comparing the variance of estimators
when we know that both estimators are unbiased, we prefer the one with a smaller variance.
so considering that we want to estimate the population mean using 5 data points, we consider the 2 estimators below:
a. Mu1 = (x1+x2+x3+x4+x5)/5
b. Mu2 = (x1+x2+x3)/3 + x4 - x5
var(Mu1) = σ²/5
var(Mu2) = (7/3)σ²
hence in comparing both estimators, estimator (a) would be better.
however I need help in understanding the math behind getting the variance of both estimators!
variance
asked Nov 25 at 13:28
darrenbarren
1
add a comment |
when we know that both estimators are unbiased, we prefer the one with a smaller variance.
so considering that we want to estimate the population mean using 5 data points, we consider the 2 estimators below:
a. Mu1 = (x1+x2+x3+x4+x5)/5
b. Mu2 = (x1+x2+x3)/3 + x4 - x5
var(Mu1) = σ²/5
var(Mu2) = (7/3)σ²
hence in comparing both estimators, estimator (a) would be better.
however I need help in understanding the math behind getting the variance of both estimators!
variance
asked Nov 25 at 13:28
darrenbarren
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 25 at 13:35
add a comment |
when we know that both estimators are unbiased, we prefer the one with a smaller variance.
so considering that we want to estimate the population mean using 5 data points, we consider the 2 estimators below:
a. Mu1 = (x1+x2+x3+x4+x5)/5
b. Mu2 = (x1+x2+x3)/3 + x4 - x5
var(Mu1) = σ²/5
var(Mu2) = (7/3)σ²
hence in comparing both estimators, estimator (a) would be better.
however I need help in understanding the math behind getting the variance of both estimators!
variance
asked Nov 25 at 13:28
darrenbarren
1
when we know that both estimators are unbiased, we prefer the one with a smaller variance.
so considering that we want to estimate the population mean using 5 data points, we consider the 2 estimators below:
a. Mu1 = (x1+x2+x3+x4+x5)/5
b. Mu2 = (x1+x2+x3)/3 + x4 - x5
var(Mu1) = σ²/5
var(Mu2) = (7/3)σ²
hence in comparing both estimators, estimator (a) would be better.
however I need help in understanding the math behind getting the variance of both estimators!
variance
variance
asked Nov 25 at 13:28
darrenbarren
1
asked Nov 25 at 13:28
darrenbarren
1
asked Nov 25 at 13:28
darrenbarren
1
asked Nov 25 at 13:28
darrenbarren
1
asked Nov 25 at 13:28
darrenbarren
1
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 25 at 13:35
add a comment |
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 25 at 13:35
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 25 at 13:35
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 25 at 13:35
add a comment |
1 Answer
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Assuming independence, we have $$Var(X pm Y) = Var(X)+Var(Y)$$
and also $$Var(kX)=k^2Var(X).$$
Hence,
$$Varleft( frac{sum_{i=1}^5X_i}5right)=frac1{25}sum_{i=1}^5Var(X_i)=frac{sigma^2}5$$
$$Varleft(frac{sum_{i=1}^3X_i}{3} +X_4-X_5right)=frac19sum_{i=1}^3Var(X_i)+Var(X_4)+Var(X_5)=left( frac13+2right)sigma^2$$
answered Nov 25 at 13:38
Siong Thye Goh
98.5k1464116
add a comment |
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1 Answer
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Assuming independence, we have $$Var(X pm Y) = Var(X)+Var(Y)$$
and also $$Var(kX)=k^2Var(X).$$
Hence,
$$Varleft( frac{sum_{i=1}^5X_i}5right)=frac1{25}sum_{i=1}^5Var(X_i)=frac{sigma^2}5$$
$$Varleft(frac{sum_{i=1}^3X_i}{3} +X_4-X_5right)=frac19sum_{i=1}^3Var(X_i)+Var(X_4)+Var(X_5)=left( frac13+2right)sigma^2$$
answered Nov 25 at 13:38
Siong Thye Goh
98.5k1464116
add a comment |
Assuming independence, we have $$Var(X pm Y) = Var(X)+Var(Y)$$
and also $$Var(kX)=k^2Var(X).$$
Hence,
$$Varleft( frac{sum_{i=1}^5X_i}5right)=frac1{25}sum_{i=1}^5Var(X_i)=frac{sigma^2}5$$
$$Varleft(frac{sum_{i=1}^3X_i}{3} +X_4-X_5right)=frac19sum_{i=1}^3Var(X_i)+Var(X_4)+Var(X_5)=left( frac13+2right)sigma^2$$
answered Nov 25 at 13:38
Siong Thye Goh
98.5k1464116
add a comment |
Assuming independence, we have $$Var(X pm Y) = Var(X)+Var(Y)$$
and also $$Var(kX)=k^2Var(X).$$
Hence,
$$Varleft( frac{sum_{i=1}^5X_i}5right)=frac1{25}sum_{i=1}^5Var(X_i)=frac{sigma^2}5$$
$$Varleft(frac{sum_{i=1}^3X_i}{3} +X_4-X_5right)=frac19sum_{i=1}^3Var(X_i)+Var(X_4)+Var(X_5)=left( frac13+2right)sigma^2$$
answered Nov 25 at 13:38
Siong Thye Goh
98.5k1464116
Assuming independence, we have $$Var(X pm Y) = Var(X)+Var(Y)$$
and also $$Var(kX)=k^2Var(X).$$
Hence,
$$Varleft( frac{sum_{i=1}^5X_i}5right)=frac1{25}sum_{i=1}^5Var(X_i)=frac{sigma^2}5$$
$$Varleft(frac{sum_{i=1}^3X_i}{3} +X_4-X_5right)=frac19sum_{i=1}^3Var(X_i)+Var(X_4)+Var(X_5)=left( frac13+2right)sigma^2$$
answered Nov 25 at 13:38
Siong Thye Goh
98.5k1464116
answered Nov 25 at 13:38
Siong Thye Goh
98.5k1464116
answered Nov 25 at 13:38
Siong Thye Goh
98.5k1464116
answered Nov 25 at 13:38
Siong Thye Goh
98.5k1464116
98.5k1464116
add a comment |
add a comment |
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 25 at 13:35