Posterior distribution as a distribution for a new random variable? So in Bayesian framework one uses observed data $X=\{x_1,...x_n\}$ to update the prior $p(\theta)$. My question is it justified to say that $p(\theta|x_1,...,x_n)$ corresponds to a new random variable itself?
 A: $\theta|x_1,...,x_n$ is a random variable. The posterior distribution is its density.
EDIT: Regarding your question in the comments, this is sort of intrinsic to conditional probability. Conditioning a RV reduces the sample space and therefore defines a new random variable. From A Course In Probability (Neil A. Weiss, 2006):

DEFINTION 5.1 Random Variable
A random variable is a real-valued function whose domain is the sample space of a random experiment. In other words, a random variable is a function $X: \Omega \rightarrow  \Re$, where $\Omega$ is the sample space of the random experiment under consideration. (p.177)

further

... conditional probabilities are no different than ordinary (unconditional) probabilities except that the sample space changes from the original one to the event being conditioned on. (p.126)

Since conditioning changes the sample space, we necessarily have defined a new random variable.
Furthermore, the fact that the posterior has a probability density is a consequence of it being a random variable, and clearly it is not the same random variable as the prior or the likelihood, otherwise it would be distributed the same way. Here's a simple example.
let $X \sim Binomial(\theta)$ and $\theta \sim Beta(a,b)$. Then, from conjugacy, we have $\theta | X \sim Beta(a+\sum X_i, b + n - \sum X_i)$. This is clearly a different random variable from $\theta$.
We could just as easily say $\theta | X = W$ and $ W \sim Beta(a+\sum X_i, b + n - \sum X_i)$, if maybe the conditional notation is what's confusing you here.
So to summarize: that the posterior is a new random variable is intrinsic to the definition of conditional probability.
