Improving on Chebyshev's inequality with normal approximation












0












$begingroup$


This problem comes from Suhov's and Kelbert's Probability and Statistics by Example I, worked example 1.6.2. It's not homework.



Let $Y_n$ be the number of spots shown cumulatively over $n$ throws of a standard six-sided die. The task is to calculate $n$ such that the following is guaranteed:



$mathbb{P}left( left|frac{Y_n}{n}-3.5right| > 0.1 right) leq 0.1$.



The authors arrive at approximately $n geq 2920$ using Chebyshev's inequality. I understand this part.



But then they note:



"Observe that using the normal approximation (the De Moivre-Laplace Theorem) yields a better lower bound: $n geq (1.96 times 10)^2 35/12 approx 794.$"



I tried approximating $Y_n$ with a Gaussian in the following way:



$mathbb{P}left( left| frac{Y_n-3.5n}{sqrt{n}sigma} right| > 0.1 frac{sqrt{n}}{sigma} right) approx 1- frac{1}{sqrt{2pi}}int_{ frac{sqrt{n}}{10sigma}}^{ frac{sqrt{n}}{10sigma}} mathrm{d}y e^{-frac{y^2}{2}} $



but I don't know how to proceed from there and would be grateful for any help!






statistics







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    This problem comes from Suhov's and Kelbert's Probability and Statistics by Example I, worked example 1.6.2. It's not homework.



    Let $Y_n$ be the number of spots shown cumulatively over $n$ throws of a standard six-sided die. The task is to calculate $n$ such that the following is guaranteed:



    $mathbb{P}left( left|frac{Y_n}{n}-3.5right| > 0.1 right) leq 0.1$.



    The authors arrive at approximately $n geq 2920$ using Chebyshev's inequality. I understand this part.



    But then they note:



    "Observe that using the normal approximation (the De Moivre-Laplace Theorem) yields a better lower bound: $n geq (1.96 times 10)^2 35/12 approx 794.$"



    I tried approximating $Y_n$ with a Gaussian in the following way:



    $mathbb{P}left( left| frac{Y_n-3.5n}{sqrt{n}sigma} right| > 0.1 frac{sqrt{n}}{sigma} right) approx 1- frac{1}{sqrt{2pi}}int_{ frac{sqrt{n}}{10sigma}}^{ frac{sqrt{n}}{10sigma}} mathrm{d}y e^{-frac{y^2}{2}} $



    but I don't know how to proceed from there and would be grateful for any help!






    statistics







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      This problem comes from Suhov's and Kelbert's Probability and Statistics by Example I, worked example 1.6.2. It's not homework.



      Let $Y_n$ be the number of spots shown cumulatively over $n$ throws of a standard six-sided die. The task is to calculate $n$ such that the following is guaranteed:



      $mathbb{P}left( left|frac{Y_n}{n}-3.5right| > 0.1 right) leq 0.1$.



      The authors arrive at approximately $n geq 2920$ using Chebyshev's inequality. I understand this part.



      But then they note:



      "Observe that using the normal approximation (the De Moivre-Laplace Theorem) yields a better lower bound: $n geq (1.96 times 10)^2 35/12 approx 794.$"



      I tried approximating $Y_n$ with a Gaussian in the following way:



      $mathbb{P}left( left| frac{Y_n-3.5n}{sqrt{n}sigma} right| > 0.1 frac{sqrt{n}}{sigma} right) approx 1- frac{1}{sqrt{2pi}}int_{ frac{sqrt{n}}{10sigma}}^{ frac{sqrt{n}}{10sigma}} mathrm{d}y e^{-frac{y^2}{2}} $



      but I don't know how to proceed from there and would be grateful for any help!






      statistics







      share|cite|improve this question











      $endgroup$




      This problem comes from Suhov's and Kelbert's Probability and Statistics by Example I, worked example 1.6.2. It's not homework.



      Let $Y_n$ be the number of spots shown cumulatively over $n$ throws of a standard six-sided die. The task is to calculate $n$ such that the following is guaranteed:



      $mathbb{P}left( left|frac{Y_n}{n}-3.5right| > 0.1 right) leq 0.1$.



      The authors arrive at approximately $n geq 2920$ using Chebyshev's inequality. I understand this part.



      But then they note:



      "Observe that using the normal approximation (the De Moivre-Laplace Theorem) yields a better lower bound: $n geq (1.96 times 10)^2 35/12 approx 794.$"



      I tried approximating $Y_n$ with a Gaussian in the following way:



      $mathbb{P}left( left| frac{Y_n-3.5n}{sqrt{n}sigma} right| > 0.1 frac{sqrt{n}}{sigma} right) approx 1- frac{1}{sqrt{2pi}}int_{ frac{sqrt{n}}{10sigma}}^{ frac{sqrt{n}}{10sigma}} mathrm{d}y e^{-frac{y^2}{2}} $



      but I don't know how to proceed from there and would be grateful for any help!






      statistics




      statistics






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 26 '18 at 15:07







      dom_miketa

















      asked Dec 25 '18 at 18:04









      dom_miketadom_miketa

      796




      796






















          1 Answer
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          1












          $begingroup$

          First of all, you're missing a factor $frac1{sqrt{2pi}}$ in the integral. Anyway, call $Phi(z)$ the CDF of a $N(0,1)$ r.v., you have to find $n$ such that
          $$Phileft(frac{sqrt n}{10sigma}right)-Phileft(-frac{sqrt n}{10sigma}right)=2Phileft(frac{sqrt n}{10sigma}right)-1ge 0.9,$$
          that is,
          $$Phileft(frac{sqrt n}{10sigma}right)ge0.95.$$
          This is equivalent to
          $$frac{sqrt n}{10sigma} ge Phi^{-1}(0.95)approx 1.96,$$
          so the approximate answer is
          $$nge 19.6^2 sigma^2,$$
          and the answer is reasonable if the bound for $n$ is large enough to use the normal approximation (it is also necessary to find the value of $sigma$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! I gave it another go today after a few more exercises which showcased this kind of approximation and arrived at a similar conclusion, except I find that $Phi^-1(0.95) approx 1.65$ (eg en.wikipedia.org/wiki/Standard_normal_table). Am I reading the table wrong?
            $endgroup$
            – dom_miketa
            Dec 26 '18 at 15:07






          • 1




            $begingroup$
            Actually not. I did not went to the table the other day so I made a mistake. You're right. $Phi^{-1}(0.95)approx1.65$. It is $Phi^{-1}(0.975)$ that is $1.96$.
            $endgroup$
            – Alejandro Nasif Salum
            Dec 27 '18 at 15:30






          • 1




            $begingroup$
            That's everything sorted then. Thanks again and have a lovely holiday if you celebrate one. :)
            $endgroup$
            – dom_miketa
            Dec 27 '18 at 18:26











          Your Answer





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          1 Answer
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          $begingroup$

          First of all, you're missing a factor $frac1{sqrt{2pi}}$ in the integral. Anyway, call $Phi(z)$ the CDF of a $N(0,1)$ r.v., you have to find $n$ such that
          $$Phileft(frac{sqrt n}{10sigma}right)-Phileft(-frac{sqrt n}{10sigma}right)=2Phileft(frac{sqrt n}{10sigma}right)-1ge 0.9,$$
          that is,
          $$Phileft(frac{sqrt n}{10sigma}right)ge0.95.$$
          This is equivalent to
          $$frac{sqrt n}{10sigma} ge Phi^{-1}(0.95)approx 1.96,$$
          so the approximate answer is
          $$nge 19.6^2 sigma^2,$$
          and the answer is reasonable if the bound for $n$ is large enough to use the normal approximation (it is also necessary to find the value of $sigma$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! I gave it another go today after a few more exercises which showcased this kind of approximation and arrived at a similar conclusion, except I find that $Phi^-1(0.95) approx 1.65$ (eg en.wikipedia.org/wiki/Standard_normal_table). Am I reading the table wrong?
            $endgroup$
            – dom_miketa
            Dec 26 '18 at 15:07






          • 1




            $begingroup$
            Actually not. I did not went to the table the other day so I made a mistake. You're right. $Phi^{-1}(0.95)approx1.65$. It is $Phi^{-1}(0.975)$ that is $1.96$.
            $endgroup$
            – Alejandro Nasif Salum
            Dec 27 '18 at 15:30






          • 1




            $begingroup$
            That's everything sorted then. Thanks again and have a lovely holiday if you celebrate one. :)
            $endgroup$
            – dom_miketa
            Dec 27 '18 at 18:26
















          1












          $begingroup$

          First of all, you're missing a factor $frac1{sqrt{2pi}}$ in the integral. Anyway, call $Phi(z)$ the CDF of a $N(0,1)$ r.v., you have to find $n$ such that
          $$Phileft(frac{sqrt n}{10sigma}right)-Phileft(-frac{sqrt n}{10sigma}right)=2Phileft(frac{sqrt n}{10sigma}right)-1ge 0.9,$$
          that is,
          $$Phileft(frac{sqrt n}{10sigma}right)ge0.95.$$
          This is equivalent to
          $$frac{sqrt n}{10sigma} ge Phi^{-1}(0.95)approx 1.96,$$
          so the approximate answer is
          $$nge 19.6^2 sigma^2,$$
          and the answer is reasonable if the bound for $n$ is large enough to use the normal approximation (it is also necessary to find the value of $sigma$).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! I gave it another go today after a few more exercises which showcased this kind of approximation and arrived at a similar conclusion, except I find that $Phi^-1(0.95) approx 1.65$ (eg en.wikipedia.org/wiki/Standard_normal_table). Am I reading the table wrong?
            $endgroup$
            – dom_miketa
            Dec 26 '18 at 15:07






          • 1




            $begingroup$
            Actually not. I did not went to the table the other day so I made a mistake. You're right. $Phi^{-1}(0.95)approx1.65$. It is $Phi^{-1}(0.975)$ that is $1.96$.
            $endgroup$
            – Alejandro Nasif Salum
            Dec 27 '18 at 15:30






          • 1




            $begingroup$
            That's everything sorted then. Thanks again and have a lovely holiday if you celebrate one. :)
            $endgroup$
            – dom_miketa
            Dec 27 '18 at 18:26














          1












          1








          1





          $begingroup$

          First of all, you're missing a factor $frac1{sqrt{2pi}}$ in the integral. Anyway, call $Phi(z)$ the CDF of a $N(0,1)$ r.v., you have to find $n$ such that
          $$Phileft(frac{sqrt n}{10sigma}right)-Phileft(-frac{sqrt n}{10sigma}right)=2Phileft(frac{sqrt n}{10sigma}right)-1ge 0.9,$$
          that is,
          $$Phileft(frac{sqrt n}{10sigma}right)ge0.95.$$
          This is equivalent to
          $$frac{sqrt n}{10sigma} ge Phi^{-1}(0.95)approx 1.96,$$
          so the approximate answer is
          $$nge 19.6^2 sigma^2,$$
          and the answer is reasonable if the bound for $n$ is large enough to use the normal approximation (it is also necessary to find the value of $sigma$).






          share|cite|improve this answer









          $endgroup$



          First of all, you're missing a factor $frac1{sqrt{2pi}}$ in the integral. Anyway, call $Phi(z)$ the CDF of a $N(0,1)$ r.v., you have to find $n$ such that
          $$Phileft(frac{sqrt n}{10sigma}right)-Phileft(-frac{sqrt n}{10sigma}right)=2Phileft(frac{sqrt n}{10sigma}right)-1ge 0.9,$$
          that is,
          $$Phileft(frac{sqrt n}{10sigma}right)ge0.95.$$
          This is equivalent to
          $$frac{sqrt n}{10sigma} ge Phi^{-1}(0.95)approx 1.96,$$
          so the approximate answer is
          $$nge 19.6^2 sigma^2,$$
          and the answer is reasonable if the bound for $n$ is large enough to use the normal approximation (it is also necessary to find the value of $sigma$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 25 '18 at 21:09









          Alejandro Nasif SalumAlejandro Nasif Salum

          4,765118




          4,765118












          • $begingroup$
            Thanks! I gave it another go today after a few more exercises which showcased this kind of approximation and arrived at a similar conclusion, except I find that $Phi^-1(0.95) approx 1.65$ (eg en.wikipedia.org/wiki/Standard_normal_table). Am I reading the table wrong?
            $endgroup$
            – dom_miketa
            Dec 26 '18 at 15:07






          • 1




            $begingroup$
            Actually not. I did not went to the table the other day so I made a mistake. You're right. $Phi^{-1}(0.95)approx1.65$. It is $Phi^{-1}(0.975)$ that is $1.96$.
            $endgroup$
            – Alejandro Nasif Salum
            Dec 27 '18 at 15:30






          • 1




            $begingroup$
            That's everything sorted then. Thanks again and have a lovely holiday if you celebrate one. :)
            $endgroup$
            – dom_miketa
            Dec 27 '18 at 18:26


















          • $begingroup$
            Thanks! I gave it another go today after a few more exercises which showcased this kind of approximation and arrived at a similar conclusion, except I find that $Phi^-1(0.95) approx 1.65$ (eg en.wikipedia.org/wiki/Standard_normal_table). Am I reading the table wrong?
            $endgroup$
            – dom_miketa
            Dec 26 '18 at 15:07






          • 1




            $begingroup$
            Actually not. I did not went to the table the other day so I made a mistake. You're right. $Phi^{-1}(0.95)approx1.65$. It is $Phi^{-1}(0.975)$ that is $1.96$.
            $endgroup$
            – Alejandro Nasif Salum
            Dec 27 '18 at 15:30






          • 1




            $begingroup$
            That's everything sorted then. Thanks again and have a lovely holiday if you celebrate one. :)
            $endgroup$
            – dom_miketa
            Dec 27 '18 at 18:26
















          $begingroup$
          Thanks! I gave it another go today after a few more exercises which showcased this kind of approximation and arrived at a similar conclusion, except I find that $Phi^-1(0.95) approx 1.65$ (eg en.wikipedia.org/wiki/Standard_normal_table). Am I reading the table wrong?
          $endgroup$
          – dom_miketa
          Dec 26 '18 at 15:07




          $begingroup$
          Thanks! I gave it another go today after a few more exercises which showcased this kind of approximation and arrived at a similar conclusion, except I find that $Phi^-1(0.95) approx 1.65$ (eg en.wikipedia.org/wiki/Standard_normal_table). Am I reading the table wrong?
          $endgroup$
          – dom_miketa
          Dec 26 '18 at 15:07




          1




          1




          $begingroup$
          Actually not. I did not went to the table the other day so I made a mistake. You're right. $Phi^{-1}(0.95)approx1.65$. It is $Phi^{-1}(0.975)$ that is $1.96$.
          $endgroup$
          – Alejandro Nasif Salum
          Dec 27 '18 at 15:30




          $begingroup$
          Actually not. I did not went to the table the other day so I made a mistake. You're right. $Phi^{-1}(0.95)approx1.65$. It is $Phi^{-1}(0.975)$ that is $1.96$.
          $endgroup$
          – Alejandro Nasif Salum
          Dec 27 '18 at 15:30




          1




          1




          $begingroup$
          That's everything sorted then. Thanks again and have a lovely holiday if you celebrate one. :)
          $endgroup$
          – dom_miketa
          Dec 27 '18 at 18:26




          $begingroup$
          That's everything sorted then. Thanks again and have a lovely holiday if you celebrate one. :)
          $endgroup$
          – dom_miketa
          Dec 27 '18 at 18:26


















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