Rao-Blackwell Theorem depends on the fact of using a sufficient statistics? Here is the standard Rao-Blackwell Theorem and its simple proof: (see Theorem 3.3).
In the proof, where do we use the fact that $T$ is a sufficient statistic? I am a bit confused here.
 A: Definition: A statistic $T(\boldsymbol{X})$ is a sufficient statistic for $\theta$ if the conditional distribution of the sample
$\boldsymbol{X}=\left(X_{1}, \ldots, X_{n}\right)$ given the value of $T(\boldsymbol{X})$ does not depend on $\theta$.

*

*That is, $T(\boldsymbol{X})$ is sufficient for $\theta$ if the $\mathrm{pdf} / \mathrm{pmf}$ $f_{\boldsymbol{X} \mid T(\boldsymbol{X})=T(\boldsymbol{x})}(\boldsymbol{x} ; \theta)$ is the same for all $\theta$.

Suppose now that our statistic $T$ is not sufficient. And consider the statement of the Rao-Blackwell theorem:
Theorem (Rao-Blackwell Theorem) Let $\hat{\theta}$ be an estimator of $\theta$ with $\mathbb{E}\left(\hat{\theta}^{2}\right)<$ $\infty$ for all $\theta$. Suppose that $T$ is sufficient for $\theta$, and let $\theta^{*}=\mathbb{E}(\hat{\theta} \mid T)$. Then for all $\theta$,
$$
\mathbb{E}\left(\theta^{*}-\theta\right)^{2} \leq \mathbb{E}(\hat{\theta}-\theta)^{2} .
$$
The inequality is strict unless $\hat{\theta}$ is a function of $T$.
If $T$ was not a sufficient statistic we wouldn't necessarily be able to have an estimator $\theta^*$ which is a function solely of $T$ (as it is this conditional expectation $\mathbb{E}[\hat{\theta}|T]$). We should instead write $\theta^* = f(\theta,T)$ for every $\theta$. This would make it uninteresting because at this point we would have an estimator $\theta^* = f(\theta,T)$ of the parameter $\theta$ that depends itself on $\theta$!! This really wouldn't solve the estimation problem of a function of the samples $g(X)$ that we wanted to use to estimate $\theta$ (this was the initial problem we were interested in the first place).
(Please, if this helps you, please leave a positive feedback and/or accept the answer; thank you).
