Bayesian Statistics -Prior and Posterior distributions Please is it ever possible for the prior distribution to contain more information about parameter(s) than the posterior distribution? If yes, when can that occur? Is it the same concept as the posterior being "diffuse with respect to the prior"? I am reading the paper on A Comparative Review of Dimension Reduction Methods in Approximate Bayesian Computation  by Blum et al. (2013) and I came across the concept of a diffuse posterior on page 6. I would appreciate an explanation.
Thanks.
 A: It is possible for the posterior variance to be strictly greater than the prior variance. For example, suppose that $X\sim Bernoulli(p)$ with a prior $p\sim Beta(1, 10)$ and you observe $X=1$. Then, since we are using a conjugate prior, the posterior is a $Beta(2, 10)$.
The prior variance is $\frac{10}{11^2\cdot 12}\approx 0.0069$. The posterior variance is $\frac{20}{12^2\cdot 13}\approx 0.0107$. Here are plots of the prior and posterior distributions.

An intuition as to why this happens: the prior is skewed towards values of $p$ close to 0, but the likelihood favours values of $p$ close to 1. The posterior distribution is torn between the two and so has a larger variance.
A: The complete quote from the paper (which should have been included in the text of the question to make it self-contained) is

Since entropy measures information and a lack of randomness (Shannon,
1948), the authors propose minimizing the entropy of the approximate
posterior, $p_\text{ABC}(θ|s^\text{obs})$, over subsets of the summary
statistics, $s$, as a proxy for determining maximal information about a
parameter of interest. High entropy results from a diffuse posterior
sample, whereas low entropy is obtained from a posterior which is more
precise in nature.

which does not mention the posterior being more diffuse than the prior. Now, as discussed in the comments, the posterior variance is on average smaller than the prior variance
$$\text{var}(\theta)=\mathbb{E}[\text{var}(\theta|X)]+\text{var}(\mathbb{E}[\theta|X])$$
but it does not mean that $$\text{var}(\theta)\ge\text{var}(\theta|X=x)$$ for all realisations $x$ of $X$, as exemplified in Robin's answer.
