Consider a sample of size $n=8$ from the $\operatorname{Uniform}(\theta,\theta+4)$ distribution where $\theta>0$. Consider two estimators of $\theta$:



(where $\bar{X}$ denotes the sample mean). By comparing the corresponding MSEs, establish whether $T_1$ is better than $T_2$ to estimate $\theta$.

I wanted to ask you an opinion about this exercise. I cannot understand where to start.


Let me provide a guide on how to compute MSE for $T_1$.

\begin{align}\mathbb{E}((T_1-\theta)^2)&=\operatorname{Var}(T_1-\theta)+\mathbb{E}^2(T_1-\theta) \\ &=\operatorname{Var}(T_1)+\mathbb{E}^2(T_1-\theta) \end{align}

Hence it suffices to compute $\operatorname{Var}(T_1)$ and $\mathbb{E}(T_1-\theta)$.

To compute $\operatorname{Var}(T_1)$, we have

$$\operatorname{Var}(T_1)=\operatorname{Var}\left( \frac{\sum_{i=1}^8X_i}{8}\right)$$

To evaluate the term above, assume $X_i$ are i.i.d from $\operatorname{Uniform}(\theta, \theta+4).$ wikipedia page about uniform distribution might be helpful to you.


$$\mathbb{E}(T_1-\theta)=\mathbb{E}\left( \frac{\sum_{i=1}^8X_i}{8}\right)-\theta$$

After you compute the two MSE value, choose the one with smaller mean square error.

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    $\begingroup$ I understand about the compute and consequently the subsequent comparison of the two MSEs, thank you so mucho for the guide. But I can not understand how evaluate and what values ​​can assume the term Xi from from Uniform(θ,θ+4). $\endgroup$ – Rick-tex Nov 25 '17 at 9:38
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    $\begingroup$ Further guide: Expection is linear. For independent samples, variance of sum is sum of variance. If $X_i \sim \operatorname{Uniform}(a,b)$, then $\operatorname{Var}(X_i) = \frac1{12}(b-a)^2$ and $\mathbb{E}(X_i)=\frac{a+b}{2}.$ $\endgroup$ – Siong Thye Goh Nov 25 '17 at 9:51
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    $\begingroup$ Ok, I know: If Xi∼Uniform(a,b), then Var(Xi)=(b−a)^2/12 and E(Xi)=(a+b)/2. So with U(θ,θ+4) Var(x)=(θ+4-θ)^2/12=4/3 And Var(T1)= (4/3)/8=1/6 Similarly For E(x)=(θ+θ+4)/2=θ+2 E((T1−θ)^2)=Var(T1−θ)+E(T1−θ)^2 = 1/6 + ((θ+2)-θ)^2 = 25/6 So I think is this MSE of T1 And The same process for The second estimator. Right? $\endgroup$ – Rick-tex Nov 27 '17 at 19:29
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    $\begingroup$ yup, seems fine. $\endgroup$ – Siong Thye Goh Nov 27 '17 at 20:16

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