Let $Y_{1},Y_{n},\ldots,Y_{n}\sim N(\mu,\sigma^{2})$ with mean know. Prove that $ \displaystyle T(Y)=\frac{1}{n}\sum_{1\leq i \leq n}Y_{i}^{2}$ is sufficient to $\sigma^{2}$.
My approach: I know that the Neyman-factorization theorem is useful here, then
\begin{eqnarray*} f(y_{1},y_{2},\ldots,y_{n};\sigma^{2})=\left( \frac{1}{\sigma \sqrt{2\pi}}\right)^{n}\exp \left\{-\frac{1}{2\sigma^{2}}\sum_{i=1}^{n}y_{i}^{2} \right\}\cdot \mathbf{1} \end{eqnarray*} but I should conclude that $\displaystyle T_{\sigma^{2}}(Y)=\sum_{1\leq i \leq n} Y_{i}^{2}$ is sufficient statistic, but is not what is requered to prove. How can I continue from here?
For the following question my post is not same of the suggested by a member of MSE. However thanks.
EDIT: If $\sum_{n}Y^{2}$ is sufficient statistic, so $\frac{1}{n} \sum_{n}Y^{2}$ is a sufficient statistic? If it's true, how can I prove this? I think that if it's true so, my approach is correct and I can conclude.
Thank you so much @tommik for your explication.
Sorry but there is an evident problem in your statement.
$$Y_i\sim N(\mu;\sigma^2)$$
with $\mu$ known parameter; $\mu \in \mathbb{R}$
Thus
$$f(\mathbf{y},\mu,\sigma^2)=(2\pi\sigma^2)^{-n/2}\exp\Bigg\{-\frac{1}{2\sigma^2}\Sigma_i(y_i-\mu)^2\Bigg\}=$$
$$=(2\pi\sigma^2)^{-n/2}\exp\Bigg\{-\frac{\Sigma_i y_i^2}{2\sigma^2}+\frac{\mu\Sigma_i y_i}{\sigma^2}-\frac{n\mu^2}{2\sigma^2}\Bigg\}$$
Now using the Factorization theorem you get
$$f(\mathbf{y},\mu,\sigma^2)=1\times\underbrace{(2\pi\sigma^2)^{-n/2}\exp\Bigg\{-\frac{\Sigma_i y_i^2}{2\sigma^2}+\frac{\mu\Sigma_i y_i}{\sigma^2}-\frac{n\mu^2}{2\sigma^2}\Bigg\}}_{g[t(\mathbf{y}),\sigma^2]}$$
Thus
$T=\Sigma_i Y_i^2$ IS NOT SUFFICIENT FOR $\sigma^2$ with $\mu$ knonw UNLESS $\mu=0$
Applying the same theorem, what is valid for $T$ is also valid for $T/n$
(the proof is trivial, just substitute the new estimator in the factorized expression)