Unbiased estimator of population variance?
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Can you help me prove whether or not $$S^2 = frac1nsum_i=1^n(X_i - mu)^2$$ is an unbiased estimator for the population variance $sigma^2$ if the mean $mu$ of the population is known?
[Before edit: Sorry about bad formula, I don't know how to type it the fancy way and I'm in a hurry, hope you'll get it.]
This my first question here.
Thanks
statistics
edited Aug 24 at 1:16
BruceET
33.7k71440
asked Aug 23 at 8:23
Nikola Kojadinović
1
 |Â
show 4 more comments
up vote
0
down vote
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Can you help me prove whether or not $$S^2 = frac1nsum_i=1^n(X_i - mu)^2$$ is an unbiased estimator for the population variance $sigma^2$ if the mean $mu$ of the population is known?
[Before edit: Sorry about bad formula, I don't know how to type it the fancy way and I'm in a hurry, hope you'll get it.]
This my first question here.
Thanks
statistics
edited Aug 24 at 1:16
BruceET
33.7k71440
asked Aug 23 at 8:23
Nikola Kojadinović
1
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 23 at 8:32
Yes, according to this formula $S^2$ is an unbiased estimator of the population variance $sigma^2.$ Hints for one method: What is $E[(fracX_i - musigma)^2]?$ What is $E[sum_i(fracX_i - musigma)^2]?$
– BruceET
Aug 24 at 1:19
@BruceET: Is it? Shouldn't it be $n-1$ in the denominator?
– joriki
Aug 24 at 4:53
1
@joriki. No, not when the population mean $mu$ is known. Here, for normal data, $nS^2/sigma^2 sim CHISQ(n),$ This is nonstandard use of notation $S^2.$ // By contrast, in the usual definition, $S^2 =frac1n-1sum_i(X_i - bar X)^2,$ but that's when $mu$ is estimated by $bar X.$ Then (for normal data) $(n-1)S^2/sigma^2 sim CHISQ(n-1).$
– BruceET
Aug 24 at 6:22
@joriki (cont.) With either def'n of $S^2.$ one has $E(S^2) = sigma^2,$ even for nonnormal data as long as $sigma^2$ exists. // Proof is somewhat easier in this case where $mu$ is known.
– BruceET
Aug 24 at 6:32
 |Â
show 4 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can you help me prove whether or not $$S^2 = frac1nsum_i=1^n(X_i - mu)^2$$ is an unbiased estimator for the population variance $sigma^2$ if the mean $mu$ of the population is known?
[Before edit: Sorry about bad formula, I don't know how to type it the fancy way and I'm in a hurry, hope you'll get it.]
This my first question here.
Thanks
statistics
edited Aug 24 at 1:16
BruceET
33.7k71440
asked Aug 23 at 8:23
Nikola Kojadinović
1
Can you help me prove whether or not $$S^2 = frac1nsum_i=1^n(X_i - mu)^2$$ is an unbiased estimator for the population variance $sigma^2$ if the mean $mu$ of the population is known?
[Before edit: Sorry about bad formula, I don't know how to type it the fancy way and I'm in a hurry, hope you'll get it.]
This my first question here.
Thanks
statistics
edited Aug 24 at 1:16
BruceET
33.7k71440
asked Aug 23 at 8:23
Nikola Kojadinović
1
edited Aug 24 at 1:16
BruceET
33.7k71440
edited Aug 24 at 1:16
BruceET
33.7k71440
edited Aug 24 at 1:16
BruceET
33.7k71440
33.7k71440
asked Aug 23 at 8:23
Nikola Kojadinović
1
asked Aug 23 at 8:23
Nikola Kojadinović
1
asked Aug 23 at 8:23
Nikola Kojadinović
1
1
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 23 at 8:32
Yes, according to this formula $S^2$ is an unbiased estimator of the population variance $sigma^2.$ Hints for one method: What is $E[(fracX_i - musigma)^2]?$ What is $E[sum_i(fracX_i - musigma)^2]?$
– BruceET
Aug 24 at 1:19
@BruceET: Is it? Shouldn't it be $n-1$ in the denominator?
– joriki
Aug 24 at 4:53
1
@joriki. No, not when the population mean $mu$ is known. Here, for normal data, $nS^2/sigma^2 sim CHISQ(n),$ This is nonstandard use of notation $S^2.$ // By contrast, in the usual definition, $S^2 =frac1n-1sum_i(X_i - bar X)^2,$ but that's when $mu$ is estimated by $bar X.$ Then (for normal data) $(n-1)S^2/sigma^2 sim CHISQ(n-1).$
– BruceET
Aug 24 at 6:22
@joriki (cont.) With either def'n of $S^2.$ one has $E(S^2) = sigma^2,$ even for nonnormal data as long as $sigma^2$ exists. // Proof is somewhat easier in this case where $mu$ is known.
– BruceET
Aug 24 at 6:32
 |Â
show 4 more comments
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 23 at 8:32
Yes, according to this formula $S^2$ is an unbiased estimator of the population variance $sigma^2.$ Hints for one method: What is $E[(fracX_i - musigma)^2]?$ What is $E[sum_i(fracX_i - musigma)^2]?$
– BruceET
Aug 24 at 1:19
@BruceET: Is it? Shouldn't it be $n-1$ in the denominator?
– joriki
Aug 24 at 4:53
1
@joriki. No, not when the population mean $mu$ is known. Here, for normal data, $nS^2/sigma^2 sim CHISQ(n),$ This is nonstandard use of notation $S^2.$ // By contrast, in the usual definition, $S^2 =frac1n-1sum_i(X_i - bar X)^2,$ but that's when $mu$ is estimated by $bar X.$ Then (for normal data) $(n-1)S^2/sigma^2 sim CHISQ(n-1).$
– BruceET
Aug 24 at 6:22
@joriki (cont.) With either def'n of $S^2.$ one has $E(S^2) = sigma^2,$ even for nonnormal data as long as $sigma^2$ exists. // Proof is somewhat easier in this case where $mu$ is known.
– BruceET
Aug 24 at 6:32
2
2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 23 at 8:32
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 23 at 8:32
Yes, according to this formula $S^2$ is an unbiased estimator of the population variance $sigma^2.$ Hints for one method: What is $E[(fracX_i - musigma)^2]?$ What is $E[sum_i(fracX_i - musigma)^2]?$
– BruceET
Aug 24 at 1:19
Yes, according to this formula $S^2$ is an unbiased estimator of the population variance $sigma^2.$ Hints for one method: What is $E[(fracX_i - musigma)^2]?$ What is $E[sum_i(fracX_i - musigma)^2]?$
– BruceET
Aug 24 at 1:19
@BruceET: Is it? Shouldn't it be $n-1$ in the denominator?
– joriki
Aug 24 at 4:53
@BruceET: Is it? Shouldn't it be $n-1$ in the denominator?
– joriki
Aug 24 at 4:53
1
1
@joriki. No, not when the population mean $mu$ is known. Here, for normal data, $nS^2/sigma^2 sim CHISQ(n),$ This is nonstandard use of notation $S^2.$ // By contrast, in the usual definition, $S^2 =frac1n-1sum_i(X_i - bar X)^2,$ but that's when $mu$ is estimated by $bar X.$ Then (for normal data) $(n-1)S^2/sigma^2 sim CHISQ(n-1).$
– BruceET
Aug 24 at 6:22
@joriki. No, not when the population mean $mu$ is known. Here, for normal data, $nS^2/sigma^2 sim CHISQ(n),$ This is nonstandard use of notation $S^2.$ // By contrast, in the usual definition, $S^2 =frac1n-1sum_i(X_i - bar X)^2,$ but that's when $mu$ is estimated by $bar X.$ Then (for normal data) $(n-1)S^2/sigma^2 sim CHISQ(n-1).$
– BruceET
Aug 24 at 6:22
@joriki (cont.) With either def'n of $S^2.$ one has $E(S^2) = sigma^2,$ even for nonnormal data as long as $sigma^2$ exists. // Proof is somewhat easier in this case where $mu$ is known.
– BruceET
Aug 24 at 6:32
@joriki (cont.) With either def'n of $S^2.$ one has $E(S^2) = sigma^2,$ even for nonnormal data as long as $sigma^2$ exists. // Proof is somewhat easier in this case where $mu$ is known.
– BruceET
Aug 24 at 6:32
 |Â
show 4 more comments
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2
Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
Aug 23 at 8:32
Yes, according to this formula $S^2$ is an unbiased estimator of the population variance $sigma^2.$ Hints for one method: What is $E[(fracX_i - musigma)^2]?$ What is $E[sum_i(fracX_i - musigma)^2]?$
– BruceET
Aug 24 at 1:19
@BruceET: Is it? Shouldn't it be $n-1$ in the denominator?
– joriki
Aug 24 at 4:53
1
@joriki. No, not when the population mean $mu$ is known. Here, for normal data, $nS^2/sigma^2 sim CHISQ(n),$ This is nonstandard use of notation $S^2.$ // By contrast, in the usual definition, $S^2 =frac1n-1sum_i(X_i - bar X)^2,$ but that's when $mu$ is estimated by $bar X.$ Then (for normal data) $(n-1)S^2/sigma^2 sim CHISQ(n-1).$
– BruceET
Aug 24 at 6:22
@joriki (cont.) With either def'n of $S^2.$ one has $E(S^2) = sigma^2,$ even for nonnormal data as long as $sigma^2$ exists. // Proof is somewhat easier in this case where $mu$ is known.
– BruceET
Aug 24 at 6:32