3) Consider a sample of size $n=8$ from the Uniform $(θ,θ+4)$ distribution where $θ>0.$
Consider two estimators of $θ$: $T_1=\overline X$ and $T_2=5\overline X$
(where $\overline X$ denotes the sample mean). By comparing the corresponding MSEs, establish whether $T_1$ is better than $T_2$ to estimate $θ.$
Even this exercise looks kind of impossible for me. Can you help me?
They're both lousy estimators of $\theta,$ and even $\overline X-2,$ isn't very good. If it were really desired to estimate $\theta,$ I'd use $-2$ plus the average of the maximum and minimum observed values.
However, we have this: $$ \text{mean squared error} = (\text{bias}^2) + \text{variance}. $$ The variance of an obsersvation from this distribution is $4^2$ times the variance of the uniform distribution on $(0,1).$ The latter is $1/12,$ so the variance of one observation from this distribution is $4^2/12 = 16/12 = 4/3.$ Therefore the variance of the mean of eight of them is $(4/3)/8 = 1/6.$ So we have $\operatorname{var}(\,\overline X\,) = 1/6$ and $\operatorname{var}(5\overline X\,) = 25\times(1/6).$
Since we also have $\operatorname{E}(\,\overline X\,) = \theta+2,$ the bias of your first estimator is $\operatorname{E}(\,\overline X\,) -\theta = 2,$ and that of your second can be similarly found.