Let $ X_{1}, X_{2}, \dots ,X_{n} $ be linearly independent random variables with the same distribution and probability density function given by $$ f_{\theta}(x)=\frac{1}{\theta+1}e^{-\frac{x}{\theta+1}} $$ if $ x >0 $ and $ 0 $ otherwise where $ \theta \in [1,2] $. Let $ T_{1} $ be an estimator for $ \theta $ given by $$ T_{1}(X_{1}, \dots , X_{n})=\frac{X_{1}+ \cdots +X_{n}}{n}-1 $$ and let $ T_{2} $ be another estimator given by $$T_{2}(X_{1}, \dots , X_{n})=\min\{{2,\max\{{1,T_{1}(X_{1}, \dots , X_{n})}}\}\} $$
I am asked to say whether or not $ T_{2} $ is an unbiased estimator for $ \theta $ and to compare the mean squared errors of $ T_{1} $ and $ T_{2} $.
What I did is the following:
We can see that the $ X_{i}'s $ have exponential distribution with parameter $ \frac{1}{\theta+1} $ thus $$ \Bbb{E}[X_{i}]=\theta+1 $$ so $$ \Bbb{E}[T_{1}]=\theta $$ thus $ T_{1} $ is unbiased. I also computed $$ MSE[T_{1}]=\frac{(\theta+1)^{2}}{n} $$but I don't know what to do about $ \Bbb{E}[T_{2}] $ and $ MSE[T_{2}] $ as well as how $ MSE[T_{2}] $ compares to $ MSE[T_{1}] $.
Any help would me much appreciated. Thank you!
It is convinient to write $T_2$ as $$ T_2=\begin{cases} 1, & T_1\leq 1\cr 2, & T_1\geq 2\cr T_1, & 1<T_1 <2\end{cases} = \begin{cases} 1, & \overline X \leq 2\cr 2, & \overline X\geq 3\cr \overline X-1, & 2<\overline X <3\end{cases} $$
Note that $T_2\geq 1$ and $\mathbb P(T_2>1)>0$. Therefore, $\mathbb E[T_2]>1$. If $\theta=1$, then $\mathbb E[T_2]\neq \theta$, so $T_2$ is biased.
To compare MSE, it is sufficient to compare random variables: $$|T_2-\theta|\leq |T_1-\theta|$$ for any $1\leq \theta\leq 2$ and for any elementary event. Therefore $$ \mathbb E[(T_2-\theta)^2] < \mathbb E[(T_1-\theta)^2] $$ We have strict inequality since $$\mathbb P\bigl(|T_2-\theta| < |T_1-\theta|\bigr)>0.$$