# Likelihood function. Comparing estimator MSE.

#### Erik!

##### New Member
Consider a sample of size n=8 from the Uniform(θ,θ+4) distribution where θ>0.
Consider two estimators of θ:

T1=X¯ T2=5X¯

(where X¯ denotes the sample mean). By comparing the corresponding MSEs, establish whether T1 is better than T2 to estimate θ.

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

Ok, I know that for compute MSE for T1 for example,

E((T1−θ)^2)=Var(T1−θ)+E2(T1−θ)
=Var(T1)+E2(T1−θ)

Hence it suffices to compute Var(T1) and E(T1−θ)
To compute Var(T1):

Var(T1)=Var(∑Xi/8)

To evaluate the term above, assume Xi
are i.i.d from Uniform(θ,θ+4).
Similarly,

E(T1−θ)=E(∑Xi/8)−θ

After you compute the two MSE value, I choice the one with smaller mean square error.
I don't undetstand how evaluate and what values ​​can assume the term Xi from Uniform(θ,θ+4) for compute my MSE. Can you help me?

#### Dason

##### Ambassador to the humans
I don't understand your question exactly. Xi can take any value between θ and θ+4.

#### Erik!

##### New Member
Ah! Il thought I understood the exercise ... because I did not understand how it proceeds for the comparison of both MSEs.

#### Erik!

##### New Member
Let two estimators of theta T1 And t2, I have to compare their mean error square.

#### Dason

##### Ambassador to the humans
So which part is giving you trouble? What have you tried?

#### Erik!

##### New Member
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 for The second estimator.
Anyway T1 is better than t2 because the T2 mean is 5X right?
Well this is how I compute It...