How to work out the variance of this estimator?
$begingroup$
We want the variance of this biased estimator.
The answer is: $$σ^2/4n$$
Why is it not this? $$σ^2/2n$$
I.e. wouldn't it be $$1/2n^2 * n * σ^2$$
Any clarification appreciated.
Thanks!
statistics variance parameter-estimation
asked May 20 '18 at 17:09
MathsHelpMathsHelp
16810
$endgroup$
add a comment |
$begingroup$
We want the variance of this biased estimator.
The answer is: $$σ^2/4n$$
Why is it not this? $$σ^2/2n$$
I.e. wouldn't it be $$1/2n^2 * n * σ^2$$
Any clarification appreciated.
Thanks!
statistics variance parameter-estimation
asked May 20 '18 at 17:09
MathsHelpMathsHelp
16810
$endgroup$
add a comment |
$begingroup$
We want the variance of this biased estimator.
The answer is: $$σ^2/4n$$
Why is it not this? $$σ^2/2n$$
I.e. wouldn't it be $$1/2n^2 * n * σ^2$$
Any clarification appreciated.
Thanks!
statistics variance parameter-estimation
asked May 20 '18 at 17:09
MathsHelpMathsHelp
16810
$endgroup$
We want the variance of this biased estimator.
The answer is: $$σ^2/4n$$
Why is it not this? $$σ^2/2n$$
I.e. wouldn't it be $$1/2n^2 * n * σ^2$$
Any clarification appreciated.
Thanks!
statistics variance parameter-estimation
statistics variance parameter-estimation
asked May 20 '18 at 17:09
MathsHelpMathsHelp
16810
asked May 20 '18 at 17:09
MathsHelpMathsHelp
16810
asked May 20 '18 at 17:09
MathsHelpMathsHelp
16810
asked May 20 '18 at 17:09
MathsHelpMathsHelp
16810
asked May 20 '18 at 17:09
MathsHelpMathsHelp
16810
16810
add a comment |
add a comment |
1 Answer
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$begingroup$
Note that, if $lambdain Bbb R$, then $text{Var}(lambda X) = lambda^{bf{2}}text{Var}(X)$, not $lambdatext{Var}(X)$.
Thus
$$text{Var}(hat{theta}) = left(frac{1}{2n}right)^2.text{Var}left(sum_{i=1}^n X_iright) = frac{1}{4n^2}.n.sigma^2 = frac{sigma^2}{4n}.$$
answered May 20 '18 at 17:22
pafpaf
4,0281824
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Note that, if $lambdain Bbb R$, then $text{Var}(lambda X) = lambda^{bf{2}}text{Var}(X)$, not $lambdatext{Var}(X)$.
Thus
$$text{Var}(hat{theta}) = left(frac{1}{2n}right)^2.text{Var}left(sum_{i=1}^n X_iright) = frac{1}{4n^2}.n.sigma^2 = frac{sigma^2}{4n}.$$
answered May 20 '18 at 17:22
pafpaf
4,0281824
$endgroup$
add a comment |
$begingroup$
Note that, if $lambdain Bbb R$, then $text{Var}(lambda X) = lambda^{bf{2}}text{Var}(X)$, not $lambdatext{Var}(X)$.
Thus
$$text{Var}(hat{theta}) = left(frac{1}{2n}right)^2.text{Var}left(sum_{i=1}^n X_iright) = frac{1}{4n^2}.n.sigma^2 = frac{sigma^2}{4n}.$$
answered May 20 '18 at 17:22
pafpaf
4,0281824
$endgroup$
add a comment |
$begingroup$
Note that, if $lambdain Bbb R$, then $text{Var}(lambda X) = lambda^{bf{2}}text{Var}(X)$, not $lambdatext{Var}(X)$.
Thus
$$text{Var}(hat{theta}) = left(frac{1}{2n}right)^2.text{Var}left(sum_{i=1}^n X_iright) = frac{1}{4n^2}.n.sigma^2 = frac{sigma^2}{4n}.$$
answered May 20 '18 at 17:22
pafpaf
4,0281824
$endgroup$
Note that, if $lambdain Bbb R$, then $text{Var}(lambda X) = lambda^{bf{2}}text{Var}(X)$, not $lambdatext{Var}(X)$.
Thus
$$text{Var}(hat{theta}) = left(frac{1}{2n}right)^2.text{Var}left(sum_{i=1}^n X_iright) = frac{1}{4n^2}.n.sigma^2 = frac{sigma^2}{4n}.$$
answered May 20 '18 at 17:22
pafpaf
4,0281824
edited Jan 9 at 6:01
answered May 20 '18 at 17:22
pafpaf
4,0281824
answered May 20 '18 at 17:22
pafpaf
4,0281824
answered May 20 '18 at 17:22
pafpaf
4,0281824
4,0281824
add a comment |
add a comment |
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