Show that $limlimits_{ntoinfty} Bbb P (L_n le mu le R_n) = 1 - alpha$.












1
















Let $X_1,X_2, cdots$ be i.i.d. random variables with mean $mu$ and variance $1$. Show that $limlimits_{ntoinfty} Bbb P (L_n le mu le R_n) = 1 - alpha ,$



where



$$R_n = frac {1} {n} sumlimits_{i=1}^{n}X_i - frac {1} {sqrt n} Phi^{-1} left (frac {alpha} {2}right ) ,$$



$$ L_n = frac {1} {n} sumlimits_{i=1}^{n}X_i + frac {1} {sqrt n} Phi^{-1} left (frac {alpha} {2}right ) ,$$



and $Phi$ is the standard normal C.D.F. It should be noted that since $alpha<1,$



$$Phi^{-1} left (frac {alpha} {2} right ) < 0.$$





I have applied Chebychev's inequality and found that



$$Bbb P (L_n le mu le R_n) ge 1 - frac {n} {left [Phi^{-1} left (frac {alpha} {2} right )right ]^{2}}.$$



Now how do I proceed? Please help me in this regard.



Thank you very much.






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    Let $X_1,X_2, cdots$ be i.i.d. random variables with mean $mu$ and variance $1$. Show that $limlimits_{ntoinfty} Bbb P (L_n le mu le R_n) = 1 - alpha ,$



    where



    $$R_n = frac {1} {n} sumlimits_{i=1}^{n}X_i - frac {1} {sqrt n} Phi^{-1} left (frac {alpha} {2}right ) ,$$



    $$ L_n = frac {1} {n} sumlimits_{i=1}^{n}X_i + frac {1} {sqrt n} Phi^{-1} left (frac {alpha} {2}right ) ,$$



    and $Phi$ is the standard normal C.D.F. It should be noted that since $alpha<1,$



    $$Phi^{-1} left (frac {alpha} {2} right ) < 0.$$





    I have applied Chebychev's inequality and found that



    $$Bbb P (L_n le mu le R_n) ge 1 - frac {n} {left [Phi^{-1} left (frac {alpha} {2} right )right ]^{2}}.$$



    Now how do I proceed? Please help me in this regard.



    Thank you very much.






    probability probability-theory probability-distributions normal-distribution







    share|cite|improve this question



























      1












      1








      1









      Let $X_1,X_2, cdots$ be i.i.d. random variables with mean $mu$ and variance $1$. Show that $limlimits_{ntoinfty} Bbb P (L_n le mu le R_n) = 1 - alpha ,$



      where



      $$R_n = frac {1} {n} sumlimits_{i=1}^{n}X_i - frac {1} {sqrt n} Phi^{-1} left (frac {alpha} {2}right ) ,$$



      $$ L_n = frac {1} {n} sumlimits_{i=1}^{n}X_i + frac {1} {sqrt n} Phi^{-1} left (frac {alpha} {2}right ) ,$$



      and $Phi$ is the standard normal C.D.F. It should be noted that since $alpha<1,$



      $$Phi^{-1} left (frac {alpha} {2} right ) < 0.$$





      I have applied Chebychev's inequality and found that



      $$Bbb P (L_n le mu le R_n) ge 1 - frac {n} {left [Phi^{-1} left (frac {alpha} {2} right )right ]^{2}}.$$



      Now how do I proceed? Please help me in this regard.



      Thank you very much.






      probability probability-theory probability-distributions normal-distribution







      share|cite|improve this question

















      Let $X_1,X_2, cdots$ be i.i.d. random variables with mean $mu$ and variance $1$. Show that $limlimits_{ntoinfty} Bbb P (L_n le mu le R_n) = 1 - alpha ,$



      where



      $$R_n = frac {1} {n} sumlimits_{i=1}^{n}X_i - frac {1} {sqrt n} Phi^{-1} left (frac {alpha} {2}right ) ,$$



      $$ L_n = frac {1} {n} sumlimits_{i=1}^{n}X_i + frac {1} {sqrt n} Phi^{-1} left (frac {alpha} {2}right ) ,$$



      and $Phi$ is the standard normal C.D.F. It should be noted that since $alpha<1,$



      $$Phi^{-1} left (frac {alpha} {2} right ) < 0.$$





      I have applied Chebychev's inequality and found that



      $$Bbb P (L_n le mu le R_n) ge 1 - frac {n} {left [Phi^{-1} left (frac {alpha} {2} right )right ]^{2}}.$$



      Now how do I proceed? Please help me in this regard.



      Thank you very much.






      probability probability-theory probability-distributions normal-distribution




      probability probability-theory probability-distributions normal-distribution






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 6:05







      Dbchatto67

















      asked Dec 1 '18 at 6:00









      Dbchatto67Dbchatto67

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          You need to rearrange the inequalities and apply CLT (central limit theorem).
          Namely, $L_n leq mu$ is equivalent to
          $$
          frac{1}{n}sumlimits_{i=1}^n (X_i - mu) leq -frac{1}{n^{1/2}}Phi^{-1}(alpha/2) Longleftrightarrow frac{1}{n^{1/2}}sumlimits_{i=1}^n (X_i - mu) leq -Phi^{-1}(alpha/2) .
          $$
          Doing a similarly rearrangement for the inequality with $R_n $, and applying the central limit theorem to pass to the limit we get
          $$
          tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) .
          $$
          In view of the definition of $Phi^{-1}$ we have $Phi(Phi^{-1}(alpha/2)) = alpha/2$. Using the fact that $Phi(-a) + Phi(a) = 1$ when $ain mathbb{R}$ it follows that $Phi(-Phi^{-1}(alpha/2)) = 1 - alpha/2$. Thus the limit in $(1)$ equals $1-alpha$.






          share|cite|improve this answer























          • Very good answer. But unfortunately you have done a slight mistake $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{-1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$ should be replaced by $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$
            – Dbchatto67
            Dec 1 '18 at 7:08










          • Thank you very much @Hayk for your help.
            – Dbchatto67
            Dec 1 '18 at 7:09










          • Actually $Phi (a) + Phi (-a) = 1$ for all $a in Bbb R$ not for only $a<0$. I appreciate you for this observation. This is the key observation you have made which makes this problem simpler.
            – Dbchatto67
            Dec 1 '18 at 7:12










          • @Dbchatto67, thanks for pointing these out, I've edited the post accordingly.
            – Hayk
            Dec 1 '18 at 7:21











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          You need to rearrange the inequalities and apply CLT (central limit theorem).
          Namely, $L_n leq mu$ is equivalent to
          $$
          frac{1}{n}sumlimits_{i=1}^n (X_i - mu) leq -frac{1}{n^{1/2}}Phi^{-1}(alpha/2) Longleftrightarrow frac{1}{n^{1/2}}sumlimits_{i=1}^n (X_i - mu) leq -Phi^{-1}(alpha/2) .
          $$
          Doing a similarly rearrangement for the inequality with $R_n $, and applying the central limit theorem to pass to the limit we get
          $$
          tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) .
          $$
          In view of the definition of $Phi^{-1}$ we have $Phi(Phi^{-1}(alpha/2)) = alpha/2$. Using the fact that $Phi(-a) + Phi(a) = 1$ when $ain mathbb{R}$ it follows that $Phi(-Phi^{-1}(alpha/2)) = 1 - alpha/2$. Thus the limit in $(1)$ equals $1-alpha$.






          share|cite|improve this answer























          • Very good answer. But unfortunately you have done a slight mistake $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{-1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$ should be replaced by $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$
            – Dbchatto67
            Dec 1 '18 at 7:08










          • Thank you very much @Hayk for your help.
            – Dbchatto67
            Dec 1 '18 at 7:09










          • Actually $Phi (a) + Phi (-a) = 1$ for all $a in Bbb R$ not for only $a<0$. I appreciate you for this observation. This is the key observation you have made which makes this problem simpler.
            – Dbchatto67
            Dec 1 '18 at 7:12










          • @Dbchatto67, thanks for pointing these out, I've edited the post accordingly.
            – Hayk
            Dec 1 '18 at 7:21
















          2














          You need to rearrange the inequalities and apply CLT (central limit theorem).
          Namely, $L_n leq mu$ is equivalent to
          $$
          frac{1}{n}sumlimits_{i=1}^n (X_i - mu) leq -frac{1}{n^{1/2}}Phi^{-1}(alpha/2) Longleftrightarrow frac{1}{n^{1/2}}sumlimits_{i=1}^n (X_i - mu) leq -Phi^{-1}(alpha/2) .
          $$
          Doing a similarly rearrangement for the inequality with $R_n $, and applying the central limit theorem to pass to the limit we get
          $$
          tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) .
          $$
          In view of the definition of $Phi^{-1}$ we have $Phi(Phi^{-1}(alpha/2)) = alpha/2$. Using the fact that $Phi(-a) + Phi(a) = 1$ when $ain mathbb{R}$ it follows that $Phi(-Phi^{-1}(alpha/2)) = 1 - alpha/2$. Thus the limit in $(1)$ equals $1-alpha$.






          share|cite|improve this answer























          • Very good answer. But unfortunately you have done a slight mistake $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{-1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$ should be replaced by $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$
            – Dbchatto67
            Dec 1 '18 at 7:08










          • Thank you very much @Hayk for your help.
            – Dbchatto67
            Dec 1 '18 at 7:09










          • Actually $Phi (a) + Phi (-a) = 1$ for all $a in Bbb R$ not for only $a<0$. I appreciate you for this observation. This is the key observation you have made which makes this problem simpler.
            – Dbchatto67
            Dec 1 '18 at 7:12










          • @Dbchatto67, thanks for pointing these out, I've edited the post accordingly.
            – Hayk
            Dec 1 '18 at 7:21














          2












          2








          2






          You need to rearrange the inequalities and apply CLT (central limit theorem).
          Namely, $L_n leq mu$ is equivalent to
          $$
          frac{1}{n}sumlimits_{i=1}^n (X_i - mu) leq -frac{1}{n^{1/2}}Phi^{-1}(alpha/2) Longleftrightarrow frac{1}{n^{1/2}}sumlimits_{i=1}^n (X_i - mu) leq -Phi^{-1}(alpha/2) .
          $$
          Doing a similarly rearrangement for the inequality with $R_n $, and applying the central limit theorem to pass to the limit we get
          $$
          tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) .
          $$
          In view of the definition of $Phi^{-1}$ we have $Phi(Phi^{-1}(alpha/2)) = alpha/2$. Using the fact that $Phi(-a) + Phi(a) = 1$ when $ain mathbb{R}$ it follows that $Phi(-Phi^{-1}(alpha/2)) = 1 - alpha/2$. Thus the limit in $(1)$ equals $1-alpha$.






          share|cite|improve this answer














          You need to rearrange the inequalities and apply CLT (central limit theorem).
          Namely, $L_n leq mu$ is equivalent to
          $$
          frac{1}{n}sumlimits_{i=1}^n (X_i - mu) leq -frac{1}{n^{1/2}}Phi^{-1}(alpha/2) Longleftrightarrow frac{1}{n^{1/2}}sumlimits_{i=1}^n (X_i - mu) leq -Phi^{-1}(alpha/2) .
          $$
          Doing a similarly rearrangement for the inequality with $R_n $, and applying the central limit theorem to pass to the limit we get
          $$
          tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) .
          $$
          In view of the definition of $Phi^{-1}$ we have $Phi(Phi^{-1}(alpha/2)) = alpha/2$. Using the fact that $Phi(-a) + Phi(a) = 1$ when $ain mathbb{R}$ it follows that $Phi(-Phi^{-1}(alpha/2)) = 1 - alpha/2$. Thus the limit in $(1)$ equals $1-alpha$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 1 '18 at 7:17

























          answered Dec 1 '18 at 6:22









          HaykHayk

          2,1171213




          2,1171213












          • Very good answer. But unfortunately you have done a slight mistake $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{-1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$ should be replaced by $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$
            – Dbchatto67
            Dec 1 '18 at 7:08










          • Thank you very much @Hayk for your help.
            – Dbchatto67
            Dec 1 '18 at 7:09










          • Actually $Phi (a) + Phi (-a) = 1$ for all $a in Bbb R$ not for only $a<0$. I appreciate you for this observation. This is the key observation you have made which makes this problem simpler.
            – Dbchatto67
            Dec 1 '18 at 7:12










          • @Dbchatto67, thanks for pointing these out, I've edited the post accordingly.
            – Hayk
            Dec 1 '18 at 7:21


















          • Very good answer. But unfortunately you have done a slight mistake $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{-1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$ should be replaced by $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$
            – Dbchatto67
            Dec 1 '18 at 7:08










          • Thank you very much @Hayk for your help.
            – Dbchatto67
            Dec 1 '18 at 7:09










          • Actually $Phi (a) + Phi (-a) = 1$ for all $a in Bbb R$ not for only $a<0$. I appreciate you for this observation. This is the key observation you have made which makes this problem simpler.
            – Dbchatto67
            Dec 1 '18 at 7:12










          • @Dbchatto67, thanks for pointing these out, I've edited the post accordingly.
            – Hayk
            Dec 1 '18 at 7:21
















          Very good answer. But unfortunately you have done a slight mistake $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{-1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$ should be replaced by $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$
          – Dbchatto67
          Dec 1 '18 at 7:08




          Very good answer. But unfortunately you have done a slight mistake $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{-1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$ should be replaced by $$ tag{1} mathbb{P}(L_n leq mu leq R_n) = mathbb{P}left(Phi^{-1}(alpha/2) leq frac{1}{n^{1/2}} sumlimits_{i=1}^n (X_i - mu) leq - Phi^{-1}(alpha/2) right) \ to Phi( -Phi^{-1}(alpha/2) ) -Phi( Phi^{-1}(alpha/2) ) . $$
          – Dbchatto67
          Dec 1 '18 at 7:08












          Thank you very much @Hayk for your help.
          – Dbchatto67
          Dec 1 '18 at 7:09




          Thank you very much @Hayk for your help.
          – Dbchatto67
          Dec 1 '18 at 7:09












          Actually $Phi (a) + Phi (-a) = 1$ for all $a in Bbb R$ not for only $a<0$. I appreciate you for this observation. This is the key observation you have made which makes this problem simpler.
          – Dbchatto67
          Dec 1 '18 at 7:12




          Actually $Phi (a) + Phi (-a) = 1$ for all $a in Bbb R$ not for only $a<0$. I appreciate you for this observation. This is the key observation you have made which makes this problem simpler.
          – Dbchatto67
          Dec 1 '18 at 7:12












          @Dbchatto67, thanks for pointing these out, I've edited the post accordingly.
          – Hayk
          Dec 1 '18 at 7:21




          @Dbchatto67, thanks for pointing these out, I've edited the post accordingly.
          – Hayk
          Dec 1 '18 at 7:21


















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          Grease: Live!

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          A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
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