Distribution of $bar{X}$ of n Bernoulli's












0












$begingroup$


I am trying to derive the distribution of $bar{X}_n$ where $X_1, X_2,...,X_n$ are iid $sim mathrm{Bern}(p)$. I used two approaches but I am debating myself and questioning which one would be correct (if any).



Method 1: Using MGF
I used the moment generating function and ended up with $bar{X}_n sim mathrm{Bern}(p^n)$



Method 2:
I used the CLT and ended up with $bar{X}_n sim N(p, sqrt{pq}/n$) for n being large.



I am not sure which one is correct (if any).



Can someone tell me if I am doing this right or not?



Thank you,
I appreciate your help










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am trying to derive the distribution of $bar{X}_n$ where $X_1, X_2,...,X_n$ are iid $sim mathrm{Bern}(p)$. I used two approaches but I am debating myself and questioning which one would be correct (if any).



    Method 1: Using MGF
    I used the moment generating function and ended up with $bar{X}_n sim mathrm{Bern}(p^n)$



    Method 2:
    I used the CLT and ended up with $bar{X}_n sim N(p, sqrt{pq}/n$) for n being large.



    I am not sure which one is correct (if any).



    Can someone tell me if I am doing this right or not?



    Thank you,
    I appreciate your help










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am trying to derive the distribution of $bar{X}_n$ where $X_1, X_2,...,X_n$ are iid $sim mathrm{Bern}(p)$. I used two approaches but I am debating myself and questioning which one would be correct (if any).



      Method 1: Using MGF
      I used the moment generating function and ended up with $bar{X}_n sim mathrm{Bern}(p^n)$



      Method 2:
      I used the CLT and ended up with $bar{X}_n sim N(p, sqrt{pq}/n$) for n being large.



      I am not sure which one is correct (if any).



      Can someone tell me if I am doing this right or not?



      Thank you,
      I appreciate your help










      share|cite|improve this question











      $endgroup$




      I am trying to derive the distribution of $bar{X}_n$ where $X_1, X_2,...,X_n$ are iid $sim mathrm{Bern}(p)$. I used two approaches but I am debating myself and questioning which one would be correct (if any).



      Method 1: Using MGF
      I used the moment generating function and ended up with $bar{X}_n sim mathrm{Bern}(p^n)$



      Method 2:
      I used the CLT and ended up with $bar{X}_n sim N(p, sqrt{pq}/n$) for n being large.



      I am not sure which one is correct (if any).



      Can someone tell me if I am doing this right or not?



      Thank you,
      I appreciate your help







      probability-theory probability-distributions






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      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 5 '18 at 5:29









      epimorphic

      2,74231533




      2,74231533










      asked Dec 4 '18 at 0:56









      S AS A

      1




      1






















          1 Answer
          1






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          0












          $begingroup$

          Neither.




          • For the first one, you are claiming that the average only take binary value.


          • For the second one, what if $n$ is small.



          Guide:




          • Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.

          • Average is simply dividing the sum by $n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
            $endgroup$
            – S A
            Dec 4 '18 at 1:10










          • $begingroup$
            Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
            $endgroup$
            – Siong Thye Goh
            Dec 4 '18 at 1:12











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          1 Answer
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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Neither.




          • For the first one, you are claiming that the average only take binary value.


          • For the second one, what if $n$ is small.



          Guide:




          • Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.

          • Average is simply dividing the sum by $n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
            $endgroup$
            – S A
            Dec 4 '18 at 1:10










          • $begingroup$
            Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
            $endgroup$
            – Siong Thye Goh
            Dec 4 '18 at 1:12
















          0












          $begingroup$

          Neither.




          • For the first one, you are claiming that the average only take binary value.


          • For the second one, what if $n$ is small.



          Guide:




          • Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.

          • Average is simply dividing the sum by $n$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
            $endgroup$
            – S A
            Dec 4 '18 at 1:10










          • $begingroup$
            Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
            $endgroup$
            – Siong Thye Goh
            Dec 4 '18 at 1:12














          0












          0








          0





          $begingroup$

          Neither.




          • For the first one, you are claiming that the average only take binary value.


          • For the second one, what if $n$ is small.



          Guide:




          • Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.

          • Average is simply dividing the sum by $n$.






          share|cite|improve this answer









          $endgroup$



          Neither.




          • For the first one, you are claiming that the average only take binary value.


          • For the second one, what if $n$ is small.



          Guide:




          • Recall that sum of IID Bernoulli follows a binomial distribution $Bin(n, p)$.

          • Average is simply dividing the sum by $n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 1:04









          Siong Thye GohSiong Thye Goh

          100k1466117




          100k1466117












          • $begingroup$
            Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
            $endgroup$
            – S A
            Dec 4 '18 at 1:10










          • $begingroup$
            Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
            $endgroup$
            – Siong Thye Goh
            Dec 4 '18 at 1:12


















          • $begingroup$
            Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
            $endgroup$
            – S A
            Dec 4 '18 at 1:10










          • $begingroup$
            Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
            $endgroup$
            – Siong Thye Goh
            Dec 4 '18 at 1:12
















          $begingroup$
          Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
          $endgroup$
          – S A
          Dec 4 '18 at 1:10




          $begingroup$
          Won't I get a Bernoulli again if I simply divided by n? Bin(n/n, p/n) => Bin(1, p/n) (one trial) => Bern(p/n)?
          $endgroup$
          – S A
          Dec 4 '18 at 1:10












          $begingroup$
          Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
          $endgroup$
          – Siong Thye Goh
          Dec 4 '18 at 1:12




          $begingroup$
          Suppose $n=2$, the values that you can get are $0, 0.5, 1$. It is possible to get value $0.5$, something that is not modelled by the Bernoulli distribution.
          $endgroup$
          – Siong Thye Goh
          Dec 4 '18 at 1:12


















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