which of the two estimators is better?
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Given a random size 3 sample of a population with mean $μ$ and variance $s^2$, it was choose two estimators from $μ$:
$bar{theta_1}=bar{x}$ and $bar{theta_2}=frac{x_1+3x_2+x_3}{5}$
in terms of risk. which of the two estimators is better?
My attempt:
We know
$E[bar{x}]=E[sum_{i=1}^nfrac{x_i}{n}]=mu$
$E[frac{X_1+3X_2+X_3}{5}]=frac{1}{5}E[X_1]+3E[X_2]+E[X_3]=mu$
So our two estimators are both unbiased estimators of $μ.$
Moreover,
$V[bar{x}]=V[sum_{i=1}^nfrac{x_i}{n}]=frac{sigma^2}{n}$
$V[frac{X_1+3X_2+X_3}{5}]=frac{1}{25}[sigma^2+9sigma^2+sigma^2]=frac{11sigma^2}{25}$
Here my doubt. depend of valor of $n$ the answer to this question. no?
probability
asked Nov 25 at 20:31
Bvss12
1,724617
add a comment |
up vote
0
down vote
favorite
Given a random size 3 sample of a population with mean $μ$ and variance $s^2$, it was choose two estimators from $μ$:
$bar{theta_1}=bar{x}$ and $bar{theta_2}=frac{x_1+3x_2+x_3}{5}$
in terms of risk. which of the two estimators is better?
My attempt:
We know
$E[bar{x}]=E[sum_{i=1}^nfrac{x_i}{n}]=mu$
$E[frac{X_1+3X_2+X_3}{5}]=frac{1}{5}E[X_1]+3E[X_2]+E[X_3]=mu$
So our two estimators are both unbiased estimators of $μ.$
Moreover,
$V[bar{x}]=V[sum_{i=1}^nfrac{x_i}{n}]=frac{sigma^2}{n}$
$V[frac{X_1+3X_2+X_3}{5}]=frac{1}{25}[sigma^2+9sigma^2+sigma^2]=frac{11sigma^2}{25}$
Here my doubt. depend of valor of $n$ the answer to this question. no?
probability
asked Nov 25 at 20:31
Bvss12
1,724617
1
I thought $n=3$.
– Ben W
Nov 25 at 20:37
oh, true true thanks :D @BenW
– Bvss12
Nov 25 at 20:42
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given a random size 3 sample of a population with mean $μ$ and variance $s^2$, it was choose two estimators from $μ$:
$bar{theta_1}=bar{x}$ and $bar{theta_2}=frac{x_1+3x_2+x_3}{5}$
in terms of risk. which of the two estimators is better?
My attempt:
We know
$E[bar{x}]=E[sum_{i=1}^nfrac{x_i}{n}]=mu$
$E[frac{X_1+3X_2+X_3}{5}]=frac{1}{5}E[X_1]+3E[X_2]+E[X_3]=mu$
So our two estimators are both unbiased estimators of $μ.$
Moreover,
$V[bar{x}]=V[sum_{i=1}^nfrac{x_i}{n}]=frac{sigma^2}{n}$
$V[frac{X_1+3X_2+X_3}{5}]=frac{1}{25}[sigma^2+9sigma^2+sigma^2]=frac{11sigma^2}{25}$
Here my doubt. depend of valor of $n$ the answer to this question. no?
probability
asked Nov 25 at 20:31
Bvss12
1,724617
Given a random size 3 sample of a population with mean $μ$ and variance $s^2$, it was choose two estimators from $μ$:
$bar{theta_1}=bar{x}$ and $bar{theta_2}=frac{x_1+3x_2+x_3}{5}$
in terms of risk. which of the two estimators is better?
My attempt:
We know
$E[bar{x}]=E[sum_{i=1}^nfrac{x_i}{n}]=mu$
$E[frac{X_1+3X_2+X_3}{5}]=frac{1}{5}E[X_1]+3E[X_2]+E[X_3]=mu$
So our two estimators are both unbiased estimators of $μ.$
Moreover,
$V[bar{x}]=V[sum_{i=1}^nfrac{x_i}{n}]=frac{sigma^2}{n}$
$V[frac{X_1+3X_2+X_3}{5}]=frac{1}{25}[sigma^2+9sigma^2+sigma^2]=frac{11sigma^2}{25}$
Here my doubt. depend of valor of $n$ the answer to this question. no?
probability
probability
asked Nov 25 at 20:31
Bvss12
1,724617
asked Nov 25 at 20:31
Bvss12
1,724617
asked Nov 25 at 20:31
Bvss12
1,724617
asked Nov 25 at 20:31
Bvss12
1,724617
asked Nov 25 at 20:31
Bvss12
1,724617
1,724617
1
I thought $n=3$.
– Ben W
Nov 25 at 20:37
oh, true true thanks :D @BenW
– Bvss12
Nov 25 at 20:42
add a comment |
1
I thought $n=3$.
– Ben W
Nov 25 at 20:37
oh, true true thanks :D @BenW
– Bvss12
Nov 25 at 20:42
1
1
I thought $n=3$.
– Ben W
Nov 25 at 20:37
I thought $n=3$.
– Ben W
Nov 25 at 20:37
oh, true true thanks :D @BenW
– Bvss12
Nov 25 at 20:42
oh, true true thanks :D @BenW
– Bvss12
Nov 25 at 20:42
add a comment |
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1
I thought $n=3$.
– Ben W
Nov 25 at 20:37
oh, true true thanks :D @BenW
– Bvss12
Nov 25 at 20:42