UMVUE Geometric Distribution












2












$begingroup$


I am trying to find the UMVUE for the parameter $p$ for an n i.i.d geometric distribution:



$(1-p)^{x-1}p$ for $x=1,2,…$ and $0<p<1$



and found that:



$P(X_1=1)$ is an unbiased estimator , so let $w=I[X_1=1]$ be my unbiased estimator and since $sum_i X_i=t$ is complete and sufficient statistic for geometric distribution, I can improve my unbiased estimator as follows:



$E[wmidsum_i X_i=t] = P(X_1=1midsum_i X_i=t) = frac{P(X_1=1,sum_i X_i=t-1)}{P(sum_i X_i=t)}$



So I have two questions now:
what is the pdf for $sum_i X_i=t-1$ ? .. I know it is negative binomial but can't write it correctly
and my second question is what is the variance of this modified unbiased estimator and does it achieve the Cramer-Rao lower bound ?






self-learning statistical-inference







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I am trying to find the UMVUE for the parameter $p$ for an n i.i.d geometric distribution:



    $(1-p)^{x-1}p$ for $x=1,2,…$ and $0<p<1$



    and found that:



    $P(X_1=1)$ is an unbiased estimator , so let $w=I[X_1=1]$ be my unbiased estimator and since $sum_i X_i=t$ is complete and sufficient statistic for geometric distribution, I can improve my unbiased estimator as follows:



    $E[wmidsum_i X_i=t] = P(X_1=1midsum_i X_i=t) = frac{P(X_1=1,sum_i X_i=t-1)}{P(sum_i X_i=t)}$



    So I have two questions now:
    what is the pdf for $sum_i X_i=t-1$ ? .. I know it is negative binomial but can't write it correctly
    and my second question is what is the variance of this modified unbiased estimator and does it achieve the Cramer-Rao lower bound ?






    self-learning statistical-inference







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I am trying to find the UMVUE for the parameter $p$ for an n i.i.d geometric distribution:



      $(1-p)^{x-1}p$ for $x=1,2,…$ and $0<p<1$



      and found that:



      $P(X_1=1)$ is an unbiased estimator , so let $w=I[X_1=1]$ be my unbiased estimator and since $sum_i X_i=t$ is complete and sufficient statistic for geometric distribution, I can improve my unbiased estimator as follows:



      $E[wmidsum_i X_i=t] = P(X_1=1midsum_i X_i=t) = frac{P(X_1=1,sum_i X_i=t-1)}{P(sum_i X_i=t)}$



      So I have two questions now:
      what is the pdf for $sum_i X_i=t-1$ ? .. I know it is negative binomial but can't write it correctly
      and my second question is what is the variance of this modified unbiased estimator and does it achieve the Cramer-Rao lower bound ?






      self-learning statistical-inference







      share|cite|improve this question











      $endgroup$




      I am trying to find the UMVUE for the parameter $p$ for an n i.i.d geometric distribution:



      $(1-p)^{x-1}p$ for $x=1,2,…$ and $0<p<1$



      and found that:



      $P(X_1=1)$ is an unbiased estimator , so let $w=I[X_1=1]$ be my unbiased estimator and since $sum_i X_i=t$ is complete and sufficient statistic for geometric distribution, I can improve my unbiased estimator as follows:



      $E[wmidsum_i X_i=t] = P(X_1=1midsum_i X_i=t) = frac{P(X_1=1,sum_i X_i=t-1)}{P(sum_i X_i=t)}$



      So I have two questions now:
      what is the pdf for $sum_i X_i=t-1$ ? .. I know it is negative binomial but can't write it correctly
      and my second question is what is the variance of this modified unbiased estimator and does it achieve the Cramer-Rao lower bound ?






      self-learning statistical-inference




      self-learning statistical-inference






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '16 at 1:46









      Momo

      12k21430




      12k21430










      asked Dec 1 '16 at 1:10









      BassemBassem

      478




      478






















          1 Answer
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          0












          $begingroup$

          For the first question only:



          $P(X_1=1)=p$



          $P(sum_{i=2}^{n}X_i=t-1)={t-2choose n-2}p^{n-1}(1-p)^{t-n}$, $t=n,n+1...$



          $P(sum_{i=1}^{n}X_i=t)={t-1choose n-1}p^{n}(1-p)^{t-n}$, $t=n,n+1...$



          So the UMVUE is $hat p=frac{n-1}{sum_{i=1}^{n} X_i-1}$



          For CRLB you may look here.



          But for the variance of the UMVUE:



          $Var(hat p)=sum_{t=n}^infty left(frac{n-1}{t-1}-pright)^2 {t-1choose n-1}p^n(1-p)^{t-n}$



          I'm afraid I was not able to get a closed form. Neither it worked for $E(hat p^2)$



          Maybe somebody else can step in.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can we find a closed form for the variance or E(p^2) when n=2 ? that is $ hat p=frac{n-1}{sum_{i=1}^{2} X_i-1}$ and how did you develop the variance summation? thanks
            $endgroup$
            – Bassem
            Dec 1 '16 at 12:24










          • $begingroup$
            $Y=sum_{i=1}^n X_i$ is Negative Binomial, $T=frac{n-1}{Y-1}$ is unbiased ($E[T]=p$), so $Var(T)=E[(T-E[T])^2]=Eleft[left(frac{n-1}{Y-1}-pright)^2right]=sum_{t=n}^inftyleft(frac{n-1}{t-1}-pright)^2 P(Y=t)$
            $endgroup$
            – Momo
            Dec 1 '16 at 14:33










          • $begingroup$
            Also, for $n=2$ you have $E[hat{p}^2]=frac{p^2log(1/p)}{1-p}$ so $Var(hat p)=E[hat{p}^2]-p^2$
            $endgroup$
            – Momo
            Dec 1 '16 at 14:39












          • $begingroup$
            Great .. thanks for the help .. but could you show what form did you use to find $E[hat{p}^2]$ ?
            $endgroup$
            – Bassem
            Dec 1 '16 at 23:37










          • $begingroup$
            $E(hat p^2)=sum_{t=n}^infty left(frac{n-1}{t-1}right)^2 {t-1choose n-1}p^n(1-p)^{t-n}$ So for $n=2$ $E(hat p^2)=sum_{t=2}^infty frac{1}{t-1} p^2(1-p)^{t-2}=frac{p^2}{1-p}sum_{t=2}^infty frac{1}{t-1} (1-p)^{t-1}=frac{p^2}{1-p}sum_{i=1}^infty frac{1}{i} (1-p)^i$ The last series needs $sum_{i=1}^infty frac{x^i}{i}$, which is obtained by integrating $sum_{i=1}^infty u^{i-1}=frac{1}{1-u}$ term by term from $u=0$ to $x$ You might consider upvoting and accepting the answer, if it was useful for you.
            $endgroup$
            – Momo
            Dec 1 '16 at 23:50













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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          0












          $begingroup$

          For the first question only:



          $P(X_1=1)=p$



          $P(sum_{i=2}^{n}X_i=t-1)={t-2choose n-2}p^{n-1}(1-p)^{t-n}$, $t=n,n+1...$



          $P(sum_{i=1}^{n}X_i=t)={t-1choose n-1}p^{n}(1-p)^{t-n}$, $t=n,n+1...$



          So the UMVUE is $hat p=frac{n-1}{sum_{i=1}^{n} X_i-1}$



          For CRLB you may look here.



          But for the variance of the UMVUE:



          $Var(hat p)=sum_{t=n}^infty left(frac{n-1}{t-1}-pright)^2 {t-1choose n-1}p^n(1-p)^{t-n}$



          I'm afraid I was not able to get a closed form. Neither it worked for $E(hat p^2)$



          Maybe somebody else can step in.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can we find a closed form for the variance or E(p^2) when n=2 ? that is $ hat p=frac{n-1}{sum_{i=1}^{2} X_i-1}$ and how did you develop the variance summation? thanks
            $endgroup$
            – Bassem
            Dec 1 '16 at 12:24










          • $begingroup$
            $Y=sum_{i=1}^n X_i$ is Negative Binomial, $T=frac{n-1}{Y-1}$ is unbiased ($E[T]=p$), so $Var(T)=E[(T-E[T])^2]=Eleft[left(frac{n-1}{Y-1}-pright)^2right]=sum_{t=n}^inftyleft(frac{n-1}{t-1}-pright)^2 P(Y=t)$
            $endgroup$
            – Momo
            Dec 1 '16 at 14:33










          • $begingroup$
            Also, for $n=2$ you have $E[hat{p}^2]=frac{p^2log(1/p)}{1-p}$ so $Var(hat p)=E[hat{p}^2]-p^2$
            $endgroup$
            – Momo
            Dec 1 '16 at 14:39












          • $begingroup$
            Great .. thanks for the help .. but could you show what form did you use to find $E[hat{p}^2]$ ?
            $endgroup$
            – Bassem
            Dec 1 '16 at 23:37










          • $begingroup$
            $E(hat p^2)=sum_{t=n}^infty left(frac{n-1}{t-1}right)^2 {t-1choose n-1}p^n(1-p)^{t-n}$ So for $n=2$ $E(hat p^2)=sum_{t=2}^infty frac{1}{t-1} p^2(1-p)^{t-2}=frac{p^2}{1-p}sum_{t=2}^infty frac{1}{t-1} (1-p)^{t-1}=frac{p^2}{1-p}sum_{i=1}^infty frac{1}{i} (1-p)^i$ The last series needs $sum_{i=1}^infty frac{x^i}{i}$, which is obtained by integrating $sum_{i=1}^infty u^{i-1}=frac{1}{1-u}$ term by term from $u=0$ to $x$ You might consider upvoting and accepting the answer, if it was useful for you.
            $endgroup$
            – Momo
            Dec 1 '16 at 23:50


















          0












          $begingroup$

          For the first question only:



          $P(X_1=1)=p$



          $P(sum_{i=2}^{n}X_i=t-1)={t-2choose n-2}p^{n-1}(1-p)^{t-n}$, $t=n,n+1...$



          $P(sum_{i=1}^{n}X_i=t)={t-1choose n-1}p^{n}(1-p)^{t-n}$, $t=n,n+1...$



          So the UMVUE is $hat p=frac{n-1}{sum_{i=1}^{n} X_i-1}$



          For CRLB you may look here.



          But for the variance of the UMVUE:



          $Var(hat p)=sum_{t=n}^infty left(frac{n-1}{t-1}-pright)^2 {t-1choose n-1}p^n(1-p)^{t-n}$



          I'm afraid I was not able to get a closed form. Neither it worked for $E(hat p^2)$



          Maybe somebody else can step in.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Can we find a closed form for the variance or E(p^2) when n=2 ? that is $ hat p=frac{n-1}{sum_{i=1}^{2} X_i-1}$ and how did you develop the variance summation? thanks
            $endgroup$
            – Bassem
            Dec 1 '16 at 12:24










          • $begingroup$
            $Y=sum_{i=1}^n X_i$ is Negative Binomial, $T=frac{n-1}{Y-1}$ is unbiased ($E[T]=p$), so $Var(T)=E[(T-E[T])^2]=Eleft[left(frac{n-1}{Y-1}-pright)^2right]=sum_{t=n}^inftyleft(frac{n-1}{t-1}-pright)^2 P(Y=t)$
            $endgroup$
            – Momo
            Dec 1 '16 at 14:33










          • $begingroup$
            Also, for $n=2$ you have $E[hat{p}^2]=frac{p^2log(1/p)}{1-p}$ so $Var(hat p)=E[hat{p}^2]-p^2$
            $endgroup$
            – Momo
            Dec 1 '16 at 14:39












          • $begingroup$
            Great .. thanks for the help .. but could you show what form did you use to find $E[hat{p}^2]$ ?
            $endgroup$
            – Bassem
            Dec 1 '16 at 23:37










          • $begingroup$
            $E(hat p^2)=sum_{t=n}^infty left(frac{n-1}{t-1}right)^2 {t-1choose n-1}p^n(1-p)^{t-n}$ So for $n=2$ $E(hat p^2)=sum_{t=2}^infty frac{1}{t-1} p^2(1-p)^{t-2}=frac{p^2}{1-p}sum_{t=2}^infty frac{1}{t-1} (1-p)^{t-1}=frac{p^2}{1-p}sum_{i=1}^infty frac{1}{i} (1-p)^i$ The last series needs $sum_{i=1}^infty frac{x^i}{i}$, which is obtained by integrating $sum_{i=1}^infty u^{i-1}=frac{1}{1-u}$ term by term from $u=0$ to $x$ You might consider upvoting and accepting the answer, if it was useful for you.
            $endgroup$
            – Momo
            Dec 1 '16 at 23:50
















          0












          0








          0





          $begingroup$

          For the first question only:



          $P(X_1=1)=p$



          $P(sum_{i=2}^{n}X_i=t-1)={t-2choose n-2}p^{n-1}(1-p)^{t-n}$, $t=n,n+1...$



          $P(sum_{i=1}^{n}X_i=t)={t-1choose n-1}p^{n}(1-p)^{t-n}$, $t=n,n+1...$



          So the UMVUE is $hat p=frac{n-1}{sum_{i=1}^{n} X_i-1}$



          For CRLB you may look here.



          But for the variance of the UMVUE:



          $Var(hat p)=sum_{t=n}^infty left(frac{n-1}{t-1}-pright)^2 {t-1choose n-1}p^n(1-p)^{t-n}$



          I'm afraid I was not able to get a closed form. Neither it worked for $E(hat p^2)$



          Maybe somebody else can step in.






          share|cite|improve this answer











          $endgroup$



          For the first question only:



          $P(X_1=1)=p$



          $P(sum_{i=2}^{n}X_i=t-1)={t-2choose n-2}p^{n-1}(1-p)^{t-n}$, $t=n,n+1...$



          $P(sum_{i=1}^{n}X_i=t)={t-1choose n-1}p^{n}(1-p)^{t-n}$, $t=n,n+1...$



          So the UMVUE is $hat p=frac{n-1}{sum_{i=1}^{n} X_i-1}$



          For CRLB you may look here.



          But for the variance of the UMVUE:



          $Var(hat p)=sum_{t=n}^infty left(frac{n-1}{t-1}-pright)^2 {t-1choose n-1}p^n(1-p)^{t-n}$



          I'm afraid I was not able to get a closed form. Neither it worked for $E(hat p^2)$



          Maybe somebody else can step in.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Apr 13 '17 at 12:21









          Community♦

          1




          1










          answered Dec 1 '16 at 1:33









          MomoMomo

          12k21430




          12k21430












          • $begingroup$
            Can we find a closed form for the variance or E(p^2) when n=2 ? that is $ hat p=frac{n-1}{sum_{i=1}^{2} X_i-1}$ and how did you develop the variance summation? thanks
            $endgroup$
            – Bassem
            Dec 1 '16 at 12:24










          • $begingroup$
            $Y=sum_{i=1}^n X_i$ is Negative Binomial, $T=frac{n-1}{Y-1}$ is unbiased ($E[T]=p$), so $Var(T)=E[(T-E[T])^2]=Eleft[left(frac{n-1}{Y-1}-pright)^2right]=sum_{t=n}^inftyleft(frac{n-1}{t-1}-pright)^2 P(Y=t)$
            $endgroup$
            – Momo
            Dec 1 '16 at 14:33










          • $begingroup$
            Also, for $n=2$ you have $E[hat{p}^2]=frac{p^2log(1/p)}{1-p}$ so $Var(hat p)=E[hat{p}^2]-p^2$
            $endgroup$
            – Momo
            Dec 1 '16 at 14:39












          • $begingroup$
            Great .. thanks for the help .. but could you show what form did you use to find $E[hat{p}^2]$ ?
            $endgroup$
            – Bassem
            Dec 1 '16 at 23:37










          • $begingroup$
            $E(hat p^2)=sum_{t=n}^infty left(frac{n-1}{t-1}right)^2 {t-1choose n-1}p^n(1-p)^{t-n}$ So for $n=2$ $E(hat p^2)=sum_{t=2}^infty frac{1}{t-1} p^2(1-p)^{t-2}=frac{p^2}{1-p}sum_{t=2}^infty frac{1}{t-1} (1-p)^{t-1}=frac{p^2}{1-p}sum_{i=1}^infty frac{1}{i} (1-p)^i$ The last series needs $sum_{i=1}^infty frac{x^i}{i}$, which is obtained by integrating $sum_{i=1}^infty u^{i-1}=frac{1}{1-u}$ term by term from $u=0$ to $x$ You might consider upvoting and accepting the answer, if it was useful for you.
            $endgroup$
            – Momo
            Dec 1 '16 at 23:50




















          • $begingroup$
            Can we find a closed form for the variance or E(p^2) when n=2 ? that is $ hat p=frac{n-1}{sum_{i=1}^{2} X_i-1}$ and how did you develop the variance summation? thanks
            $endgroup$
            – Bassem
            Dec 1 '16 at 12:24










          • $begingroup$
            $Y=sum_{i=1}^n X_i$ is Negative Binomial, $T=frac{n-1}{Y-1}$ is unbiased ($E[T]=p$), so $Var(T)=E[(T-E[T])^2]=Eleft[left(frac{n-1}{Y-1}-pright)^2right]=sum_{t=n}^inftyleft(frac{n-1}{t-1}-pright)^2 P(Y=t)$
            $endgroup$
            – Momo
            Dec 1 '16 at 14:33










          • $begingroup$
            Also, for $n=2$ you have $E[hat{p}^2]=frac{p^2log(1/p)}{1-p}$ so $Var(hat p)=E[hat{p}^2]-p^2$
            $endgroup$
            – Momo
            Dec 1 '16 at 14:39












          • $begingroup$
            Great .. thanks for the help .. but could you show what form did you use to find $E[hat{p}^2]$ ?
            $endgroup$
            – Bassem
            Dec 1 '16 at 23:37










          • $begingroup$
            $E(hat p^2)=sum_{t=n}^infty left(frac{n-1}{t-1}right)^2 {t-1choose n-1}p^n(1-p)^{t-n}$ So for $n=2$ $E(hat p^2)=sum_{t=2}^infty frac{1}{t-1} p^2(1-p)^{t-2}=frac{p^2}{1-p}sum_{t=2}^infty frac{1}{t-1} (1-p)^{t-1}=frac{p^2}{1-p}sum_{i=1}^infty frac{1}{i} (1-p)^i$ The last series needs $sum_{i=1}^infty frac{x^i}{i}$, which is obtained by integrating $sum_{i=1}^infty u^{i-1}=frac{1}{1-u}$ term by term from $u=0$ to $x$ You might consider upvoting and accepting the answer, if it was useful for you.
            $endgroup$
            – Momo
            Dec 1 '16 at 23:50


















          $begingroup$
          Can we find a closed form for the variance or E(p^2) when n=2 ? that is $ hat p=frac{n-1}{sum_{i=1}^{2} X_i-1}$ and how did you develop the variance summation? thanks
          $endgroup$
          – Bassem
          Dec 1 '16 at 12:24




          $begingroup$
          Can we find a closed form for the variance or E(p^2) when n=2 ? that is $ hat p=frac{n-1}{sum_{i=1}^{2} X_i-1}$ and how did you develop the variance summation? thanks
          $endgroup$
          – Bassem
          Dec 1 '16 at 12:24












          $begingroup$
          $Y=sum_{i=1}^n X_i$ is Negative Binomial, $T=frac{n-1}{Y-1}$ is unbiased ($E[T]=p$), so $Var(T)=E[(T-E[T])^2]=Eleft[left(frac{n-1}{Y-1}-pright)^2right]=sum_{t=n}^inftyleft(frac{n-1}{t-1}-pright)^2 P(Y=t)$
          $endgroup$
          – Momo
          Dec 1 '16 at 14:33




          $begingroup$
          $Y=sum_{i=1}^n X_i$ is Negative Binomial, $T=frac{n-1}{Y-1}$ is unbiased ($E[T]=p$), so $Var(T)=E[(T-E[T])^2]=Eleft[left(frac{n-1}{Y-1}-pright)^2right]=sum_{t=n}^inftyleft(frac{n-1}{t-1}-pright)^2 P(Y=t)$
          $endgroup$
          – Momo
          Dec 1 '16 at 14:33












          $begingroup$
          Also, for $n=2$ you have $E[hat{p}^2]=frac{p^2log(1/p)}{1-p}$ so $Var(hat p)=E[hat{p}^2]-p^2$
          $endgroup$
          – Momo
          Dec 1 '16 at 14:39






          $begingroup$
          Also, for $n=2$ you have $E[hat{p}^2]=frac{p^2log(1/p)}{1-p}$ so $Var(hat p)=E[hat{p}^2]-p^2$
          $endgroup$
          – Momo
          Dec 1 '16 at 14:39














          $begingroup$
          Great .. thanks for the help .. but could you show what form did you use to find $E[hat{p}^2]$ ?
          $endgroup$
          – Bassem
          Dec 1 '16 at 23:37




          $begingroup$
          Great .. thanks for the help .. but could you show what form did you use to find $E[hat{p}^2]$ ?
          $endgroup$
          – Bassem
          Dec 1 '16 at 23:37












          $begingroup$
          $E(hat p^2)=sum_{t=n}^infty left(frac{n-1}{t-1}right)^2 {t-1choose n-1}p^n(1-p)^{t-n}$ So for $n=2$ $E(hat p^2)=sum_{t=2}^infty frac{1}{t-1} p^2(1-p)^{t-2}=frac{p^2}{1-p}sum_{t=2}^infty frac{1}{t-1} (1-p)^{t-1}=frac{p^2}{1-p}sum_{i=1}^infty frac{1}{i} (1-p)^i$ The last series needs $sum_{i=1}^infty frac{x^i}{i}$, which is obtained by integrating $sum_{i=1}^infty u^{i-1}=frac{1}{1-u}$ term by term from $u=0$ to $x$ You might consider upvoting and accepting the answer, if it was useful for you.
          $endgroup$
          – Momo
          Dec 1 '16 at 23:50






          $begingroup$
          $E(hat p^2)=sum_{t=n}^infty left(frac{n-1}{t-1}right)^2 {t-1choose n-1}p^n(1-p)^{t-n}$ So for $n=2$ $E(hat p^2)=sum_{t=2}^infty frac{1}{t-1} p^2(1-p)^{t-2}=frac{p^2}{1-p}sum_{t=2}^infty frac{1}{t-1} (1-p)^{t-1}=frac{p^2}{1-p}sum_{i=1}^infty frac{1}{i} (1-p)^i$ The last series needs $sum_{i=1}^infty frac{x^i}{i}$, which is obtained by integrating $sum_{i=1}^infty u^{i-1}=frac{1}{1-u}$ term by term from $u=0$ to $x$ You might consider upvoting and accepting the answer, if it was useful for you.
          $endgroup$
          – Momo
          Dec 1 '16 at 23:50




















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