Mixing
Determine the convergence of $overline{X^2}_{n}$
$begingroup$
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
asked Jan 29 at 5:20
JosephJoseph
827
$endgroup$
add a comment |
$begingroup$
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
asked Jan 29 at 5:20
JosephJoseph
827
$endgroup$
$begingroup$
Existence of finite moments is usually an assumption made here.
$endgroup$
– StubbornAtom
Jan 29 at 6:01
add a comment |
$begingroup$
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
asked Jan 29 at 5:20
JosephJoseph
827
$endgroup$
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
asked Jan 29 at 5:20
JosephJoseph
827
asked Jan 29 at 5:20
JosephJoseph
827
asked Jan 29 at 5:20
JosephJoseph
827
asked Jan 29 at 5:20
JosephJoseph
827
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $mathbb{E}[X_{i}^{2}]$
answered Jan 29 at 5:23
Ekesh KumarEkesh Kumar
1,08428
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $mathbb{E}[X_{i}^{2}]$
answered Jan 29 at 5:23
Ekesh KumarEkesh Kumar
1,08428
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $mathbb{E}[X_{i}^{2}]$
answered Jan 29 at 5:23
Ekesh KumarEkesh Kumar
1,08428
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $mathbb{E}[X_{i}^{2}]$
answered Jan 29 at 5:23
Ekesh KumarEkesh Kumar
1,08428
$endgroup$
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 $mathbb{E}[X_{i}^{2}]$
answered Jan 29 at 5:23
Ekesh KumarEkesh Kumar
1,08428
answered Jan 29 at 5:23
Ekesh KumarEkesh Kumar
1,08428
answered Jan 29 at 5:23
Ekesh KumarEkesh Kumar
1,08428
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
Existence of finite moments is usually an assumption made here.
$endgroup$
– StubbornAtom
Jan 29 at 6:01