Show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant $2p$












1














Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.



I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.



My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$



I can also use the following criterion:



Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$



To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?






probability convergence







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  • Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
    – dem0nakos
    Jan 4 at 20:38










  • Can you use the law of large numbers?
    – Lundborg
    Jan 4 at 20:54










  • @dem0nakos I didn't know that inequality, super useful.
    – qcc101
    Jan 4 at 20:58
















1














Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.



I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.



My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$



I can also use the following criterion:



Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$



To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?






probability convergence







share|cite|improve this question
























  • Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
    – dem0nakos
    Jan 4 at 20:38










  • Can you use the law of large numbers?
    – Lundborg
    Jan 4 at 20:54










  • @dem0nakos I didn't know that inequality, super useful.
    – qcc101
    Jan 4 at 20:58














1












1








1







Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.



I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.



My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$



I can also use the following criterion:



Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$



To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?






probability convergence







share|cite|improve this question















Let $X_n$ ~ Bernoulli(p). Let $Y_n = X_n + X_{n+1}$.
Let $T_n = frac{1}{n}sum_{i=1}^{n} Y_i$.
I want to show that the sequence $(T_n)_{ngeq 1}$ converges in probability to the constant 2p.



I found that $E[T_n] = 2p$ and that $operatorname{Var}[T_n] = 2p(1-p)frac{2n-1}{n^2}$.



My definition of convergence in probability is the following:
$$forall epsilon > 0 spacemathbb{P}(vert T_n - 2p vert > epsilon) to 0$$



I can also use the following criterion:



Convergence in probability iff $$lim_{ntoinfty} mathbb{E}Big[frac{vert T_n - 2pvert}{vert T_n - 2pvert + 1}Big] = 0$$



To me using the criterion here seems smart because I already know that the expected value is $2p$, but I am not sure how to proceed. Any hints?






probability convergence




probability convergence






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edited Jan 4 at 21:57









Davide Giraudo

125k16150261




125k16150261










asked Jan 4 at 20:30









qcc101qcc101

499113




499113












  • Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
    – dem0nakos
    Jan 4 at 20:38










  • Can you use the law of large numbers?
    – Lundborg
    Jan 4 at 20:54










  • @dem0nakos I didn't know that inequality, super useful.
    – qcc101
    Jan 4 at 20:58


















  • Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
    – dem0nakos
    Jan 4 at 20:38










  • Can you use the law of large numbers?
    – Lundborg
    Jan 4 at 20:54










  • @dem0nakos I didn't know that inequality, super useful.
    – qcc101
    Jan 4 at 20:58
















Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
– dem0nakos
Jan 4 at 20:38




Hint: You can try to use Chebyshev's inequality $mathbb{P}bigl(|X-EX|>epsilonbigr)leq frac{Var(X)}{epsilon^2}.$
– dem0nakos
Jan 4 at 20:38












Can you use the law of large numbers?
– Lundborg
Jan 4 at 20:54




Can you use the law of large numbers?
– Lundborg
Jan 4 at 20:54












@dem0nakos I didn't know that inequality, super useful.
– qcc101
Jan 4 at 20:58




@dem0nakos I didn't know that inequality, super useful.
– qcc101
Jan 4 at 20:58










2 Answers
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Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.



Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.






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    1














    Note that



    $$
    T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
    $$



    Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that



    $$
    2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
    $$



    and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding



    $$
    T_n overset{P}{to} 2 p
    $$



    as desired.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      4














      Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.



      Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.






      share|cite|improve this answer


























        4














        Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.



        Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.






        share|cite|improve this answer
























          4












          4








          4






          Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.



          Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.






          share|cite|improve this answer












          Claim. If $mu_n = mathbf{E}(T_n) to mu$ and $sigma_n^2 = mathbf{V}mathrm{ar}(T_n) to 0$ then $T_n to mu$ in $mathscr{L}^2$ and, hence, in probability too.



          Proof. We have $mathbf{E}(|T_n - mu|^2) = mathbf{E}(|T_n - mu_n|^2) + 2(mu_n - mu) mathbf{E}(T_n - mu_n) + (mu_n - mu)^2 to 0.$ Q.E.D.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 20:39









          Will M.Will M.

          2,442314




          2,442314























              1














              Note that



              $$
              T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
              $$



              Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that



              $$
              2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
              $$



              and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding



              $$
              T_n overset{P}{to} 2 p
              $$



              as desired.






              share|cite|improve this answer


























                1














                Note that



                $$
                T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
                $$



                Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that



                $$
                2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
                $$



                and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding



                $$
                T_n overset{P}{to} 2 p
                $$



                as desired.






                share|cite|improve this answer
























                  1












                  1








                  1






                  Note that



                  $$
                  T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
                  $$



                  Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that



                  $$
                  2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
                  $$



                  and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding



                  $$
                  T_n overset{P}{to} 2 p
                  $$



                  as desired.






                  share|cite|improve this answer












                  Note that



                  $$
                  T_n = frac{1}{n} sum_{i=1}^n Y_i = frac{1}{n} sum_{i=1}^n (X_i + X_{i+1}) = 2 cdot frac{1}{n} sum_{i=1}^n X_i - frac{1}{n}X_1
                  $$



                  Now $X_n$ is an i.i.d sequence of random variables with mean $E(X_i)=p$, thus the law of large numbers states that



                  $$
                  2 cdot frac{1}{n} sum_{i=1}^n X_i overset{P}{to} 2 p
                  $$



                  and since $X_1$ is constant, clearly $frac{1}{n} X_1 overset{P}{to} 0$ thus yielding



                  $$
                  T_n overset{P}{to} 2 p
                  $$



                  as desired.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 20:58









                  LundborgLundborg

                  746414




                  746414






























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