Inverse Gaussian Distribution and the Central Limit Theorem












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$begingroup$


Let the random variables $Y_1,ldots,Y_n$ be independent and identically distributed (i.i.d.) (standard) Inverse Gaussian random variables with parameters $mu$ and $lambda$.



Then, let the random variables $tilde{A}$ and $A$ be given, which are defined as follows:
$A = sqrt n cdot tilde{A} = sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right] - thetaright)$ ($b_i in mathbb{R}$ ($i=1,ldots,n$) are constants (with $b_i neq b_j$ for any $i neq j$) and $theta$ is a parameter).



Goal: Determine what ultimately follows if one uses the (Lindeberg-Lévy) Central Limit Theorem on $e^{b_i Y_i}$.



The CLT informs us that $sqrt n (overline{Y} - E(Y_i)) xrightarrow{D} mathcal{N}(0,Var(Y_i))$, i.e. $sqrt n left(frac{1}{n} cdot sum_{i=1}^n [Y_i] - mu right) xrightarrow{D} mathcal{N}(0,frac{mu^3}{lambda})$.



It seems to me that it follows that $sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right]- e^{b_i mu}right) xrightarrow{D} ldots$.



Question: How to proceed from here (i.e. how to determine the limit distribution of $sqrt n cdot tilde{A}$)?






probability-theory central-limit-theorem probability-limit-theorems







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Let the random variables $Y_1,ldots,Y_n$ be independent and identically distributed (i.i.d.) (standard) Inverse Gaussian random variables with parameters $mu$ and $lambda$.



    Then, let the random variables $tilde{A}$ and $A$ be given, which are defined as follows:
    $A = sqrt n cdot tilde{A} = sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right] - thetaright)$ ($b_i in mathbb{R}$ ($i=1,ldots,n$) are constants (with $b_i neq b_j$ for any $i neq j$) and $theta$ is a parameter).



    Goal: Determine what ultimately follows if one uses the (Lindeberg-Lévy) Central Limit Theorem on $e^{b_i Y_i}$.



    The CLT informs us that $sqrt n (overline{Y} - E(Y_i)) xrightarrow{D} mathcal{N}(0,Var(Y_i))$, i.e. $sqrt n left(frac{1}{n} cdot sum_{i=1}^n [Y_i] - mu right) xrightarrow{D} mathcal{N}(0,frac{mu^3}{lambda})$.



    It seems to me that it follows that $sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right]- e^{b_i mu}right) xrightarrow{D} ldots$.



    Question: How to proceed from here (i.e. how to determine the limit distribution of $sqrt n cdot tilde{A}$)?






    probability-theory central-limit-theorem probability-limit-theorems







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let the random variables $Y_1,ldots,Y_n$ be independent and identically distributed (i.i.d.) (standard) Inverse Gaussian random variables with parameters $mu$ and $lambda$.



      Then, let the random variables $tilde{A}$ and $A$ be given, which are defined as follows:
      $A = sqrt n cdot tilde{A} = sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right] - thetaright)$ ($b_i in mathbb{R}$ ($i=1,ldots,n$) are constants (with $b_i neq b_j$ for any $i neq j$) and $theta$ is a parameter).



      Goal: Determine what ultimately follows if one uses the (Lindeberg-Lévy) Central Limit Theorem on $e^{b_i Y_i}$.



      The CLT informs us that $sqrt n (overline{Y} - E(Y_i)) xrightarrow{D} mathcal{N}(0,Var(Y_i))$, i.e. $sqrt n left(frac{1}{n} cdot sum_{i=1}^n [Y_i] - mu right) xrightarrow{D} mathcal{N}(0,frac{mu^3}{lambda})$.



      It seems to me that it follows that $sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right]- e^{b_i mu}right) xrightarrow{D} ldots$.



      Question: How to proceed from here (i.e. how to determine the limit distribution of $sqrt n cdot tilde{A}$)?






      probability-theory central-limit-theorem probability-limit-theorems







      share|cite|improve this question









      $endgroup$




      Let the random variables $Y_1,ldots,Y_n$ be independent and identically distributed (i.i.d.) (standard) Inverse Gaussian random variables with parameters $mu$ and $lambda$.



      Then, let the random variables $tilde{A}$ and $A$ be given, which are defined as follows:
      $A = sqrt n cdot tilde{A} = sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right] - thetaright)$ ($b_i in mathbb{R}$ ($i=1,ldots,n$) are constants (with $b_i neq b_j$ for any $i neq j$) and $theta$ is a parameter).



      Goal: Determine what ultimately follows if one uses the (Lindeberg-Lévy) Central Limit Theorem on $e^{b_i Y_i}$.



      The CLT informs us that $sqrt n (overline{Y} - E(Y_i)) xrightarrow{D} mathcal{N}(0,Var(Y_i))$, i.e. $sqrt n left(frac{1}{n} cdot sum_{i=1}^n [Y_i] - mu right) xrightarrow{D} mathcal{N}(0,frac{mu^3}{lambda})$.



      It seems to me that it follows that $sqrt n left(frac{1}{n} cdot sum_{i=1}^n left[e^{b_i Y_i}right]- e^{b_i mu}right) xrightarrow{D} ldots$.



      Question: How to proceed from here (i.e. how to determine the limit distribution of $sqrt n cdot tilde{A}$)?






      probability-theory central-limit-theorem probability-limit-theorems




      probability-theory central-limit-theorem probability-limit-theorems






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 28 at 19:54









      AnnaAnna

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      7619






















          1 Answer
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          $begingroup$

          You may consider the random variable
          begin{align}
          frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu})
          end{align}
          where
          $s_n^2 = sum_{i=1}^n sigma_i^2$ and $sigma_i$ is the variance of $e^{b_iY_i}$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT
          begin{align}
          frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu}) xrightarrow{D} mathcal{N}(0,1)
          end{align}
          as $ntoinfty$.






          share|cite|improve this answer









          $endgroup$














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            0












            $begingroup$

            You may consider the random variable
            begin{align}
            frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu})
            end{align}
            where
            $s_n^2 = sum_{i=1}^n sigma_i^2$ and $sigma_i$ is the variance of $e^{b_iY_i}$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT
            begin{align}
            frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu}) xrightarrow{D} mathcal{N}(0,1)
            end{align}
            as $ntoinfty$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              You may consider the random variable
              begin{align}
              frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu})
              end{align}
              where
              $s_n^2 = sum_{i=1}^n sigma_i^2$ and $sigma_i$ is the variance of $e^{b_iY_i}$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT
              begin{align}
              frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu}) xrightarrow{D} mathcal{N}(0,1)
              end{align}
              as $ntoinfty$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                You may consider the random variable
                begin{align}
                frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu})
                end{align}
                where
                $s_n^2 = sum_{i=1}^n sigma_i^2$ and $sigma_i$ is the variance of $e^{b_iY_i}$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT
                begin{align}
                frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu}) xrightarrow{D} mathcal{N}(0,1)
                end{align}
                as $ntoinfty$.






                share|cite|improve this answer









                $endgroup$



                You may consider the random variable
                begin{align}
                frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu})
                end{align}
                where
                $s_n^2 = sum_{i=1}^n sigma_i^2$ and $sigma_i$ is the variance of $e^{b_iY_i}$. Then, you can check whether the Lindeberg's condition is satisfied. If it is satisfied, using the Lindeberg CLT
                begin{align}
                frac{1}{s_n} sum_{i=1}^n (e^{b_i Y_i} - e^{b_i mu}) xrightarrow{D} mathcal{N}(0,1)
                end{align}
                as $ntoinfty$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 31 at 15:19









                user52227user52227

                1,038512




                1,038512






























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