Question about limiting distribution for standard normal distribution












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I am studying CLT and meet a question but I have no idea how to solve it,
could you please show me how to prove this question (b)?



Thank you so much!
Wang



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statistics central-limit-theorem







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    0












    $begingroup$


    I am studying CLT and meet a question but I have no idea how to solve it,
    could you please show me how to prove this question (b)?



    Thank you so much!
    Wang



    Question picture






    statistics central-limit-theorem







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am studying CLT and meet a question but I have no idea how to solve it,
      could you please show me how to prove this question (b)?



      Thank you so much!
      Wang



      Question picture






      statistics central-limit-theorem







      share|cite|improve this question











      $endgroup$




      I am studying CLT and meet a question but I have no idea how to solve it,
      could you please show me how to prove this question (b)?



      Thank you so much!
      Wang



      Question picture






      statistics central-limit-theorem




      statistics central-limit-theorem






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 2 '18 at 22:22









      Henry

      99.5k479165




      99.5k479165










      asked Dec 2 '18 at 22:16









      Wang WangWang Wang

      31




      31






















          2 Answers
          2






          active

          oldest

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          0












          $begingroup$

          Hint for (b):




          • In this case you can separate the sum into $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}+dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$


          • Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $sumlimits_{i=1}^n Z_i$ and so of $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}$, and thus find the distribution to which this converges in the limit


          • $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit


          • Finally you can combine these to find the limiting distribution of $dfrac{sumlimits_{i=1}^n left(Z_i+frac1nright)}{sqrt{n}}$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much for your comment!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 1:48










          • $begingroup$
            Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 2:41










          • $begingroup$
            $sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
            $endgroup$
            – Henry
            Dec 3 '18 at 9:04



















          0












          $begingroup$

          (a) $P(|1/nsum (X_i-p_i)|>epsilon)leq 1/(n^2epsilon^2)Var(sum X_i)=1/(n^2epsilon^2)sum p_i(1-p_i)leq 1/(n^2epsilon^2)sum 1/4=frac{1}{4nepsilon^2}rightarrow 0.$
          (b) $sum (Z_i+1/n)sim N(1,n)Rightarrow 1/sqrt{n}sum (Z_i+1/n)sim N(1/sqrt{n},1)Rightarrow1/sqrt{n}sum (Z_i+1/n)rightarrow_d N(0,1)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 1:47










          • $begingroup$
            For $0<x<1$, $x(1-x)leq 1/4.$
            $endgroup$
            – John_Wick
            Dec 3 '18 at 2:08










          • $begingroup$
            Ok, I understand now! Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 2:13











          Your Answer





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          2 Answers
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          2 Answers
          2






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          active

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          0












          $begingroup$

          Hint for (b):




          • In this case you can separate the sum into $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}+dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$


          • Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $sumlimits_{i=1}^n Z_i$ and so of $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}$, and thus find the distribution to which this converges in the limit


          • $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit


          • Finally you can combine these to find the limiting distribution of $dfrac{sumlimits_{i=1}^n left(Z_i+frac1nright)}{sqrt{n}}$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much for your comment!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 1:48










          • $begingroup$
            Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 2:41










          • $begingroup$
            $sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
            $endgroup$
            – Henry
            Dec 3 '18 at 9:04
















          0












          $begingroup$

          Hint for (b):




          • In this case you can separate the sum into $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}+dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$


          • Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $sumlimits_{i=1}^n Z_i$ and so of $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}$, and thus find the distribution to which this converges in the limit


          • $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit


          • Finally you can combine these to find the limiting distribution of $dfrac{sumlimits_{i=1}^n left(Z_i+frac1nright)}{sqrt{n}}$







          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much for your comment!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 1:48










          • $begingroup$
            Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 2:41










          • $begingroup$
            $sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
            $endgroup$
            – Henry
            Dec 3 '18 at 9:04














          0












          0








          0





          $begingroup$

          Hint for (b):




          • In this case you can separate the sum into $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}+dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$


          • Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $sumlimits_{i=1}^n Z_i$ and so of $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}$, and thus find the distribution to which this converges in the limit


          • $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit


          • Finally you can combine these to find the limiting distribution of $dfrac{sumlimits_{i=1}^n left(Z_i+frac1nright)}{sqrt{n}}$







          share|cite|improve this answer









          $endgroup$



          Hint for (b):




          • In this case you can separate the sum into $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}+dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$


          • Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $sumlimits_{i=1}^n Z_i$ and so of $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}$, and thus find the distribution to which this converges in the limit


          • $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit


          • Finally you can combine these to find the limiting distribution of $dfrac{sumlimits_{i=1}^n left(Z_i+frac1nright)}{sqrt{n}}$








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 22:39









          HenryHenry

          99.5k479165




          99.5k479165












          • $begingroup$
            Thank you so much for your comment!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 1:48










          • $begingroup$
            Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 2:41










          • $begingroup$
            $sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
            $endgroup$
            – Henry
            Dec 3 '18 at 9:04


















          • $begingroup$
            Thank you so much for your comment!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 1:48










          • $begingroup$
            Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 2:41










          • $begingroup$
            $sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
            $endgroup$
            – Henry
            Dec 3 '18 at 9:04
















          $begingroup$
          Thank you so much for your comment!
          $endgroup$
          – Wang Wang
          Dec 3 '18 at 1:48




          $begingroup$
          Thank you so much for your comment!
          $endgroup$
          – Wang Wang
          Dec 3 '18 at 1:48












          $begingroup$
          Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
          $endgroup$
          – Wang Wang
          Dec 3 '18 at 2:41




          $begingroup$
          Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
          $endgroup$
          – Wang Wang
          Dec 3 '18 at 2:41












          $begingroup$
          $sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
          $endgroup$
          – Henry
          Dec 3 '18 at 9:04




          $begingroup$
          $sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
          $endgroup$
          – Henry
          Dec 3 '18 at 9:04











          0












          $begingroup$

          (a) $P(|1/nsum (X_i-p_i)|>epsilon)leq 1/(n^2epsilon^2)Var(sum X_i)=1/(n^2epsilon^2)sum p_i(1-p_i)leq 1/(n^2epsilon^2)sum 1/4=frac{1}{4nepsilon^2}rightarrow 0.$
          (b) $sum (Z_i+1/n)sim N(1,n)Rightarrow 1/sqrt{n}sum (Z_i+1/n)sim N(1/sqrt{n},1)Rightarrow1/sqrt{n}sum (Z_i+1/n)rightarrow_d N(0,1)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 1:47










          • $begingroup$
            For $0<x<1$, $x(1-x)leq 1/4.$
            $endgroup$
            – John_Wick
            Dec 3 '18 at 2:08










          • $begingroup$
            Ok, I understand now! Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 2:13
















          0












          $begingroup$

          (a) $P(|1/nsum (X_i-p_i)|>epsilon)leq 1/(n^2epsilon^2)Var(sum X_i)=1/(n^2epsilon^2)sum p_i(1-p_i)leq 1/(n^2epsilon^2)sum 1/4=frac{1}{4nepsilon^2}rightarrow 0.$
          (b) $sum (Z_i+1/n)sim N(1,n)Rightarrow 1/sqrt{n}sum (Z_i+1/n)sim N(1/sqrt{n},1)Rightarrow1/sqrt{n}sum (Z_i+1/n)rightarrow_d N(0,1)$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 1:47










          • $begingroup$
            For $0<x<1$, $x(1-x)leq 1/4.$
            $endgroup$
            – John_Wick
            Dec 3 '18 at 2:08










          • $begingroup$
            Ok, I understand now! Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 2:13














          0












          0








          0





          $begingroup$

          (a) $P(|1/nsum (X_i-p_i)|>epsilon)leq 1/(n^2epsilon^2)Var(sum X_i)=1/(n^2epsilon^2)sum p_i(1-p_i)leq 1/(n^2epsilon^2)sum 1/4=frac{1}{4nepsilon^2}rightarrow 0.$
          (b) $sum (Z_i+1/n)sim N(1,n)Rightarrow 1/sqrt{n}sum (Z_i+1/n)sim N(1/sqrt{n},1)Rightarrow1/sqrt{n}sum (Z_i+1/n)rightarrow_d N(0,1)$






          share|cite|improve this answer









          $endgroup$



          (a) $P(|1/nsum (X_i-p_i)|>epsilon)leq 1/(n^2epsilon^2)Var(sum X_i)=1/(n^2epsilon^2)sum p_i(1-p_i)leq 1/(n^2epsilon^2)sum 1/4=frac{1}{4nepsilon^2}rightarrow 0.$
          (b) $sum (Z_i+1/n)sim N(1,n)Rightarrow 1/sqrt{n}sum (Z_i+1/n)sim N(1/sqrt{n},1)Rightarrow1/sqrt{n}sum (Z_i+1/n)rightarrow_d N(0,1)$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 22:42









          John_WickJohn_Wick

          1,536111




          1,536111












          • $begingroup$
            Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 1:47










          • $begingroup$
            For $0<x<1$, $x(1-x)leq 1/4.$
            $endgroup$
            – John_Wick
            Dec 3 '18 at 2:08










          • $begingroup$
            Ok, I understand now! Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 2:13


















          • $begingroup$
            Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 1:47










          • $begingroup$
            For $0<x<1$, $x(1-x)leq 1/4.$
            $endgroup$
            – John_Wick
            Dec 3 '18 at 2:08










          • $begingroup$
            Ok, I understand now! Thanks!
            $endgroup$
            – Wang Wang
            Dec 3 '18 at 2:13
















          $begingroup$
          Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
          $endgroup$
          – Wang Wang
          Dec 3 '18 at 1:47




          $begingroup$
          Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
          $endgroup$
          – Wang Wang
          Dec 3 '18 at 1:47












          $begingroup$
          For $0<x<1$, $x(1-x)leq 1/4.$
          $endgroup$
          – John_Wick
          Dec 3 '18 at 2:08




          $begingroup$
          For $0<x<1$, $x(1-x)leq 1/4.$
          $endgroup$
          – John_Wick
          Dec 3 '18 at 2:08












          $begingroup$
          Ok, I understand now! Thanks!
          $endgroup$
          – Wang Wang
          Dec 3 '18 at 2:13




          $begingroup$
          Ok, I understand now! Thanks!
          $endgroup$
          – Wang Wang
          Dec 3 '18 at 2:13


















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          Musa

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          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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