Estimation in multivariate normal
Clash Royale CLAN TAG#URR8PPP
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I need to show $underlinehatmu=underlinebary$:
$$L(y_1, y_2, ...) = prod _i=1^n f(y_i,underlinemu,underlineSigma)$$
$$ = prod _i=1^n frac1^1/2e^-frac12(y_i -mu)'Sigma ^-1(y_i -mu) $$
Taking log:
$$= -np logbiggl(sqrt2pibiggl)-fracn2log(Sigma)-frac12sum_i=1^n(y_i -mu)'Sigma ^-1(y_i -mu)$$
Maximising:
$$ 0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
Then, I do not know how to proceed.
I presume
$$ nmu =sum^n_i=1 y_i $$
probability multivariable-calculus normal-distribution
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1
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I need to show $underlinehatmu=underlinebary$:
$$L(y_1, y_2, ...) = prod _i=1^n f(y_i,underlinemu,underlineSigma)$$
$$ = prod _i=1^n frac1^1/2e^-frac12(y_i -mu)'Sigma ^-1(y_i -mu) $$
Taking log:
$$= -np logbiggl(sqrt2pibiggl)-fracn2log(Sigma)-frac12sum_i=1^n(y_i -mu)'Sigma ^-1(y_i -mu)$$
Maximising:
$$ 0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
Then, I do not know how to proceed.
I presume
$$ nmu =sum^n_i=1 y_i $$
probability multivariable-calculus normal-distribution
1
After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g.log
, oroperatornamename
for operators that don't have a command of their own.
â joriki
Sep 3 at 11:47
1
Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
â Tos Hina
Sep 3 at 18:34
2
It looks like the last statement immediately follows from the previous. What are you looking for?
â herb steinberg
Sep 3 at 18:37
2
It would help if you explicitly defined $hatmu$ and $bary$.
â herb steinberg
Sep 3 at 19:25
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to show $underlinehatmu=underlinebary$:
$$L(y_1, y_2, ...) = prod _i=1^n f(y_i,underlinemu,underlineSigma)$$
$$ = prod _i=1^n frac1^1/2e^-frac12(y_i -mu)'Sigma ^-1(y_i -mu) $$
Taking log:
$$= -np logbiggl(sqrt2pibiggl)-fracn2log(Sigma)-frac12sum_i=1^n(y_i -mu)'Sigma ^-1(y_i -mu)$$
Maximising:
$$ 0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
Then, I do not know how to proceed.
I presume
$$ nmu =sum^n_i=1 y_i $$
probability multivariable-calculus normal-distribution
I need to show $underlinehatmu=underlinebary$:
$$L(y_1, y_2, ...) = prod _i=1^n f(y_i,underlinemu,underlineSigma)$$
$$ = prod _i=1^n frac1^1/2e^-frac12(y_i -mu)'Sigma ^-1(y_i -mu) $$
Taking log:
$$= -np logbiggl(sqrt2pibiggl)-fracn2log(Sigma)-frac12sum_i=1^n(y_i -mu)'Sigma ^-1(y_i -mu)$$
Maximising:
$$ 0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
Then, I do not know how to proceed.
I presume
$$ nmu =sum^n_i=1 y_i $$
probability multivariable-calculus normal-distribution
probability multivariable-calculus normal-distribution
edited Sep 3 at 18:32
asked Sep 3 at 10:45
Tos Hina
1,027518
1,027518
1
After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g.log
, oroperatornamename
for operators that don't have a command of their own.
â joriki
Sep 3 at 11:47
1
Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
â Tos Hina
Sep 3 at 18:34
2
It looks like the last statement immediately follows from the previous. What are you looking for?
â herb steinberg
Sep 3 at 18:37
2
It would help if you explicitly defined $hatmu$ and $bary$.
â herb steinberg
Sep 3 at 19:25
add a comment |Â
1
After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g.log
, oroperatornamename
for operators that don't have a command of their own.
â joriki
Sep 3 at 11:47
1
Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
â Tos Hina
Sep 3 at 18:34
2
It looks like the last statement immediately follows from the previous. What are you looking for?
â herb steinberg
Sep 3 at 18:37
2
It would help if you explicitly defined $hatmu$ and $bary$.
â herb steinberg
Sep 3 at 19:25
1
1
After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g.
log
, or operatornamename
for operators that don't have a command of their own.â joriki
Sep 3 at 11:47
After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g.
log
, or operatornamename
for operators that don't have a command of their own.â joriki
Sep 3 at 11:47
1
1
Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
â Tos Hina
Sep 3 at 18:34
Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
â Tos Hina
Sep 3 at 18:34
2
2
It looks like the last statement immediately follows from the previous. What are you looking for?
â herb steinberg
Sep 3 at 18:37
It looks like the last statement immediately follows from the previous. What are you looking for?
â herb steinberg
Sep 3 at 18:37
2
2
It would help if you explicitly defined $hatmu$ and $bary$.
â herb steinberg
Sep 3 at 19:25
It would help if you explicitly defined $hatmu$ and $bary$.
â herb steinberg
Sep 3 at 19:25
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
You have done everything correct. You've reached
$$0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
or more formally (since your maximum is attained at the MLE)
$$0 = frac-12 times 2 sum^n_i=1 (y_i - hatmu)$$
where $hatmu$ is your MLE. Then
$$0 = sum^n_i=1 (y_i - hatmu)$$
i.e.
$$0 = sum^n_i=1 y_i -sum^n_i=1 hatmu$$
But $hatmu$ is a constant being summed up $n$ times, so
$$0 = sum^n_i=1 y_i - n hatmu$$
Finally,
$$hatmu = frac1n sum^n_i=1 y_i = bary$$
is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $mu$.
add a comment |Â
up vote
1
down vote
Recall that for a vector $x in mathbbR ^ p$, and real symmetric matrix $A$,
$$
fracpartialpartial x x ^ T A x = 2 A x,
$$
hence,
$$
fracpartialpartial mu ln L ( cdot ; mu, Sigma )
=
sum frac12fracpartialpartial mu(y_i-mu)^TSigma^-1 (y_i-mu) = - Sigma^-1 sum_i=1^n(y_i-mu) = 0,
$$
multiplying by $Sigma$ and re-arranging the equation you get
$$
hatmu = frac1nsum_i=1^n y_i.
$$
Don't forget that $y_i$ are vectors, hence $hatmu$ is $p times 1$ vector of sample means.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have done everything correct. You've reached
$$0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
or more formally (since your maximum is attained at the MLE)
$$0 = frac-12 times 2 sum^n_i=1 (y_i - hatmu)$$
where $hatmu$ is your MLE. Then
$$0 = sum^n_i=1 (y_i - hatmu)$$
i.e.
$$0 = sum^n_i=1 y_i -sum^n_i=1 hatmu$$
But $hatmu$ is a constant being summed up $n$ times, so
$$0 = sum^n_i=1 y_i - n hatmu$$
Finally,
$$hatmu = frac1n sum^n_i=1 y_i = bary$$
is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $mu$.
add a comment |Â
up vote
1
down vote
accepted
You have done everything correct. You've reached
$$0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
or more formally (since your maximum is attained at the MLE)
$$0 = frac-12 times 2 sum^n_i=1 (y_i - hatmu)$$
where $hatmu$ is your MLE. Then
$$0 = sum^n_i=1 (y_i - hatmu)$$
i.e.
$$0 = sum^n_i=1 y_i -sum^n_i=1 hatmu$$
But $hatmu$ is a constant being summed up $n$ times, so
$$0 = sum^n_i=1 y_i - n hatmu$$
Finally,
$$hatmu = frac1n sum^n_i=1 y_i = bary$$
is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $mu$.
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have done everything correct. You've reached
$$0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
or more formally (since your maximum is attained at the MLE)
$$0 = frac-12 times 2 sum^n_i=1 (y_i - hatmu)$$
where $hatmu$ is your MLE. Then
$$0 = sum^n_i=1 (y_i - hatmu)$$
i.e.
$$0 = sum^n_i=1 y_i -sum^n_i=1 hatmu$$
But $hatmu$ is a constant being summed up $n$ times, so
$$0 = sum^n_i=1 y_i - n hatmu$$
Finally,
$$hatmu = frac1n sum^n_i=1 y_i = bary$$
is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $mu$.
You have done everything correct. You've reached
$$0 = frac-12 times 2 sum^n_i=1 (y_i -mu)$$
or more formally (since your maximum is attained at the MLE)
$$0 = frac-12 times 2 sum^n_i=1 (y_i - hatmu)$$
where $hatmu$ is your MLE. Then
$$0 = sum^n_i=1 (y_i - hatmu)$$
i.e.
$$0 = sum^n_i=1 y_i -sum^n_i=1 hatmu$$
But $hatmu$ is a constant being summed up $n$ times, so
$$0 = sum^n_i=1 y_i - n hatmu$$
Finally,
$$hatmu = frac1n sum^n_i=1 y_i = bary$$
is the MLE estimate, i.e. the sample mean of the data is the MLE estimate of $mu$.
answered Sep 3 at 19:52
Ahmad Bazzi
5,7791623
5,7791623
add a comment |Â
add a comment |Â
up vote
1
down vote
Recall that for a vector $x in mathbbR ^ p$, and real symmetric matrix $A$,
$$
fracpartialpartial x x ^ T A x = 2 A x,
$$
hence,
$$
fracpartialpartial mu ln L ( cdot ; mu, Sigma )
=
sum frac12fracpartialpartial mu(y_i-mu)^TSigma^-1 (y_i-mu) = - Sigma^-1 sum_i=1^n(y_i-mu) = 0,
$$
multiplying by $Sigma$ and re-arranging the equation you get
$$
hatmu = frac1nsum_i=1^n y_i.
$$
Don't forget that $y_i$ are vectors, hence $hatmu$ is $p times 1$ vector of sample means.
add a comment |Â
up vote
1
down vote
Recall that for a vector $x in mathbbR ^ p$, and real symmetric matrix $A$,
$$
fracpartialpartial x x ^ T A x = 2 A x,
$$
hence,
$$
fracpartialpartial mu ln L ( cdot ; mu, Sigma )
=
sum frac12fracpartialpartial mu(y_i-mu)^TSigma^-1 (y_i-mu) = - Sigma^-1 sum_i=1^n(y_i-mu) = 0,
$$
multiplying by $Sigma$ and re-arranging the equation you get
$$
hatmu = frac1nsum_i=1^n y_i.
$$
Don't forget that $y_i$ are vectors, hence $hatmu$ is $p times 1$ vector of sample means.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Recall that for a vector $x in mathbbR ^ p$, and real symmetric matrix $A$,
$$
fracpartialpartial x x ^ T A x = 2 A x,
$$
hence,
$$
fracpartialpartial mu ln L ( cdot ; mu, Sigma )
=
sum frac12fracpartialpartial mu(y_i-mu)^TSigma^-1 (y_i-mu) = - Sigma^-1 sum_i=1^n(y_i-mu) = 0,
$$
multiplying by $Sigma$ and re-arranging the equation you get
$$
hatmu = frac1nsum_i=1^n y_i.
$$
Don't forget that $y_i$ are vectors, hence $hatmu$ is $p times 1$ vector of sample means.
Recall that for a vector $x in mathbbR ^ p$, and real symmetric matrix $A$,
$$
fracpartialpartial x x ^ T A x = 2 A x,
$$
hence,
$$
fracpartialpartial mu ln L ( cdot ; mu, Sigma )
=
sum frac12fracpartialpartial mu(y_i-mu)^TSigma^-1 (y_i-mu) = - Sigma^-1 sum_i=1^n(y_i-mu) = 0,
$$
multiplying by $Sigma$ and re-arranging the equation you get
$$
hatmu = frac1nsum_i=1^n y_i.
$$
Don't forget that $y_i$ are vectors, hence $hatmu$ is $p times 1$ vector of sample means.
answered Sep 3 at 19:45
V. Vancak
10.1k2926
10.1k2926
add a comment |Â
add a comment |Â
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1
After $116$ questions and $65$ answers, I would thought you'd have had enough opportunity to familiarize yourself with how we typeset math on this site. Please see this tutorial and reference. In particular, double dollar signs yield displayed equations, which are much easier to read, especially when they heavily mix fractions, subscripts, superscripts and radicals as in your case; and operator names should be in Roman font, using e.g.
log
, oroperatornamename
for operators that don't have a command of their own.â joriki
Sep 3 at 11:47
1
Advice is well received, and editing done also. Sometimes a bit of sarcasm can change so much. :P
â Tos Hina
Sep 3 at 18:34
2
It looks like the last statement immediately follows from the previous. What are you looking for?
â herb steinberg
Sep 3 at 18:37
2
It would help if you explicitly defined $hatmu$ and $bary$.
â herb steinberg
Sep 3 at 19:25