents of $\theta$ and $(1-\theta)$ in the PDF and its kernel.
