do you know what it means for an estimator to be unbiased?
do you know what the efficiency of an estimator is?
You should have some idea how to start these problems or they wouldn't give them to you. Right?
Let Y1, Y2, ... Yn denote a random sample from uniform distribution
on the interval (0, theta) ...
let YBAR = sample mean, MAX = sample maximum
Consider:
estimator1 = 2(YBAR)
estimator2 = ([n+1]/n)MAX
Show that both estimators are unbiased estimators of theta.
Find the efficiency of estimator1 relative to estimator2.
Which estimator is preferable? Explain.
I have no idea how to start this. Any help would be appreciated!!
do you know what it means for an estimator to be unbiased?
do you know what the efficiency of an estimator is?
You should have some idea how to start these problems or they wouldn't give them to you. Right?
I know that the Uniform distribution from (0, θ) is:
f(x) = {1/θ for 0 ≤ x ≤ θ and 0 elsewhere
E(x) = θ/2 and this would be the 1st population moment
x̅ would be the first sample moment
setting E(x) = x̅ you would get: theta hat = 2*x̅
Note that E(2*x̅ ) = 2*E(x̅ ) = 2*(θ/2) = θ making it an unbiased estimator
I'm not even sure if that is correct or what I am supposed to be doing.
so much terminology
You have a bunch of i.i.d. samples $Y_n$ all distributed $U[0, \theta]$
You take their sample mean. i.e.
$\bar{Y}=\dfrac 1 N \displaystyle{\sum_{n=1}^N}Y_n$
The expected value of the sample mean is
$E[\bar{Y}]=\dfrac 1 N \displaystyle{\sum_{n=1}^N}E[Y_n]$
$E[\bar{Y}]=\dfrac 1 N \cdot N \left(\dfrac \theta 2\right) = \dfrac \theta 2$
also
$E[Y_n] = \dfrac \theta 2$
so $E[\bar{Y}]=E[Y_n]~\forall n$ and thus the sample mean is an unbiased estimator of $\dfrac \theta 2$ and thus of $\theta$
see if you can show the 2nd estimator is unbiased.
For the second estimator, I have:
Var(x) = (θ^2)/12
E(x^2) = Var(x) + (E(x))^2 = (θ^2)/12 + θ/2 = [7*θ^2]/12 and this would be the 2nd population moment
The 2nd sample moment would be: [Σ(Xi)^2]/n
Setting the 2nd population moment and sample moment equal to each other you get:
[7*θ^2]/12 = [Σ(Xi)^2]/n
solving for θ you get: θ = sqrt(12*Σ(Xi)^2]/7n)
you lost me on this. You just need to find the expected value of the estimate $\dfrac{n+1} n \max(Y_n)$
Can you determine the probability distribution of the max of $n~ U[0,\theta]$ random variables?
hint: $Pr[\max(Y_n) < y] = Pr[\text{all }Y_n < y] = \left(\dfrac y \theta \right)^n$
This is what I did:
P[MAX(Yn) ≤ y] = P[y1, y2, ... , yn ≤ y] = (y/θ)^n
f(MAX(Yn)) = n/[θ*(y/θ)^(n-1)]
E[((n+1)/n)*MAX ] = integral from 0 to θ of [(y*(n+1))/n] * f(MAX(Yn)) dy
=> integral from 0 to θ of (n+1)*(y/θ)^n dy
=> (1/θ^n)*θ^(n+1) = θ
So estimator 2 would be an unbiased estimator
To calculate var(y), we need to calculate E(y^2)
Integral[0, θ] (1/θ)* y^2 dy = (1/(3θ)) y^3 * E[0,θ] = 1/(3θ)( θ^3 - 0) = θ^(2/3)
Then, var(y) = E(y^2) - (E(y))^2 = theta^2/3 - (theta/2)^2 = theta^2/3 - theta^2/4 = theta^2/12
var(y-bar) = theta^2/12n
Then, as estimator 1 = 2 y-bar, var(2 y-bar) = 4 theta^2/12 n = theta^2/3n
For the var(estimator2), first we calculate E(estimator^2) =
I[0, theta] y^2((n+1)/n)^2 n/theta(y/theta)^n-1 dy =
(n+1)^2/(n theta^n) I[0, theta] y^n+1 dy =
(n+1)^2/(n(n+2) theta^n) y^(n+2) E[0, theta] =
(n+1)^2/(n(n+2) theta^n)(theta^(n+2)-0) =
(n+1)^2/(n(n+2) theta^2
Then, var(estimator 2) = (n+1)^2/(n(n+2) theta^2 - theta^2 = theta^2/(n(n+2))
Then, eff estimator 1/eff estimator 2 for unbiased estimators = var estimator 2/var estimator 1 =
theta^2/(n(n+2))/(theta^2/3n) =
3/(n+2)
3/(n+2) < 1 for n > 1
As the efficiency of estimator 1 relative to estimator 2 is less than 1 for all n > 1, estimator 2 is preferred for n > 1
it's so hard to read your math.
I get
$E[\left(2\bar{Y}-\theta\right)^2] = \dfrac {n }{3}\theta^2$
$E[\left(\max(Y_n)-\theta\right)^2] =\left(\dfrac{n+1}{n}\right)^2 \dfrac{n \theta^2}{n+2}=\dfrac{(n+1)^2}{n(n+2)}\theta^2$
$e=\dfrac{\dfrac{(n+1)^2}{n(n+2)}}{\dfrac {n}{3}}= \dfrac{3(n+1)^2}{n^2(n+2)}$
$e < 1~\forall n>4$
So as $n>4$ the statistic $\dfrac{n+1}{n}\max(Y_n)$ becomes more efficient than $2\bar{Y}$.
You should probably double check this.