The Fisher information of a random variable $Y$ about a parameter $\theta$ upon which the probability of $Y$ depends is:

$\mathcal{I}_Y(\theta)= -E\left[\left.\strut \frac{\partial^2}{\partial \theta^2} \,\log f\left(Y;\theta\right)\right | \theta \right]\enspace ,$

where $f\left(Y;\theta\right)$ denotes the probability density of $Y$ conditional on the value of $\theta$. Roughly speaking, Fisher information measures the extent to which $Y$ can be used to estimate $\theta$, the precise statement given by the Cramér-Rao Theorem.

In the case of a $\{0,1\}$ -valued Bernoulli random variable $X$ with mean $\theta\in(0,1)$, a sample of $X$ (considered itself as a random variable) has Fisher information $(\theta(1-\theta))^{-1}$ about $\theta$, as computed for example on Wikipedia. My question is whether this is known to be optimal for a single bit.

More precisely:

Question:Let $x_1,x_2,\ldots,x_k$ be iid $\{0,1\}$-valued Bernoulli random variables with mean $\theta$ (I'm happy taking $k=2$), and let $g$ be a function mapping $(x_1,x_2,\ldots,x_k)$ to another $\{0,1\}$-valued Bernoulli random variable $Y$. Is it guaranteed that $\mathcal{I}_{g(x_1,x_2,\ldots,x_k)}(\theta)\leq (\theta(1-\theta))^{-1}= \mathcal{I}_{x_1}(\theta)$ ?

My naïve intuition is that this is true, because no single bit ought to be able to carry more information about $\theta$ than a sample. In this sense, if you want to communicate a parameter of a random variable you can't do better than sending samples. If there were any way to compress the information in Bernoulli samples, everyone would be doing it! But I don't know where such problems are discussed and I can't see a proof.