Why does the function VarianceMLE give a different result from Variance?
$begingroup$
Why does the function VarianceMLE give a different result from Variance?
And what is it in Mathematica 11.3?
Please see the picture above t665he MLE is 621 and the other is 665.
probability-or-statistics
edited Jan 14 at 8:01
m_goldberg
85k872196
asked Jan 14 at 7:24
FacetFacet
254
$endgroup$
add a comment |
$begingroup$
Why does the function VarianceMLE give a different result from Variance?
And what is it in Mathematica 11.3?
Please see the picture above t665he MLE is 621 and the other is 665.
probability-or-statistics
edited Jan 14 at 8:01
m_goldberg
85k872196
asked Jan 14 at 7:24
FacetFacet
254
$endgroup$
$begingroup$
In Mathematica 11.3,<< Statistics`produces an error message andVariance@data // Ngives 665.524
$endgroup$
– m_goldberg
Jan 14 at 7:47
2
$begingroup$
I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is thatVariancedivides byN-1(N == Length[data]) while the MLE estimator divides byN.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:22
1
$begingroup$
And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:23
$begingroup$
Which book did you see this in? It looks like a scan.
$endgroup$
– Szabolcs
Jan 14 at 11:45
add a comment |
$begingroup$
Why does the function VarianceMLE give a different result from Variance?
And what is it in Mathematica 11.3?
Please see the picture above t665he MLE is 621 and the other is 665.
probability-or-statistics
edited Jan 14 at 8:01
m_goldberg
85k872196
asked Jan 14 at 7:24
FacetFacet
254
$endgroup$
Why does the function VarianceMLE give a different result from Variance?
And what is it in Mathematica 11.3?
Please see the picture above t665he MLE is 621 and the other is 665.
probability-or-statistics
probability-or-statistics
edited Jan 14 at 8:01
m_goldberg
85k872196
asked Jan 14 at 7:24
FacetFacet
254
edited Jan 14 at 8:01
m_goldberg
85k872196
asked Jan 14 at 7:24
FacetFacet
254
edited Jan 14 at 8:01
m_goldberg
85k872196
edited Jan 14 at 8:01
m_goldberg
85k872196
edited Jan 14 at 8:01
m_goldberg
85k872196
85k872196
asked Jan 14 at 7:24
FacetFacet
254
asked Jan 14 at 7:24
FacetFacet
254
asked Jan 14 at 7:24
FacetFacet
254
254
$begingroup$
In Mathematica 11.3,<< Statistics`produces an error message andVariance@data // Ngives 665.524
$endgroup$
– m_goldberg
Jan 14 at 7:47
2
$begingroup$
I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is thatVariancedivides byN-1(N == Length[data]) while the MLE estimator divides byN.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:22
1
$begingroup$
And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:23
$begingroup$
Which book did you see this in? It looks like a scan.
$endgroup$
– Szabolcs
Jan 14 at 11:45
add a comment |
$begingroup$
In Mathematica 11.3,<< Statistics`produces an error message andVariance@data // Ngives 665.524
$endgroup$
– m_goldberg
Jan 14 at 7:47
2
$begingroup$
I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is thatVariancedivides byN-1(N == Length[data]) while the MLE estimator divides byN.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:22
1
$begingroup$
And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:23
$begingroup$
Which book did you see this in? It looks like a scan.
$endgroup$
– Szabolcs
Jan 14 at 11:45
$begingroup$
In Mathematica 11.3,
<< Statistics` produces an error message and Variance@data // N gives 665.524$endgroup$
– m_goldberg
Jan 14 at 7:47
$begingroup$
In Mathematica 11.3,
<< Statistics` produces an error message and Variance@data // N gives 665.524$endgroup$
– m_goldberg
Jan 14 at 7:47
2
2
$begingroup$
I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is that
Variance divides by N-1 (N == Length[data]) while the MLE estimator divides by N.$endgroup$
– Sjoerd Smit
Jan 14 at 9:22
$begingroup$
I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is that
Variance divides by N-1 (N == Length[data]) while the MLE estimator divides by N.$endgroup$
– Sjoerd Smit
Jan 14 at 9:22
1
1
$begingroup$
And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:23
$begingroup$
And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:23
$begingroup$
Which book did you see this in? It looks like a scan.
$endgroup$
– Szabolcs
Jan 14 at 11:45
$begingroup$
Which book did you see this in? It looks like a scan.
$endgroup$
– Szabolcs
Jan 14 at 11:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).
data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};
Variance[data]
13976/21
myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);
myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];
myVariance[data]
myVarianceMLE[data]
13976/21
27952/45
answered Jan 14 at 9:28
Sjoerd SmitSjoerd Smit
3,410715
$endgroup$
$begingroup$
Thank you for your kind help, I will remember to post code next time.
$endgroup$
– Facet
Jan 14 at 14:47
$begingroup$
Wolfram Research article also mentions a replacement forVarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
$endgroup$
– kirma
Jan 18 at 17:31
add a comment |
$begingroup$
VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.
Let ${y_i : 1 leq i leq n}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is
$$ sigma_y^2 = frac{1}{n} sum_{i=1}^n (y_i - bar{y}) text{,} $$
where $bar{y} = frac{1}{n} sum_{i=1}^n y_i$ is the sample mean. This $sigma_y^2$ is computed by VarianceMLE.
If $sigma^2$ is the population variance, with some work, one can show that the expected value of $sigma_y^2$ is $frac{n-1}{n} sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via
$$ s^2 = frac{n}{n-1} sigma_y^2 = frac{1}{n-1}sum_{i=1}^n (y_i - bar{y}) text{.} $$
This $s^2$ is computed by Variance. From the documentation (in the Details):
"Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."
The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:
"VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"
answered Jan 14 at 15:15
Eric TowersEric Towers
2,306613
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).
data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};
Variance[data]
13976/21
myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);
myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];
myVariance[data]
myVarianceMLE[data]
13976/21
27952/45
answered Jan 14 at 9:28
Sjoerd SmitSjoerd Smit
3,410715
$endgroup$
$begingroup$
Thank you for your kind help, I will remember to post code next time.
$endgroup$
– Facet
Jan 14 at 14:47
$begingroup$
Wolfram Research article also mentions a replacement forVarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
$endgroup$
– kirma
Jan 18 at 17:31
add a comment |
$begingroup$
I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).
data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};
Variance[data]
13976/21
myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);
myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];
myVariance[data]
myVarianceMLE[data]
13976/21
27952/45
answered Jan 14 at 9:28
Sjoerd SmitSjoerd Smit
3,410715
$endgroup$
$begingroup$
Thank you for your kind help, I will remember to post code next time.
$endgroup$
– Facet
Jan 14 at 14:47
$begingroup$
Wolfram Research article also mentions a replacement forVarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
$endgroup$
– kirma
Jan 18 at 17:31
add a comment |
$begingroup$
I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).
data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};
Variance[data]
13976/21
myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);
myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];
myVariance[data]
myVarianceMLE[data]
13976/21
27952/45
answered Jan 14 at 9:28
Sjoerd SmitSjoerd Smit
3,410715
$endgroup$
I just checked my guess in my comment and I was right. VarianceMLE is the maximum likelihood variance estimator (see, e.g. here).
data = {34, 56, 28, 62, 32, 90, 20, 10, 12, 35, 63, 78, 12, 25, 68};
Variance[data]
13976/21
myVariance[lst_List] := Total[(lst - Mean[lst])^2]/(Length[lst] - 1);
myVarianceMLE[lst_List] := Total[(lst - Mean[lst])^2]/Length[lst];
myVariance[data]
myVarianceMLE[data]
13976/21
27952/45
answered Jan 14 at 9:28
Sjoerd SmitSjoerd Smit
3,410715
answered Jan 14 at 9:28
Sjoerd SmitSjoerd Smit
3,410715
answered Jan 14 at 9:28
Sjoerd SmitSjoerd Smit
3,410715
answered Jan 14 at 9:28
Sjoerd SmitSjoerd Smit
3,410715
3,410715
$begingroup$
Thank you for your kind help, I will remember to post code next time.
$endgroup$
– Facet
Jan 14 at 14:47
$begingroup$
Wolfram Research article also mentions a replacement forVarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
$endgroup$
– kirma
Jan 18 at 17:31
add a comment |
$begingroup$
Thank you for your kind help, I will remember to post code next time.
$endgroup$
– Facet
Jan 14 at 14:47
$begingroup$
Wolfram Research article also mentions a replacement forVarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .
$endgroup$
– kirma
Jan 18 at 17:31
$begingroup$
Thank you for your kind help, I will remember to post code next time.
$endgroup$
– Facet
Jan 14 at 14:47
$begingroup$
Thank you for your kind help, I will remember to post code next time.
$endgroup$
– Facet
Jan 14 at 14:47
$begingroup$
Wolfram Research article also mentions a replacement for
VarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .$endgroup$
– kirma
Jan 18 at 17:31
$begingroup$
Wolfram Research article also mentions a replacement for
VarianceMLE: reference.wolfram.com/language/Compatibility/tutorial/… .$endgroup$
– kirma
Jan 18 at 17:31
add a comment |
$begingroup$
VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.
Let ${y_i : 1 leq i leq n}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is
$$ sigma_y^2 = frac{1}{n} sum_{i=1}^n (y_i - bar{y}) text{,} $$
where $bar{y} = frac{1}{n} sum_{i=1}^n y_i$ is the sample mean. This $sigma_y^2$ is computed by VarianceMLE.
If $sigma^2$ is the population variance, with some work, one can show that the expected value of $sigma_y^2$ is $frac{n-1}{n} sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via
$$ s^2 = frac{n}{n-1} sigma_y^2 = frac{1}{n-1}sum_{i=1}^n (y_i - bar{y}) text{.} $$
This $s^2$ is computed by Variance. From the documentation (in the Details):
"Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."
The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:
"VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"
answered Jan 14 at 15:15
Eric TowersEric Towers
2,306613
$endgroup$
add a comment |
$begingroup$
VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.
Let ${y_i : 1 leq i leq n}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is
$$ sigma_y^2 = frac{1}{n} sum_{i=1}^n (y_i - bar{y}) text{,} $$
where $bar{y} = frac{1}{n} sum_{i=1}^n y_i$ is the sample mean. This $sigma_y^2$ is computed by VarianceMLE.
If $sigma^2$ is the population variance, with some work, one can show that the expected value of $sigma_y^2$ is $frac{n-1}{n} sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via
$$ s^2 = frac{n}{n-1} sigma_y^2 = frac{1}{n-1}sum_{i=1}^n (y_i - bar{y}) text{.} $$
This $s^2$ is computed by Variance. From the documentation (in the Details):
"Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."
The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:
"VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"
answered Jan 14 at 15:15
Eric TowersEric Towers
2,306613
$endgroup$
add a comment |
$begingroup$
VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.
Let ${y_i : 1 leq i leq n}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is
$$ sigma_y^2 = frac{1}{n} sum_{i=1}^n (y_i - bar{y}) text{,} $$
where $bar{y} = frac{1}{n} sum_{i=1}^n y_i$ is the sample mean. This $sigma_y^2$ is computed by VarianceMLE.
If $sigma^2$ is the population variance, with some work, one can show that the expected value of $sigma_y^2$ is $frac{n-1}{n} sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via
$$ s^2 = frac{n}{n-1} sigma_y^2 = frac{1}{n-1}sum_{i=1}^n (y_i - bar{y}) text{.} $$
This $s^2$ is computed by Variance. From the documentation (in the Details):
"Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."
The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:
"VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"
answered Jan 14 at 15:15
Eric TowersEric Towers
2,306613
$endgroup$
VarianceMLE computes a biased, maximum likelihood estimate of the population variance. Variance computes an unbiased estimate of the population variance. It can be shown that VarianceMLE underestimates the variance of the population.
Let ${y_i : 1 leq i leq n}$ be a sample of $n$ values from a population. The variance (central second moment) of the sample is
$$ sigma_y^2 = frac{1}{n} sum_{i=1}^n (y_i - bar{y}) text{,} $$
where $bar{y} = frac{1}{n} sum_{i=1}^n y_i$ is the sample mean. This $sigma_y^2$ is computed by VarianceMLE.
If $sigma^2$ is the population variance, with some work, one can show that the expected value of $sigma_y^2$ is $frac{n-1}{n} sigma^2$, so the sample variance is a biased estimator of the population variance. We can make this an unbiased estimator via
$$ s^2 = frac{n}{n-1} sigma_y^2 = frac{1}{n-1}sum_{i=1}^n (y_i - bar{y}) text{.} $$
This $s^2$ is computed by Variance. From the documentation (in the Details):
"Variance[list] is equivalent to Total[(list-Mean[list])^2]/(Length[list]-1) for real-valued data."
The very sparse documentation for VarianceMLE indicates that it is implemented in terms of Variance:
"VarianceMLE[data_] := Variance[data] (Length[data] - 1)/Length[data]"
answered Jan 14 at 15:15
Eric TowersEric Towers
2,306613
answered Jan 14 at 15:15
Eric TowersEric Towers
2,306613
answered Jan 14 at 15:15
Eric TowersEric Towers
2,306613
answered Jan 14 at 15:15
Eric TowersEric Towers
2,306613
2,306613
add a comment |
add a comment |
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$begingroup$
In Mathematica 11.3,
<< Statistics`produces an error message andVariance@data // Ngives 665.524$endgroup$
– m_goldberg
Jan 14 at 7:47
2
$begingroup$
I'm going to guess that VarianceMLE is the maximum likelihood variance estimator rather than the unbiased one (assuming normally distributed data). The difference between the two is that
Variancedivides byN-1(N == Length[data]) while the MLE estimator divides byN.$endgroup$
– Sjoerd Smit
Jan 14 at 9:22
1
$begingroup$
And for future reference: please post copyable code in your question rather than a screenshot. This makes it much easier for someone else to copy your code and try things out.
$endgroup$
– Sjoerd Smit
Jan 14 at 9:23
$begingroup$
Which book did you see this in? It looks like a scan.
$endgroup$
– Szabolcs
Jan 14 at 11:45