Determine the convergence of $overline{X^2}_{n}$












0












$begingroup$


Let's say I sample $X_{1},X_{2},dots,X_{n}$ from a random variable X with a distribution. It is not important to know what the distribution is at this point.



I am trying to determine whether $overline{X^2}_{n}$ converges given that $overline{X^2}_{n}:= frac{1}{n}sum_{i = 1}^{n}X^{2}_{i}$.



I am thinking of applying Law of Large number in this case, but I have not figured out exactly to determine whether it converges.



Maybe I do not even need LLN.



Any tip would be appreciated.






probability statistics convergence law-of-large-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Existence of finite moments is usually an assumption made here.
    $endgroup$
    – StubbornAtom
    Jan 29 at 6:01
















0












$begingroup$


Let's say I sample $X_{1},X_{2},dots,X_{n}$ from a random variable X with a distribution. It is not important to know what the distribution is at this point.



I am trying to determine whether $overline{X^2}_{n}$ converges given that $overline{X^2}_{n}:= frac{1}{n}sum_{i = 1}^{n}X^{2}_{i}$.



I am thinking of applying Law of Large number in this case, but I have not figured out exactly to determine whether it converges.



Maybe I do not even need LLN.



Any tip would be appreciated.






probability statistics convergence law-of-large-numbers







share|cite|improve this question









$endgroup$












  • $begingroup$
    Existence of finite moments is usually an assumption made here.
    $endgroup$
    – StubbornAtom
    Jan 29 at 6:01














0












0








0





$begingroup$


Let's say I sample $X_{1},X_{2},dots,X_{n}$ from a random variable X with a distribution. It is not important to know what the distribution is at this point.



I am trying to determine whether $overline{X^2}_{n}$ converges given that $overline{X^2}_{n}:= frac{1}{n}sum_{i = 1}^{n}X^{2}_{i}$.



I am thinking of applying Law of Large number in this case, but I have not figured out exactly to determine whether it converges.



Maybe I do not even need LLN.



Any tip would be appreciated.






probability statistics convergence law-of-large-numbers







share|cite|improve this question









$endgroup$




Let's say I sample $X_{1},X_{2},dots,X_{n}$ from a random variable X with a distribution. It is not important to know what the distribution is at this point.



I am trying to determine whether $overline{X^2}_{n}$ converges given that $overline{X^2}_{n}:= frac{1}{n}sum_{i = 1}^{n}X^{2}_{i}$.



I am thinking of applying Law of Large number in this case, but I have not figured out exactly to determine whether it converges.



Maybe I do not even need LLN.



Any tip would be appreciated.






probability statistics convergence law-of-large-numbers




probability statistics convergence law-of-large-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 29 at 5:20









JosephJoseph

827




827












  • $begingroup$
    Existence of finite moments is usually an assumption made here.
    $endgroup$
    – StubbornAtom
    Jan 29 at 6:01


















  • $begingroup$
    Existence of finite moments is usually an assumption made here.
    $endgroup$
    – StubbornAtom
    Jan 29 at 6:01
















$begingroup$
Existence of finite moments is usually an assumption made here.
$endgroup$
– StubbornAtom
Jan 29 at 6:01




$begingroup$
Existence of finite moments is usually an assumption made here.
$endgroup$
– StubbornAtom
Jan 29 at 6:01










1 Answer
1






active

oldest

votes


















1












$begingroup$

For each index $i$, define $y_{i} = X_{i}^{2}$. Then, the sequence ${y_{i}}$ forms a new sequence of i.i.d random variables. We can rewrite our sum as follows:



$$overline{X^{2}_{n}} = frac{1}{n} sum_{i=1}^{n} X_{i}^{2} = frac{1}{n} sum_{i = 1}^{n} y_{i},$$



which, by the Weak Law of Large Numbers converges to $mathbb{E}[y_{i}].$ So, we have that our sequence converges to $mathbb{E}[y_{i}] = mathbb{E}[X_{i}^{2}].$



Thus, the series converges to $mathbb{E}[X_{i}^{2}]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks there! I appreciate the input.
    $endgroup$
    – Joseph
    Jan 29 at 5:30










  • $begingroup$
    You're welcome @BolenRoss
    $endgroup$
    – Ekesh Kumar
    Jan 29 at 6:43






  • 1




    $begingroup$
    Need to assume that $mathbb{E}X_1^2<infty$. Also one may use the SLLN to get a.s. convergence.
    $endgroup$
    – d.k.o.
    Jan 29 at 7:36












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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









1












$begingroup$

For each index $i$, define $y_{i} = X_{i}^{2}$. Then, the sequence ${y_{i}}$ forms a new sequence of i.i.d random variables. We can rewrite our sum as follows:



$$overline{X^{2}_{n}} = frac{1}{n} sum_{i=1}^{n} X_{i}^{2} = frac{1}{n} sum_{i = 1}^{n} y_{i},$$



which, by the Weak Law of Large Numbers converges to $mathbb{E}[y_{i}].$ So, we have that our sequence converges to $mathbb{E}[y_{i}] = mathbb{E}[X_{i}^{2}].$



Thus, the series converges to $mathbb{E}[X_{i}^{2}]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks there! I appreciate the input.
    $endgroup$
    – Joseph
    Jan 29 at 5:30










  • $begingroup$
    You're welcome @BolenRoss
    $endgroup$
    – Ekesh Kumar
    Jan 29 at 6:43






  • 1




    $begingroup$
    Need to assume that $mathbb{E}X_1^2<infty$. Also one may use the SLLN to get a.s. convergence.
    $endgroup$
    – d.k.o.
    Jan 29 at 7:36
















1












$begingroup$

For each index $i$, define $y_{i} = X_{i}^{2}$. Then, the sequence ${y_{i}}$ forms a new sequence of i.i.d random variables. We can rewrite our sum as follows:



$$overline{X^{2}_{n}} = frac{1}{n} sum_{i=1}^{n} X_{i}^{2} = frac{1}{n} sum_{i = 1}^{n} y_{i},$$



which, by the Weak Law of Large Numbers converges to $mathbb{E}[y_{i}].$ So, we have that our sequence converges to $mathbb{E}[y_{i}] = mathbb{E}[X_{i}^{2}].$



Thus, the series converges to $mathbb{E}[X_{i}^{2}]$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks there! I appreciate the input.
    $endgroup$
    – Joseph
    Jan 29 at 5:30










  • $begingroup$
    You're welcome @BolenRoss
    $endgroup$
    – Ekesh Kumar
    Jan 29 at 6:43






  • 1




    $begingroup$
    Need to assume that $mathbb{E}X_1^2<infty$. Also one may use the SLLN to get a.s. convergence.
    $endgroup$
    – d.k.o.
    Jan 29 at 7:36














1












1








1





$begingroup$

For each index $i$, define $y_{i} = X_{i}^{2}$. Then, the sequence ${y_{i}}$ forms a new sequence of i.i.d random variables. We can rewrite our sum as follows:



$$overline{X^{2}_{n}} = frac{1}{n} sum_{i=1}^{n} X_{i}^{2} = frac{1}{n} sum_{i = 1}^{n} y_{i},$$



which, by the Weak Law of Large Numbers converges to $mathbb{E}[y_{i}].$ So, we have that our sequence converges to $mathbb{E}[y_{i}] = mathbb{E}[X_{i}^{2}].$



Thus, the series converges to $mathbb{E}[X_{i}^{2}]$






share|cite|improve this answer









$endgroup$



For each index $i$, define $y_{i} = X_{i}^{2}$. Then, the sequence ${y_{i}}$ forms a new sequence of i.i.d random variables. We can rewrite our sum as follows:



$$overline{X^{2}_{n}} = frac{1}{n} sum_{i=1}^{n} X_{i}^{2} = frac{1}{n} sum_{i = 1}^{n} y_{i},$$



which, by the Weak Law of Large Numbers converges to $mathbb{E}[y_{i}].$ So, we have that our sequence converges to $mathbb{E}[y_{i}] = mathbb{E}[X_{i}^{2}].$



Thus, the series converges to $mathbb{E}[X_{i}^{2}]$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 29 at 5:23









Ekesh KumarEkesh Kumar

1,08428




1,08428












  • $begingroup$
    Thanks there! I appreciate the input.
    $endgroup$
    – Joseph
    Jan 29 at 5:30










  • $begingroup$
    You're welcome @BolenRoss
    $endgroup$
    – Ekesh Kumar
    Jan 29 at 6:43






  • 1




    $begingroup$
    Need to assume that $mathbb{E}X_1^2<infty$. Also one may use the SLLN to get a.s. convergence.
    $endgroup$
    – d.k.o.
    Jan 29 at 7:36


















  • $begingroup$
    Thanks there! I appreciate the input.
    $endgroup$
    – Joseph
    Jan 29 at 5:30










  • $begingroup$
    You're welcome @BolenRoss
    $endgroup$
    – Ekesh Kumar
    Jan 29 at 6:43






  • 1




    $begingroup$
    Need to assume that $mathbb{E}X_1^2<infty$. Also one may use the SLLN to get a.s. convergence.
    $endgroup$
    – d.k.o.
    Jan 29 at 7:36
















$begingroup$
Thanks there! I appreciate the input.
$endgroup$
– Joseph
Jan 29 at 5:30




$begingroup$
Thanks there! I appreciate the input.
$endgroup$
– Joseph
Jan 29 at 5:30












$begingroup$
You're welcome @BolenRoss
$endgroup$
– Ekesh Kumar
Jan 29 at 6:43




$begingroup$
You're welcome @BolenRoss
$endgroup$
– Ekesh Kumar
Jan 29 at 6:43




1




1




$begingroup$
Need to assume that $mathbb{E}X_1^2<infty$. Also one may use the SLLN to get a.s. convergence.
$endgroup$
– d.k.o.
Jan 29 at 7:36




$begingroup$
Need to assume that $mathbb{E}X_1^2<infty$. Also one may use the SLLN to get a.s. convergence.
$endgroup$
– d.k.o.
Jan 29 at 7:36


















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