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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 $PX_i=1$ 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













0












$begingroup$


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



$hatlambda = barX = frac1n sumlimits_i=1^n X_i$



a) Show that $fracbarX-musqrtlambda/nsim N(0,1)$ 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



$$fracbarX-musigma/sqrtn=fracbarX-musqrtlambda/n$$



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...










share|cite|improve this question











$endgroup$











  • $begingroup$
    The result in part a) should be that $fracsqrtn(overline X-lambda)sqrtlambdastackrelasim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_alpha/2sqrtfracoverline Xn<lambda<overline X+z_alpha/2sqrt fracoverline Xn$ or consider a variance stabilising transform to find the C.I.
    $endgroup$
    – StubbornAtom
    Mar 14 at 19:36
















0












$begingroup$


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



$hatlambda = barX = frac1n sumlimits_i=1^n X_i$



a) Show that $fracbarX-musqrtlambda/nsim N(0,1)$ 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



$$fracbarX-musigma/sqrtn=fracbarX-musqrtlambda/n$$



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...










share|cite|improve this question











$endgroup$











  • $begingroup$
    The result in part a) should be that $fracsqrtn(overline X-lambda)sqrtlambdastackrelasim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_alpha/2sqrtfracoverline Xn<lambda<overline X+z_alpha/2sqrt fracoverline Xn$ or consider a variance stabilising transform to find the C.I.
    $endgroup$
    – StubbornAtom
    Mar 14 at 19:36














0












0








0





$begingroup$


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



$hatlambda = barX = frac1n sumlimits_i=1^n X_i$



a) Show that $fracbarX-musqrtlambda/nsim N(0,1)$ 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



$$fracbarX-musigma/sqrtn=fracbarX-musqrtlambda/n$$



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...










share|cite|improve this question











$endgroup$




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



$hatlambda = barX = frac1n sumlimits_i=1^n X_i$



a) Show that $fracbarX-musqrtlambda/nsim N(0,1)$ 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



$$fracbarX-musigma/sqrtn=fracbarX-musqrtlambda/n$$



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 15 at 17:52









StubbornAtom

6,29831339




6,29831339










asked Mar 14 at 18:31









CruZCruZ

638




638











  • $begingroup$
    The result in part a) should be that $fracsqrtn(overline X-lambda)sqrtlambdastackrelasim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_alpha/2sqrtfracoverline Xn<lambda<overline X+z_alpha/2sqrt fracoverline Xn$ 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 $fracsqrtn(overline X-lambda)sqrtlambdastackrelasim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_alpha/2sqrtfracoverline Xn<lambda<overline X+z_alpha/2sqrt fracoverline Xn$ 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 $fracsqrtn(overline X-lambda)sqrtlambdastackrelasim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_alpha/2sqrtfracoverline Xn<lambda<overline X+z_alpha/2sqrt fracoverline Xn$ 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 $fracsqrtn(overline X-lambda)sqrtlambdastackrelasim N(0,1)$, which follows from Lindeberg-Lévy CLT. Using this in part b), you can get a confidence interval for $lambda$ from $overline X-z_alpha/2sqrtfracoverline Xn<lambda<overline X+z_alpha/2sqrt fracoverline Xn$ or consider a variance stabilising transform to find the C.I.
$endgroup$
– StubbornAtom
Mar 14 at 19:36











1 Answer
1






active

oldest

votes


















0












$begingroup$

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
$$
sqrtnfrac(barX - mathbbEX)sqrt Var(X) xrightarrowD N(0,1),
$$

in your case $mathbbEX = Var(X) = lambda$. Thus for any finite $n$,
$$
sqrtnfrac(barX - lambda )sqrt lambda approx N(0,1).
$$

For (b), using (a) you know that
$$
mathbbPleft(z_a/2 le sqrtnfrac(barX - lambda )sqrt lambda le z_1-a/2 right) approx 1 - a
$$

re-arranging the inequality you have
$$
mathbbPleft( barX - z_1-a/2sqrtlambda/n le lambda le barX + z_1-a/2sqrtlambda/n right) approx 1 - a,
$$

replace the $lambda$ with its estimator $barX$ in $sqrtlambda/n$ and you have the CI.






share|cite|improve this answer









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    active

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    0












    $begingroup$

    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
    $$
    sqrtnfrac(barX - mathbbEX)sqrt Var(X) xrightarrowD N(0,1),
    $$

    in your case $mathbbEX = Var(X) = lambda$. Thus for any finite $n$,
    $$
    sqrtnfrac(barX - lambda )sqrt lambda approx N(0,1).
    $$

    For (b), using (a) you know that
    $$
    mathbbPleft(z_a/2 le sqrtnfrac(barX - lambda )sqrt lambda le z_1-a/2 right) approx 1 - a
    $$

    re-arranging the inequality you have
    $$
    mathbbPleft( barX - z_1-a/2sqrtlambda/n le lambda le barX + z_1-a/2sqrtlambda/n right) approx 1 - a,
    $$

    replace the $lambda$ with its estimator $barX$ in $sqrtlambda/n$ and you have the CI.






    share|cite|improve this answer









    $endgroup$

















      0












      $begingroup$

      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
      $$
      sqrtnfrac(barX - mathbbEX)sqrt Var(X) xrightarrowD N(0,1),
      $$

      in your case $mathbbEX = Var(X) = lambda$. Thus for any finite $n$,
      $$
      sqrtnfrac(barX - lambda )sqrt lambda approx N(0,1).
      $$

      For (b), using (a) you know that
      $$
      mathbbPleft(z_a/2 le sqrtnfrac(barX - lambda )sqrt lambda le z_1-a/2 right) approx 1 - a
      $$

      re-arranging the inequality you have
      $$
      mathbbPleft( barX - z_1-a/2sqrtlambda/n le lambda le barX + z_1-a/2sqrtlambda/n right) approx 1 - a,
      $$

      replace the $lambda$ with its estimator $barX$ in $sqrtlambda/n$ and you have the CI.






      share|cite|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$

        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
        $$
        sqrtnfrac(barX - mathbbEX)sqrt Var(X) xrightarrowD N(0,1),
        $$

        in your case $mathbbEX = Var(X) = lambda$. Thus for any finite $n$,
        $$
        sqrtnfrac(barX - lambda )sqrt lambda approx N(0,1).
        $$

        For (b), using (a) you know that
        $$
        mathbbPleft(z_a/2 le sqrtnfrac(barX - lambda )sqrt lambda le z_1-a/2 right) approx 1 - a
        $$

        re-arranging the inequality you have
        $$
        mathbbPleft( barX - z_1-a/2sqrtlambda/n le lambda le barX + z_1-a/2sqrtlambda/n right) approx 1 - a,
        $$

        replace the $lambda$ with its estimator $barX$ in $sqrtlambda/n$ and you have the CI.






        share|cite|improve this answer









        $endgroup$



        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
        $$
        sqrtnfrac(barX - mathbbEX)sqrt Var(X) xrightarrowD N(0,1),
        $$

        in your case $mathbbEX = Var(X) = lambda$. Thus for any finite $n$,
        $$
        sqrtnfrac(barX - lambda )sqrt lambda approx N(0,1).
        $$

        For (b), using (a) you know that
        $$
        mathbbPleft(z_a/2 le sqrtnfrac(barX - lambda )sqrt lambda le z_1-a/2 right) approx 1 - a
        $$

        re-arranging the inequality you have
        $$
        mathbbPleft( barX - z_1-a/2sqrtlambda/n le lambda le barX + z_1-a/2sqrtlambda/n right) approx 1 - a,
        $$

        replace the $lambda$ with its estimator $barX$ in $sqrtlambda/n$ and you have the CI.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 15 at 19:22









        V. VancakV. Vancak

        11.4k3926




        11.4k3926



























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