Derive the Cramer-Rao lower bound (CRLB) for the variance of any unbiased for estimator of $lambda$ for random sample from Poisson ($lambda$)
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Let $Y_1,...,Y_n$ be a random sample from Poisson ($lambda$).
Derive the Cramer-Rao lower bound (CRLB) for the variance of any unbiased for estimator of $lambda$.
I first set
$$L(lambda) = prod_i=1^n fraclambda^y e^-lambday! = fraclambda^prod_i = 1^n y_i e^-lambda nprod_i=1^n y_i ! .$$
Then setting it to $ln$ form:
$$ln L(lambda) = fracsum_i=1^n y_i ln(lambda) - lambda nsum_i=1^n y_i !.$$
Can anyone please confirm what I am doing is correct before I move forward?
probability statistics probability-theory probability-distributions
edited Sep 9 at 9:07
Jendrik Stelzner
7,69121137
asked Apr 26 '14 at 20:17
afsdf dfsaf
59611234
add a comment |Â
up vote
0
down vote
favorite
Let $Y_1,...,Y_n$ be a random sample from Poisson ($lambda$).
Derive the Cramer-Rao lower bound (CRLB) for the variance of any unbiased for estimator of $lambda$.
I first set
$$L(lambda) = prod_i=1^n fraclambda^y e^-lambday! = fraclambda^prod_i = 1^n y_i e^-lambda nprod_i=1^n y_i ! .$$
Then setting it to $ln$ form:
$$ln L(lambda) = fracsum_i=1^n y_i ln(lambda) - lambda nsum_i=1^n y_i !.$$
Can anyone please confirm what I am doing is correct before I move forward?
probability statistics probability-theory probability-distributions
edited Sep 9 at 9:07
Jendrik Stelzner
7,69121137
asked Apr 26 '14 at 20:17
afsdf dfsaf
59611234
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $Y_1,...,Y_n$ be a random sample from Poisson ($lambda$).
Derive the Cramer-Rao lower bound (CRLB) for the variance of any unbiased for estimator of $lambda$.
I first set
$$L(lambda) = prod_i=1^n fraclambda^y e^-lambday! = fraclambda^prod_i = 1^n y_i e^-lambda nprod_i=1^n y_i ! .$$
Then setting it to $ln$ form:
$$ln L(lambda) = fracsum_i=1^n y_i ln(lambda) - lambda nsum_i=1^n y_i !.$$
Can anyone please confirm what I am doing is correct before I move forward?
probability statistics probability-theory probability-distributions
edited Sep 9 at 9:07
Jendrik Stelzner
7,69121137
asked Apr 26 '14 at 20:17
afsdf dfsaf
59611234
Let $Y_1,...,Y_n$ be a random sample from Poisson ($lambda$).
Derive the Cramer-Rao lower bound (CRLB) for the variance of any unbiased for estimator of $lambda$.
I first set
$$L(lambda) = prod_i=1^n fraclambda^y e^-lambday! = fraclambda^prod_i = 1^n y_i e^-lambda nprod_i=1^n y_i ! .$$
Then setting it to $ln$ form:
$$ln L(lambda) = fracsum_i=1^n y_i ln(lambda) - lambda nsum_i=1^n y_i !.$$
Can anyone please confirm what I am doing is correct before I move forward?
probability statistics probability-theory probability-distributions
probability statistics probability-theory probability-distributions
edited Sep 9 at 9:07
Jendrik Stelzner
7,69121137
asked Apr 26 '14 at 20:17
afsdf dfsaf
59611234
edited Sep 9 at 9:07
Jendrik Stelzner
7,69121137
asked Apr 26 '14 at 20:17
afsdf dfsaf
59611234
edited Sep 9 at 9:07
Jendrik Stelzner
7,69121137
edited Sep 9 at 9:07
Jendrik Stelzner
7,69121137
edited Sep 9 at 9:07
Jendrik Stelzner
7,69121137
7,69121137
asked Apr 26 '14 at 20:17
afsdf dfsaf
59611234
asked Apr 26 '14 at 20:17
afsdf dfsaf
59611234
asked Apr 26 '14 at 20:17
afsdf dfsaf
59611234
59611234
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The likelihood function is
$$
L(lambda)=left(prod_i=1^n frac1y_i!right)lambda^sum_i=1^n y_ie^-lambda n
$$
and hence the log-likelihood function becomes
$$
l(lambda)=-lambda n+log(lambda)sum_i=1^n y_i-sum_i=1^n log(y_i!).
$$
answered Apr 26 '14 at 20:47
Stefan Hansen
20.3k73460
are the following correct? $$fracd ln L(lambda)lambda = frac1lambda sum_i=1^n y_i - n$$ $$fracd^2 ln L(lambda)lambda = frac-1lambda^2 sum_i=1^n y_i $$ So $$E[frac-1lambda^2 sum_i=1^n y_i] = frac-1lambda^2 nbary$$
– afsdf dfsaf
Apr 26 '14 at 23:37
The mean of $y_i$ is $lambda$ not $y_i$ and hence the last is not correct.
– Stefan Hansen
Apr 27 '14 at 7:47
For my last equation: $$Eleft[frac-1lambda^2 sum_i=1^n y_iright] = frac-1lambda^2 n lambda = frac-nlambda$$. Therefore, the CRLB is $frac-lambdan$, right?
– afsdf dfsaf
Apr 27 '14 at 12:03
No, the Cramer-Rao lower bound for the variance is certainly not negative. You're missing a minus-sign in the expression for the Fisher information.
– Stefan Hansen
Apr 27 '14 at 12:12
so my answer is only off by a minus sign right?
– afsdf dfsaf
Apr 27 '14 at 12:16
 |Â
show 13 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The likelihood function is
$$
L(lambda)=left(prod_i=1^n frac1y_i!right)lambda^sum_i=1^n y_ie^-lambda n
$$
and hence the log-likelihood function becomes
$$
l(lambda)=-lambda n+log(lambda)sum_i=1^n y_i-sum_i=1^n log(y_i!).
$$
answered Apr 26 '14 at 20:47
Stefan Hansen
20.3k73460
are the following correct? $$fracd ln L(lambda)lambda = frac1lambda sum_i=1^n y_i - n$$ $$fracd^2 ln L(lambda)lambda = frac-1lambda^2 sum_i=1^n y_i $$ So $$E[frac-1lambda^2 sum_i=1^n y_i] = frac-1lambda^2 nbary$$
– afsdf dfsaf
Apr 26 '14 at 23:37
The mean of $y_i$ is $lambda$ not $y_i$ and hence the last is not correct.
– Stefan Hansen
Apr 27 '14 at 7:47
For my last equation: $$Eleft[frac-1lambda^2 sum_i=1^n y_iright] = frac-1lambda^2 n lambda = frac-nlambda$$. Therefore, the CRLB is $frac-lambdan$, right?
– afsdf dfsaf
Apr 27 '14 at 12:03
No, the Cramer-Rao lower bound for the variance is certainly not negative. You're missing a minus-sign in the expression for the Fisher information.
– Stefan Hansen
Apr 27 '14 at 12:12
so my answer is only off by a minus sign right?
– afsdf dfsaf
Apr 27 '14 at 12:16
 |Â
show 13 more comments
up vote
2
down vote
accepted
The likelihood function is
$$
L(lambda)=left(prod_i=1^n frac1y_i!right)lambda^sum_i=1^n y_ie^-lambda n
$$
and hence the log-likelihood function becomes
$$
l(lambda)=-lambda n+log(lambda)sum_i=1^n y_i-sum_i=1^n log(y_i!).
$$
answered Apr 26 '14 at 20:47
Stefan Hansen
20.3k73460
are the following correct? $$fracd ln L(lambda)lambda = frac1lambda sum_i=1^n y_i - n$$ $$fracd^2 ln L(lambda)lambda = frac-1lambda^2 sum_i=1^n y_i $$ So $$E[frac-1lambda^2 sum_i=1^n y_i] = frac-1lambda^2 nbary$$
– afsdf dfsaf
Apr 26 '14 at 23:37
The mean of $y_i$ is $lambda$ not $y_i$ and hence the last is not correct.
– Stefan Hansen
Apr 27 '14 at 7:47
For my last equation: $$Eleft[frac-1lambda^2 sum_i=1^n y_iright] = frac-1lambda^2 n lambda = frac-nlambda$$. Therefore, the CRLB is $frac-lambdan$, right?
– afsdf dfsaf
Apr 27 '14 at 12:03
No, the Cramer-Rao lower bound for the variance is certainly not negative. You're missing a minus-sign in the expression for the Fisher information.
– Stefan Hansen
Apr 27 '14 at 12:12
so my answer is only off by a minus sign right?
– afsdf dfsaf
Apr 27 '14 at 12:16
 |Â
show 13 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The likelihood function is
$$
L(lambda)=left(prod_i=1^n frac1y_i!right)lambda^sum_i=1^n y_ie^-lambda n
$$
and hence the log-likelihood function becomes
$$
l(lambda)=-lambda n+log(lambda)sum_i=1^n y_i-sum_i=1^n log(y_i!).
$$
answered Apr 26 '14 at 20:47
Stefan Hansen
20.3k73460
The likelihood function is
$$
L(lambda)=left(prod_i=1^n frac1y_i!right)lambda^sum_i=1^n y_ie^-lambda n
$$
and hence the log-likelihood function becomes
$$
l(lambda)=-lambda n+log(lambda)sum_i=1^n y_i-sum_i=1^n log(y_i!).
$$
answered Apr 26 '14 at 20:47
Stefan Hansen
20.3k73460
answered Apr 26 '14 at 20:47
Stefan Hansen
20.3k73460
answered Apr 26 '14 at 20:47
Stefan Hansen
20.3k73460
answered Apr 26 '14 at 20:47
Stefan Hansen
20.3k73460
20.3k73460
are the following correct? $$fracd ln L(lambda)lambda = frac1lambda sum_i=1^n y_i - n$$ $$fracd^2 ln L(lambda)lambda = frac-1lambda^2 sum_i=1^n y_i $$ So $$E[frac-1lambda^2 sum_i=1^n y_i] = frac-1lambda^2 nbary$$
– afsdf dfsaf
Apr 26 '14 at 23:37
The mean of $y_i$ is $lambda$ not $y_i$ and hence the last is not correct.
– Stefan Hansen
Apr 27 '14 at 7:47
For my last equation: $$Eleft[frac-1lambda^2 sum_i=1^n y_iright] = frac-1lambda^2 n lambda = frac-nlambda$$. Therefore, the CRLB is $frac-lambdan$, right?
– afsdf dfsaf
Apr 27 '14 at 12:03
No, the Cramer-Rao lower bound for the variance is certainly not negative. You're missing a minus-sign in the expression for the Fisher information.
– Stefan Hansen
Apr 27 '14 at 12:12
so my answer is only off by a minus sign right?
– afsdf dfsaf
Apr 27 '14 at 12:16
 |Â
show 13 more comments
are the following correct? $$fracd ln L(lambda)lambda = frac1lambda sum_i=1^n y_i - n$$ $$fracd^2 ln L(lambda)lambda = frac-1lambda^2 sum_i=1^n y_i $$ So $$E[frac-1lambda^2 sum_i=1^n y_i] = frac-1lambda^2 nbary$$
– afsdf dfsaf
Apr 26 '14 at 23:37
The mean of $y_i$ is $lambda$ not $y_i$ and hence the last is not correct.
– Stefan Hansen
Apr 27 '14 at 7:47
For my last equation: $$Eleft[frac-1lambda^2 sum_i=1^n y_iright] = frac-1lambda^2 n lambda = frac-nlambda$$. Therefore, the CRLB is $frac-lambdan$, right?
– afsdf dfsaf
Apr 27 '14 at 12:03
No, the Cramer-Rao lower bound for the variance is certainly not negative. You're missing a minus-sign in the expression for the Fisher information.
– Stefan Hansen
Apr 27 '14 at 12:12
so my answer is only off by a minus sign right?
– afsdf dfsaf
Apr 27 '14 at 12:16
are the following correct? $$fracd ln L(lambda)lambda = frac1lambda sum_i=1^n y_i - n$$ $$fracd^2 ln L(lambda)lambda = frac-1lambda^2 sum_i=1^n y_i $$ So $$E[frac-1lambda^2 sum_i=1^n y_i] = frac-1lambda^2 nbary$$
– afsdf dfsaf
Apr 26 '14 at 23:37
are the following correct? $$fracd ln L(lambda)lambda = frac1lambda sum_i=1^n y_i - n$$ $$fracd^2 ln L(lambda)lambda = frac-1lambda^2 sum_i=1^n y_i $$ So $$E[frac-1lambda^2 sum_i=1^n y_i] = frac-1lambda^2 nbary$$
– afsdf dfsaf
Apr 26 '14 at 23:37
The mean of $y_i$ is $lambda$ not $y_i$ and hence the last is not correct.
– Stefan Hansen
Apr 27 '14 at 7:47
The mean of $y_i$ is $lambda$ not $y_i$ and hence the last is not correct.
– Stefan Hansen
Apr 27 '14 at 7:47
For my last equation: $$Eleft[frac-1lambda^2 sum_i=1^n y_iright] = frac-1lambda^2 n lambda = frac-nlambda$$. Therefore, the CRLB is $frac-lambdan$, right?
– afsdf dfsaf
Apr 27 '14 at 12:03
For my last equation: $$Eleft[frac-1lambda^2 sum_i=1^n y_iright] = frac-1lambda^2 n lambda = frac-nlambda$$. Therefore, the CRLB is $frac-lambdan$, right?
– afsdf dfsaf
Apr 27 '14 at 12:03
No, the Cramer-Rao lower bound for the variance is certainly not negative. You're missing a minus-sign in the expression for the Fisher information.
– Stefan Hansen
Apr 27 '14 at 12:12
No, the Cramer-Rao lower bound for the variance is certainly not negative. You're missing a minus-sign in the expression for the Fisher information.
– Stefan Hansen
Apr 27 '14 at 12:12
so my answer is only off by a minus sign right?
– afsdf dfsaf
Apr 27 '14 at 12:16
so my answer is only off by a minus sign right?
– afsdf dfsaf
Apr 27 '14 at 12:16
 |Â
show 13 more comments
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