Normal approximation of MLE of Poisson distribution and confidence intervalConfidence interval for Poisson...
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Normal approximation of MLE of Poisson distribution and confidence interval
Confidence interval for Poisson distribution coefficientConfidence interval of the parameter of $exp$ and normal distribution from MLE?Determine the asymptotic distribution of $bar X_n$, properly centered and $sqrt n$ scaledConfidence interval; exponential distribution (normal or student approximation?)MLE and unbiased estimator of $P{X_{i}=1}$ given poisson distributionMLE of Poisson distribution with new observationconfidence interval when both $mu$ and $sigma^2$ are unknownMLE, Confidence Interval, and Asymptotic DistributionsConfidence interval for mean of non-normal distribution based on a small sampleIs Confidence Interval taken on one Random Sample or A Sampling Distribution
$begingroup$
Let $(X_1,ldots,X_n)$ denote a random sample from a Poisson distribution with parameter $lambda$.
Maximum Likelihood Estimate of $lambda$ is given by
$hat{lambda} = bar{X} = frac{1}{n} sumlimits_{i=1}^n X_i$
a) Show that $frac{bar{X}-mu}{sqrt{lambda/n}}sim N(0,1)$ approx.
b) Using this normal approximation find the theoretical confidence interval with the $1-alpha/2$ quantile from the $N(0,1)$ distribution.
Thoughts: I'm quite stuck on problem a). I don't quite se om the MLE which is equal the the sample mean is relevant here but I am probably (most definitely) missing something. My only idea is to use CLT in some way since
$$frac{bar{X}-mu}{sigma/sqrt{n}}=frac{bar{X}-mu}{sqrt{lambda/n}}$$
as the variance of Poisson is just $lambda$. But from here I don't know what to do.... Can someone help me?
And for problem b) I have 0 ideas...
statistics poisson-distribution central-limit-theorem confidence-interval
edited Mar 15 at 17:52
StubbornAtom
6,30831440
asked Mar 14 at 18:31
CruZCruZ
638
$endgroup$
add a comment |
$begingroup$
Let $(X_1,ldots,X_n)$ denote a random sample from a Poisson distribution with parameter $lambda$.
Maximum Likelihood Estimate of $lambda$ is given by
$hat{lambda} = bar{X} = frac{1}{n} sumlimits_{i=1}^n X_i$
a) Show that $frac{bar{X}-mu}{sqrt{lambda/n}}sim N(0,1)$ approx.
b) Using this normal approximation find the theoretical confidence interval with the $1-alpha/2$ quantile from the $N(0,1)$ distribution.
Thoughts: I'm quite stuck on problem a). I don't quite se om the MLE which is equal the the sample mean is relevant here but I am probably (most definitely) missing something. My only idea is to use CLT in some way since
$$frac{bar{X}-mu}{sigma/sqrt{n}}=frac{bar{X}-mu}{sqrt{lambda/n}}$$
as the variance of Poisson is just $lambda$. But from here I don't know what to do.... Can someone help me?
And for problem b) I have 0 ideas...
statistics poisson-distribution central-limit-theorem confidence-interval
edited Mar 15 at 17:52
StubbornAtom
6,30831440
asked Mar 14 at 18:31
CruZCruZ
638
$endgroup$
$begingroup$
The result in part a) should be that $frac{sqrt{n}(overline X-lambda)}{sqrt{lambda}}stackrel{a}sim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_{alpha/2}sqrt{frac{overline X}{n}}<lambda<overline X+z_{alpha/2}sqrt{ frac{overline X}{n}}$ or consider a variance stabilising transform to find the C.I.
$endgroup$
– StubbornAtom
Mar 14 at 19:36
add a comment |
$begingroup$
Let $(X_1,ldots,X_n)$ denote a random sample from a Poisson distribution with parameter $lambda$.
Maximum Likelihood Estimate of $lambda$ is given by
$hat{lambda} = bar{X} = frac{1}{n} sumlimits_{i=1}^n X_i$
a) Show that $frac{bar{X}-mu}{sqrt{lambda/n}}sim N(0,1)$ approx.
b) Using this normal approximation find the theoretical confidence interval with the $1-alpha/2$ quantile from the $N(0,1)$ distribution.
Thoughts: I'm quite stuck on problem a). I don't quite se om the MLE which is equal the the sample mean is relevant here but I am probably (most definitely) missing something. My only idea is to use CLT in some way since
$$frac{bar{X}-mu}{sigma/sqrt{n}}=frac{bar{X}-mu}{sqrt{lambda/n}}$$
as the variance of Poisson is just $lambda$. But from here I don't know what to do.... Can someone help me?
And for problem b) I have 0 ideas...
statistics poisson-distribution central-limit-theorem confidence-interval
edited Mar 15 at 17:52
StubbornAtom
6,30831440
asked Mar 14 at 18:31
CruZCruZ
638
$endgroup$
Let $(X_1,ldots,X_n)$ denote a random sample from a Poisson distribution with parameter $lambda$.
Maximum Likelihood Estimate of $lambda$ is given by
$hat{lambda} = bar{X} = frac{1}{n} sumlimits_{i=1}^n X_i$
a) Show that $frac{bar{X}-mu}{sqrt{lambda/n}}sim N(0,1)$ approx.
b) Using this normal approximation find the theoretical confidence interval with the $1-alpha/2$ quantile from the $N(0,1)$ distribution.
Thoughts: I'm quite stuck on problem a). I don't quite se om the MLE which is equal the the sample mean is relevant here but I am probably (most definitely) missing something. My only idea is to use CLT in some way since
$$frac{bar{X}-mu}{sigma/sqrt{n}}=frac{bar{X}-mu}{sqrt{lambda/n}}$$
as the variance of Poisson is just $lambda$. But from here I don't know what to do.... Can someone help me?
And for problem b) I have 0 ideas...
statistics poisson-distribution central-limit-theorem confidence-interval
statistics poisson-distribution central-limit-theorem confidence-interval
edited Mar 15 at 17:52
StubbornAtom
6,30831440
asked Mar 14 at 18:31
CruZCruZ
638
edited Mar 15 at 17:52
StubbornAtom
6,30831440
asked Mar 14 at 18:31
CruZCruZ
638
edited Mar 15 at 17:52
StubbornAtom
6,30831440
edited Mar 15 at 17:52
StubbornAtom
6,30831440
edited Mar 15 at 17:52
StubbornAtom
6,30831440
6,30831440
asked Mar 14 at 18:31
CruZCruZ
638
asked Mar 14 at 18:31
CruZCruZ
638
asked Mar 14 at 18:31
CruZCruZ
638
638
$begingroup$
The result in part a) should be that $frac{sqrt{n}(overline X-lambda)}{sqrt{lambda}}stackrel{a}sim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_{alpha/2}sqrt{frac{overline X}{n}}<lambda<overline X+z_{alpha/2}sqrt{ frac{overline X}{n}}$ or consider a variance stabilising transform to find the C.I.
$endgroup$
– StubbornAtom
Mar 14 at 19:36
add a comment |
$begingroup$
The result in part a) should be that $frac{sqrt{n}(overline X-lambda)}{sqrt{lambda}}stackrel{a}sim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_{alpha/2}sqrt{frac{overline X}{n}}<lambda<overline X+z_{alpha/2}sqrt{ frac{overline X}{n}}$ or consider a variance stabilising transform to find the C.I.
$endgroup$
– StubbornAtom
Mar 14 at 19:36
$begingroup$
The result in part a) should be that $frac{sqrt{n}(overline X-lambda)}{sqrt{lambda}}stackrel{a}sim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_{alpha/2}sqrt{frac{overline X}{n}}<lambda<overline X+z_{alpha/2}sqrt{ frac{overline X}{n}}$ or consider a variance stabilising transform to find the C.I.
$endgroup$
– StubbornAtom
Mar 14 at 19:36
$begingroup$
The result in part a) should be that $frac{sqrt{n}(overline X-lambda)}{sqrt{lambda}}stackrel{a}sim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_{alpha/2}sqrt{frac{overline X}{n}}<lambda<overline X+z_{alpha/2}sqrt{ frac{overline X}{n}}$ or consider a variance stabilising transform to find the C.I.
$endgroup$
– StubbornAtom
Mar 14 at 19:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have an answer in the comments, so just slightly formalizing it. For (a) you have $n$ i.i.d Poisson random variables, thus a finite second moment, hence by the CLT
$$
sqrt{n}frac{(bar{X} - mathbb{E}X)}{sqrt{ Var(X) }} xrightarrow{D} N(0,1),
$$
in your case $mathbb{E}X = Var(X) = lambda$. Thus for any finite $n$,
$$
sqrt{n}frac{(bar{X} - lambda )}{sqrt{ lambda}} approx N(0,1).
$$
For (b), using (a) you know that
$$
mathbb{P}left(z_{a/2} le sqrt{n}frac{(bar{X} - lambda )}{sqrt{ lambda}} le z_{1-a/2} right) approx 1 - a
$$
re-arranging the inequality you have
$$
mathbb{P}left( bar{X} - z_{1-a/2}sqrt{lambda/n} le lambda le bar{X} + z_{1-a/2}sqrt{lambda/n} right) approx 1 - a,
$$
replace the $lambda$ with its estimator $bar{X}$ in $sqrt{lambda/n}$ and you have the CI.
answered Mar 15 at 19:22
V. VancakV. Vancak
11.4k3926
$endgroup$
add a comment |
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$begingroup$
You have an answer in the comments, so just slightly formalizing it. For (a) you have $n$ i.i.d Poisson random variables, thus a finite second moment, hence by the CLT
$$
sqrt{n}frac{(bar{X} - mathbb{E}X)}{sqrt{ Var(X) }} xrightarrow{D} N(0,1),
$$
in your case $mathbb{E}X = Var(X) = lambda$. Thus for any finite $n$,
$$
sqrt{n}frac{(bar{X} - lambda )}{sqrt{ lambda}} approx N(0,1).
$$
For (b), using (a) you know that
$$
mathbb{P}left(z_{a/2} le sqrt{n}frac{(bar{X} - lambda )}{sqrt{ lambda}} le z_{1-a/2} right) approx 1 - a
$$
re-arranging the inequality you have
$$
mathbb{P}left( bar{X} - z_{1-a/2}sqrt{lambda/n} le lambda le bar{X} + z_{1-a/2}sqrt{lambda/n} right) approx 1 - a,
$$
replace the $lambda$ with its estimator $bar{X}$ in $sqrt{lambda/n}$ and you have the CI.
answered Mar 15 at 19:22
V. VancakV. Vancak
11.4k3926
$endgroup$
add a comment |
$begingroup$
You have an answer in the comments, so just slightly formalizing it. For (a) you have $n$ i.i.d Poisson random variables, thus a finite second moment, hence by the CLT
$$
sqrt{n}frac{(bar{X} - mathbb{E}X)}{sqrt{ Var(X) }} xrightarrow{D} N(0,1),
$$
in your case $mathbb{E}X = Var(X) = lambda$. Thus for any finite $n$,
$$
sqrt{n}frac{(bar{X} - lambda )}{sqrt{ lambda}} approx N(0,1).
$$
For (b), using (a) you know that
$$
mathbb{P}left(z_{a/2} le sqrt{n}frac{(bar{X} - lambda )}{sqrt{ lambda}} le z_{1-a/2} right) approx 1 - a
$$
re-arranging the inequality you have
$$
mathbb{P}left( bar{X} - z_{1-a/2}sqrt{lambda/n} le lambda le bar{X} + z_{1-a/2}sqrt{lambda/n} right) approx 1 - a,
$$
replace the $lambda$ with its estimator $bar{X}$ in $sqrt{lambda/n}$ and you have the CI.
answered Mar 15 at 19:22
V. VancakV. Vancak
11.4k3926
$endgroup$
add a comment |
$begingroup$
You have an answer in the comments, so just slightly formalizing it. For (a) you have $n$ i.i.d Poisson random variables, thus a finite second moment, hence by the CLT
$$
sqrt{n}frac{(bar{X} - mathbb{E}X)}{sqrt{ Var(X) }} xrightarrow{D} N(0,1),
$$
in your case $mathbb{E}X = Var(X) = lambda$. Thus for any finite $n$,
$$
sqrt{n}frac{(bar{X} - lambda )}{sqrt{ lambda}} approx N(0,1).
$$
For (b), using (a) you know that
$$
mathbb{P}left(z_{a/2} le sqrt{n}frac{(bar{X} - lambda )}{sqrt{ lambda}} le z_{1-a/2} right) approx 1 - a
$$
re-arranging the inequality you have
$$
mathbb{P}left( bar{X} - z_{1-a/2}sqrt{lambda/n} le lambda le bar{X} + z_{1-a/2}sqrt{lambda/n} right) approx 1 - a,
$$
replace the $lambda$ with its estimator $bar{X}$ in $sqrt{lambda/n}$ and you have the CI.
answered Mar 15 at 19:22
V. VancakV. Vancak
11.4k3926
$endgroup$
You have an answer in the comments, so just slightly formalizing it. For (a) you have $n$ i.i.d Poisson random variables, thus a finite second moment, hence by the CLT
$$
sqrt{n}frac{(bar{X} - mathbb{E}X)}{sqrt{ Var(X) }} xrightarrow{D} N(0,1),
$$
in your case $mathbb{E}X = Var(X) = lambda$. Thus for any finite $n$,
$$
sqrt{n}frac{(bar{X} - lambda )}{sqrt{ lambda}} approx N(0,1).
$$
For (b), using (a) you know that
$$
mathbb{P}left(z_{a/2} le sqrt{n}frac{(bar{X} - lambda )}{sqrt{ lambda}} le z_{1-a/2} right) approx 1 - a
$$
re-arranging the inequality you have
$$
mathbb{P}left( bar{X} - z_{1-a/2}sqrt{lambda/n} le lambda le bar{X} + z_{1-a/2}sqrt{lambda/n} right) approx 1 - a,
$$
replace the $lambda$ with its estimator $bar{X}$ in $sqrt{lambda/n}$ and you have the CI.
answered Mar 15 at 19:22
V. VancakV. Vancak
11.4k3926
answered Mar 15 at 19:22
V. VancakV. Vancak
11.4k3926
answered Mar 15 at 19:22
V. VancakV. Vancak
11.4k3926
answered Mar 15 at 19:22
V. VancakV. Vancak
11.4k3926
11.4k3926
add a comment |
add a comment |
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$begingroup$
The result in part a) should be that $frac{sqrt{n}(overline X-lambda)}{sqrt{lambda}}stackrel{a}sim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_{alpha/2}sqrt{frac{overline X}{n}}<lambda<overline X+z_{alpha/2}sqrt{ frac{overline X}{n}}$ or consider a variance stabilising transform to find the C.I.
$endgroup$
– StubbornAtom
Mar 14 at 19:36