Calculate the expected value of $E[(T- \theta)^2]_\theta = Var_\theta(T)$ with $ T : = \frac{2}{n} \cdot \sum_{i=0}^n{X_i}$ and $E[T] = \theta$
i realised that one way to calculate it with :
$E_\theta[(T-\theta)^2] = E_\theta[T^2-2\cdot\theta \cdot T + \theta^2] = E[T^2]_\theta - 2\cdot \theta \cdot E[T]_\theta + E[\theta^2]_\theta$
Im however stuck on how to calculate the $E[T^2]_\theta$ , my intuition suggest that i can use the Variance but I'm not sure how.
suggestions ?
Assuming $X_i$ are RVs, we can do: \begin{align} \mathbb{V}[T] &= \mathbb{E}[T^2] - \mathbb{E}[T]^2 \\ &= \mathbb{E}\left[ \left( \frac{2}{n}\sum_i X_i \right)^2 \right] - \theta^2 \\ &= \frac{4}{n^2}\mathbb{E}\left[ \sum_i \sum_j X_i X_j \right] - \theta^2 \\ &= \frac{4}{n^2}\sum_i \sum_j \mathbb{E}\left[ X_i X_j \right] - \theta^2 \\ &= \frac{4}{n^2}\sum_i \left( \mathbb{E}\left[ X_i^2 \right] + \sum_{j\ne i} \mathbb{E}\left[ X_i X_j \right] \right) - \theta^2 \\ \end{align} There's not much you can do from here, unless you make some more assumptions.
For example, let's suppose the $X_i$ are IID with $\mathbb{E}[X_i]=\theta/2$ and $\mathbb{V}[X_i]=\sigma^2$. Then: $\mathbb{E}[X_i^2] = \mathbb{E}[X_i]^2 + \mathbb{V}[X_i]=\sigma^2+\theta^2/4$. By independence, $\mathbb{E}[X_iX_j]=\mathbb{E}[X_i]\mathbb{E}[X_j]=\theta^2/4$. Then in this special case we get: $$ \mathbb{V}[T] = -\theta^2 + \frac{4}{n^2}\sum_i \sigma^2+\frac{\theta^2}{4} + (n-1)\left[ \frac{\theta^2}{4} \right] = \frac{4\sigma^2}{n} $$ which is not surprising given the central limit theorem.
One more example to avoid the IID case. Suppose $\mathbb{E}[X_i]=\mu_i, \mathbb{V}[X_k]=\sigma^2_k=c_{kk}$, and the covariance between dependent variables is $\text{cov}(X_i,X_j)=c_{ij}$. Then $\mathbb{E}[T] = (2/n)\sum_i\mu_i$, so $\sum_i\sum_j \mu_i\mu_j=(\sum_i \mu_i)^2 = (n^2/4)\theta^2 $. Then: $$ \mathbb{V}[T] = \frac{4}{n^2}\left( \sum_i\sigma^2_i + \sum_k\sum_{j\ne k} c_{ij} \right) $$
Hopefully this is actually what you are asking.