Mixing
Convergence to normal distribution without i.i.d.
Suppose that $X_n$ are independent and $mathbb{P}(X_m = m) = mathbb{P}(X_m=-m) = frac{1}{2m^2}$ and for $mgeq 2$
begin{equation*}
mathbb{P}(X_m = 1) = mathbb{P}(X_m=-1) = frac{1-m^{-2}}{2}.
end{equation*}
It is easy to see $frac{mathrm{Var}(S_n)}{n} longrightarrow 2$ but I cannot prove $frac{S_n}{sqrt{n}} Longrightarrow mathcal{N}(0,1)$. Clearly the assumptions of Feller-Lindeberg theorem doesn't hold.
probability probability-theory probability-distributions normal-distribution
asked Nov 21 '18 at 7:11
Sean
527513
add a comment |
Suppose that $X_n$ are independent and $mathbb{P}(X_m = m) = mathbb{P}(X_m=-m) = frac{1}{2m^2}$ and for $mgeq 2$
begin{equation*}
mathbb{P}(X_m = 1) = mathbb{P}(X_m=-1) = frac{1-m^{-2}}{2}.
end{equation*}
It is easy to see $frac{mathrm{Var}(S_n)}{n} longrightarrow 2$ but I cannot prove $frac{S_n}{sqrt{n}} Longrightarrow mathcal{N}(0,1)$. Clearly the assumptions of Feller-Lindeberg theorem doesn't hold.
probability probability-theory probability-distributions normal-distribution
asked Nov 21 '18 at 7:11
Sean
527513
You've defined $X_n$ but you've not defined $S_n$.
– heropup
Nov 21 '18 at 7:33
add a comment |
Suppose that $X_n$ are independent and $mathbb{P}(X_m = m) = mathbb{P}(X_m=-m) = frac{1}{2m^2}$ and for $mgeq 2$
begin{equation*}
mathbb{P}(X_m = 1) = mathbb{P}(X_m=-1) = frac{1-m^{-2}}{2}.
end{equation*}
It is easy to see $frac{mathrm{Var}(S_n)}{n} longrightarrow 2$ but I cannot prove $frac{S_n}{sqrt{n}} Longrightarrow mathcal{N}(0,1)$. Clearly the assumptions of Feller-Lindeberg theorem doesn't hold.
probability probability-theory probability-distributions normal-distribution
asked Nov 21 '18 at 7:11
Sean
527513
Suppose that $X_n$ are independent and $mathbb{P}(X_m = m) = mathbb{P}(X_m=-m) = frac{1}{2m^2}$ and for $mgeq 2$
begin{equation*}
mathbb{P}(X_m = 1) = mathbb{P}(X_m=-1) = frac{1-m^{-2}}{2}.
end{equation*}
It is easy to see $frac{mathrm{Var}(S_n)}{n} longrightarrow 2$ but I cannot prove $frac{S_n}{sqrt{n}} Longrightarrow mathcal{N}(0,1)$. Clearly the assumptions of Feller-Lindeberg theorem doesn't hold.
probability probability-theory probability-distributions normal-distribution
probability probability-theory probability-distributions normal-distribution
asked Nov 21 '18 at 7:11
Sean
527513
asked Nov 21 '18 at 7:11
Sean
527513
asked Nov 21 '18 at 7:11
Sean
527513
asked Nov 21 '18 at 7:11
Sean
527513
asked Nov 21 '18 at 7:11
Sean
527513
527513
You've defined $X_n$ but you've not defined $S_n$.
– heropup
Nov 21 '18 at 7:33
add a comment |
You've defined $X_n$ but you've not defined $S_n$.
– heropup
Nov 21 '18 at 7:33
You've defined $X_n$ but you've not defined $S_n$.
– heropup
Nov 21 '18 at 7:33
You've defined $X_n$ but you've not defined $S_n$.
– heropup
Nov 21 '18 at 7:33
add a comment |
1 Answer
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Set $Z_m = X_{m} 1_{{|X_m| le 1}} + 1_{{X_m>1}}-1_{{X_m<-1}}$. Then $(Z_i)_{i in mathbb{N}}$ is a sequence of i.i.d. random variables with $mathbb{E}(Z_i) =0$ and $mathrm{Var}(Z_i) =1$. Thus, we can conclude by the central limit law that
$$frac{1}{sqrt{n}}sum_{k=1}^n Z_k Rightarrow mathcal{N}(0,1).$$
Let $S_n^{(1)} := sum_{k=1}^n Z_k$ and $S_n^{(2)}: = sum_{k=1}^n R_k$ with $R_k = X_k - Z_k = X_m 1_{{|X_m|>1}}+(1_{{X_m<-1}}-1_{{X_m>1}})$. If we can show that $(sqrt{n})^{-1} S_n^{(2)} rightarrow 0$ e.g. in $L^1$, then this already implies convergence in distribution. By Slutsky's theorem we can can could that
$$frac{1}{sqrt{n}} sum_{k=1}^n X_k = frac{1}{sqrt{n}} S_n^{(1)} + frac{1}{sqrt{n}} S_n^{(2)} Rightarrow mathcal{N}(0,1)+0 =mathcal{N}(0,1).$$
It remains to show that $(sqrt{n})^{-1}mathbb{E}[|S_n^{(2)}|] rightarrow 0$. In fact, we have
begin{align}
frac{1}{sqrt{n}} mathbb{E}| S_n^{(2)} | &le frac{1}{sqrt{n}} sum_{k=1}^n mathbb{E}(1+|X_m|) 1_{{|X_m|>1}} le frac{2}{sqrt{n}} sum_{k=1}^n mathbb{E}|X_m| 1_{{|X_m|>1}} \
& lefrac{2}{sqrt{n}} sum_{k=1}^n mathbb{E}|X_m| 1_{{|X_m|>1}} lefrac{2}{sqrt{n}} sum_{k=1}^n frac{1}{k} ll frac{log(n)}{sqrt{n}},
end{align}
where the last step can be established by using an integral comparison argument. (The sum is at most $1+int_1^{n} frac{1}{x} , d x$.)
answered Nov 21 '18 at 7:36
p4sch
4,770217
add a comment |
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1 Answer
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1 Answer
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Set $Z_m = X_{m} 1_{{|X_m| le 1}} + 1_{{X_m>1}}-1_{{X_m<-1}}$. Then $(Z_i)_{i in mathbb{N}}$ is a sequence of i.i.d. random variables with $mathbb{E}(Z_i) =0$ and $mathrm{Var}(Z_i) =1$. Thus, we can conclude by the central limit law that
$$frac{1}{sqrt{n}}sum_{k=1}^n Z_k Rightarrow mathcal{N}(0,1).$$
Let $S_n^{(1)} := sum_{k=1}^n Z_k$ and $S_n^{(2)}: = sum_{k=1}^n R_k$ with $R_k = X_k - Z_k = X_m 1_{{|X_m|>1}}+(1_{{X_m<-1}}-1_{{X_m>1}})$. If we can show that $(sqrt{n})^{-1} S_n^{(2)} rightarrow 0$ e.g. in $L^1$, then this already implies convergence in distribution. By Slutsky's theorem we can can could that
$$frac{1}{sqrt{n}} sum_{k=1}^n X_k = frac{1}{sqrt{n}} S_n^{(1)} + frac{1}{sqrt{n}} S_n^{(2)} Rightarrow mathcal{N}(0,1)+0 =mathcal{N}(0,1).$$
It remains to show that $(sqrt{n})^{-1}mathbb{E}[|S_n^{(2)}|] rightarrow 0$. In fact, we have
begin{align}
frac{1}{sqrt{n}} mathbb{E}| S_n^{(2)} | &le frac{1}{sqrt{n}} sum_{k=1}^n mathbb{E}(1+|X_m|) 1_{{|X_m|>1}} le frac{2}{sqrt{n}} sum_{k=1}^n mathbb{E}|X_m| 1_{{|X_m|>1}} \
& lefrac{2}{sqrt{n}} sum_{k=1}^n mathbb{E}|X_m| 1_{{|X_m|>1}} lefrac{2}{sqrt{n}} sum_{k=1}^n frac{1}{k} ll frac{log(n)}{sqrt{n}},
end{align}
where the last step can be established by using an integral comparison argument. (The sum is at most $1+int_1^{n} frac{1}{x} , d x$.)
answered Nov 21 '18 at 7:36
p4sch
4,770217
add a comment |
Set $Z_m = X_{m} 1_{{|X_m| le 1}} + 1_{{X_m>1}}-1_{{X_m<-1}}$. Then $(Z_i)_{i in mathbb{N}}$ is a sequence of i.i.d. random variables with $mathbb{E}(Z_i) =0$ and $mathrm{Var}(Z_i) =1$. Thus, we can conclude by the central limit law that
$$frac{1}{sqrt{n}}sum_{k=1}^n Z_k Rightarrow mathcal{N}(0,1).$$
Let $S_n^{(1)} := sum_{k=1}^n Z_k$ and $S_n^{(2)}: = sum_{k=1}^n R_k$ with $R_k = X_k - Z_k = X_m 1_{{|X_m|>1}}+(1_{{X_m<-1}}-1_{{X_m>1}})$. If we can show that $(sqrt{n})^{-1} S_n^{(2)} rightarrow 0$ e.g. in $L^1$, then this already implies convergence in distribution. By Slutsky's theorem we can can could that
$$frac{1}{sqrt{n}} sum_{k=1}^n X_k = frac{1}{sqrt{n}} S_n^{(1)} + frac{1}{sqrt{n}} S_n^{(2)} Rightarrow mathcal{N}(0,1)+0 =mathcal{N}(0,1).$$
It remains to show that $(sqrt{n})^{-1}mathbb{E}[|S_n^{(2)}|] rightarrow 0$. In fact, we have
begin{align}
frac{1}{sqrt{n}} mathbb{E}| S_n^{(2)} | &le frac{1}{sqrt{n}} sum_{k=1}^n mathbb{E}(1+|X_m|) 1_{{|X_m|>1}} le frac{2}{sqrt{n}} sum_{k=1}^n mathbb{E}|X_m| 1_{{|X_m|>1}} \
& lefrac{2}{sqrt{n}} sum_{k=1}^n mathbb{E}|X_m| 1_{{|X_m|>1}} lefrac{2}{sqrt{n}} sum_{k=1}^n frac{1}{k} ll frac{log(n)}{sqrt{n}},
end{align}
where the last step can be established by using an integral comparison argument. (The sum is at most $1+int_1^{n} frac{1}{x} , d x$.)
answered Nov 21 '18 at 7:36
p4sch
4,770217
add a comment |
Set $Z_m = X_{m} 1_{{|X_m| le 1}} + 1_{{X_m>1}}-1_{{X_m<-1}}$. Then $(Z_i)_{i in mathbb{N}}$ is a sequence of i.i.d. random variables with $mathbb{E}(Z_i) =0$ and $mathrm{Var}(Z_i) =1$. Thus, we can conclude by the central limit law that
$$frac{1}{sqrt{n}}sum_{k=1}^n Z_k Rightarrow mathcal{N}(0,1).$$
Let $S_n^{(1)} := sum_{k=1}^n Z_k$ and $S_n^{(2)}: = sum_{k=1}^n R_k$ with $R_k = X_k - Z_k = X_m 1_{{|X_m|>1}}+(1_{{X_m<-1}}-1_{{X_m>1}})$. If we can show that $(sqrt{n})^{-1} S_n^{(2)} rightarrow 0$ e.g. in $L^1$, then this already implies convergence in distribution. By Slutsky's theorem we can can could that
$$frac{1}{sqrt{n}} sum_{k=1}^n X_k = frac{1}{sqrt{n}} S_n^{(1)} + frac{1}{sqrt{n}} S_n^{(2)} Rightarrow mathcal{N}(0,1)+0 =mathcal{N}(0,1).$$
It remains to show that $(sqrt{n})^{-1}mathbb{E}[|S_n^{(2)}|] rightarrow 0$. In fact, we have
begin{align}
frac{1}{sqrt{n}} mathbb{E}| S_n^{(2)} | &le frac{1}{sqrt{n}} sum_{k=1}^n mathbb{E}(1+|X_m|) 1_{{|X_m|>1}} le frac{2}{sqrt{n}} sum_{k=1}^n mathbb{E}|X_m| 1_{{|X_m|>1}} \
& lefrac{2}{sqrt{n}} sum_{k=1}^n mathbb{E}|X_m| 1_{{|X_m|>1}} lefrac{2}{sqrt{n}} sum_{k=1}^n frac{1}{k} ll frac{log(n)}{sqrt{n}},
end{align}
where the last step can be established by using an integral comparison argument. (The sum is at most $1+int_1^{n} frac{1}{x} , d x$.)
answered Nov 21 '18 at 7:36
p4sch
4,770217
Set $Z_m = X_{m} 1_{{|X_m| le 1}} + 1_{{X_m>1}}-1_{{X_m<-1}}$. Then $(Z_i)_{i in mathbb{N}}$ is a sequence of i.i.d. random variables with $mathbb{E}(Z_i) =0$ and $mathrm{Var}(Z_i) =1$. Thus, we can conclude by the central limit law that
$$frac{1}{sqrt{n}}sum_{k=1}^n Z_k Rightarrow mathcal{N}(0,1).$$
Let $S_n^{(1)} := sum_{k=1}^n Z_k$ and $S_n^{(2)}: = sum_{k=1}^n R_k$ with $R_k = X_k - Z_k = X_m 1_{{|X_m|>1}}+(1_{{X_m<-1}}-1_{{X_m>1}})$. If we can show that $(sqrt{n})^{-1} S_n^{(2)} rightarrow 0$ e.g. in $L^1$, then this already implies convergence in distribution. By Slutsky's theorem we can can could that
$$frac{1}{sqrt{n}} sum_{k=1}^n X_k = frac{1}{sqrt{n}} S_n^{(1)} + frac{1}{sqrt{n}} S_n^{(2)} Rightarrow mathcal{N}(0,1)+0 =mathcal{N}(0,1).$$
It remains to show that $(sqrt{n})^{-1}mathbb{E}[|S_n^{(2)}|] rightarrow 0$. In fact, we have
begin{align}
frac{1}{sqrt{n}} mathbb{E}| S_n^{(2)} | &le frac{1}{sqrt{n}} sum_{k=1}^n mathbb{E}(1+|X_m|) 1_{{|X_m|>1}} le frac{2}{sqrt{n}} sum_{k=1}^n mathbb{E}|X_m| 1_{{|X_m|>1}} \
& lefrac{2}{sqrt{n}} sum_{k=1}^n mathbb{E}|X_m| 1_{{|X_m|>1}} lefrac{2}{sqrt{n}} sum_{k=1}^n frac{1}{k} ll frac{log(n)}{sqrt{n}},
end{align}
where the last step can be established by using an integral comparison argument. (The sum is at most $1+int_1^{n} frac{1}{x} , d x$.)
answered Nov 21 '18 at 7:36
p4sch
4,770217
answered Nov 21 '18 at 7:36
p4sch
4,770217
answered Nov 21 '18 at 7:36
p4sch
4,770217
answered Nov 21 '18 at 7:36
p4sch
4,770217
4,770217
add a comment |
add a comment |
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You've defined $X_n$ but you've not defined $S_n$.
– heropup
Nov 21 '18 at 7:33