Intuitive meaning of a transition kernel in a Markov Chain in MCMC algorithms? I read that in the Metropolis-Hastings Algorithm, the transition kernel $K$ is given by:
$$
\forall \theta^{t-1} \in \Theta , \ \ \ \ \  \ \ K(\theta|\theta^{t-1}) = \alpha\left(\theta|\theta^{t-1}\right)q\left(\theta|\theta^{t-1}\right) + \left(1-\alpha\left(\theta^{t-1}\right)\right)\delta_{\theta^{t-1}}(\theta)
$$
Here, $\alpha\left(\theta|\theta^{t-1}\right)$ denotes the probability of acceptance given $\theta^{t-1}$, 
$q\left(\theta|\theta^{t-1}\right)$ denotes the proposal distribution and $\alpha\left(\theta^{t-1}\right)$ denotes the overall probability of accepting a candidate, given that the current state is $\theta^{t-1}$:
$$
\int_{\Theta} \alpha\left(\theta|\theta^{t-1}\right)q\left(\theta|\theta^{t-1}\right)d\theta
$$
and finally, $\delta_{\theta^{t-1}}(\theta)$ is the dirac delta mass.
I am wondering if there is any intuitive meaning behind defining the transition kernel like this and why it is called a kernel and what it is used for. Thanks!
 A: The function$$K(\theta|\theta^{t-1}) = \alpha\left(\theta|\theta^{t-1}\right)q\left(\theta|\theta^{t-1}\right) + \left(1-\alpha\left(\theta^{t-1}\right)\right)\delta_{\theta^{t-1}}(\theta)$$is


*

*a probability density in $\theta$ for a given value of $\theta^{t-1}$

*measurable in $\theta^{t-1}$ for a given value of $\theta$


hence qualifies as a Markov kernel. It allows for the generation of $\theta^t$ given $\theta^{t-1}$ and hence for the production of a Markov chain $(\theta^t)_t$. The mixture structure of $K(\cdot|\theta^{t-1})$ means that, with probability $\left(1-\alpha\left(\theta^{t-1}\right) \right)$ the chain repeats itself and with probability $\alpha\left(\theta^{t-1}\right)$ it moves to another value. Since, in most cases, this probability$$\alpha(\theta^{t-1})=\int_{\Theta} \alpha\left(\theta|\theta^{t-1}\right)q\left(\theta|\theta^{t-1}\right)d\theta$$cannot be computed in closed form, it is often replaced with the unbiased estimate $\alpha\left(\theta|\theta^{t-1}\right)$ when $\theta\sim q\left(\theta|\theta^{t-1}\right)$.
