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






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      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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