Mixing
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$mathbb E[S_n'^4]le Ccdot n^2$
Let $X_1,...,X_n$ be iid random variables on $(Omega,mathcal A,mathbb P), mathbb E[X_1]=muinmathbb R, mathbb E[X_1]^4<infty, X_i':=X_i-mu, S_n'=X_1'+...+X_n'.$
Prove that $mathbb E[S_n'^4]le Ccdot n^2$ for $C>0 $ constant.
I have tried everything I could think of already, but it didn't lead anywhere. My best shot was
$$mathbb E[|S_n'|^4]=int_0^infty4t^3 mathbb P(|S_n|ge t) text{d}t le 4int_0^{infty} t^3cdot t^{-2} Var(S_n) text{d}t= 4int_0^{infty} tcdot Var(S_n) text{d}t\=4int_0^{infty} t n Var(X_1) text{d}t$$ with Tchebychef. Can anyone help me with that?
stochastic-integrals integral-inequality expected-value
asked Nov 21 '18 at 16:40
newbie
318212
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Let $X_1,...,X_n$ be iid random variables on $(Omega,mathcal A,mathbb P), mathbb E[X_1]=muinmathbb R, mathbb E[X_1]^4<infty, X_i':=X_i-mu, S_n'=X_1'+...+X_n'.$
Prove that $mathbb E[S_n'^4]le Ccdot n^2$ for $C>0 $ constant.
I have tried everything I could think of already, but it didn't lead anywhere. My best shot was
$$mathbb E[|S_n'|^4]=int_0^infty4t^3 mathbb P(|S_n|ge t) text{d}t le 4int_0^{infty} t^3cdot t^{-2} Var(S_n) text{d}t= 4int_0^{infty} tcdot Var(S_n) text{d}t\=4int_0^{infty} t n Var(X_1) text{d}t$$ with Tchebychef. Can anyone help me with that?
stochastic-integrals integral-inequality expected-value
asked Nov 21 '18 at 16:40
newbie
318212
add a comment |
Let $X_1,...,X_n$ be iid random variables on $(Omega,mathcal A,mathbb P), mathbb E[X_1]=muinmathbb R, mathbb E[X_1]^4<infty, X_i':=X_i-mu, S_n'=X_1'+...+X_n'.$
Prove that $mathbb E[S_n'^4]le Ccdot n^2$ for $C>0 $ constant.
I have tried everything I could think of already, but it didn't lead anywhere. My best shot was
$$mathbb E[|S_n'|^4]=int_0^infty4t^3 mathbb P(|S_n|ge t) text{d}t le 4int_0^{infty} t^3cdot t^{-2} Var(S_n) text{d}t= 4int_0^{infty} tcdot Var(S_n) text{d}t\=4int_0^{infty} t n Var(X_1) text{d}t$$ with Tchebychef. Can anyone help me with that?
stochastic-integrals integral-inequality expected-value
asked Nov 21 '18 at 16:40
newbie
318212
Let $X_1,...,X_n$ be iid random variables on $(Omega,mathcal A,mathbb P), mathbb E[X_1]=muinmathbb R, mathbb E[X_1]^4<infty, X_i':=X_i-mu, S_n'=X_1'+...+X_n'.$
Prove that $mathbb E[S_n'^4]le Ccdot n^2$ for $C>0 $ constant.
I have tried everything I could think of already, but it didn't lead anywhere. My best shot was
$$mathbb E[|S_n'|^4]=int_0^infty4t^3 mathbb P(|S_n|ge t) text{d}t le 4int_0^{infty} t^3cdot t^{-2} Var(S_n) text{d}t= 4int_0^{infty} tcdot Var(S_n) text{d}t\=4int_0^{infty} t n Var(X_1) text{d}t$$ with Tchebychef. Can anyone help me with that?
stochastic-integrals integral-inequality expected-value
stochastic-integrals integral-inequality expected-value
asked Nov 21 '18 at 16:40
newbie
318212
asked Nov 21 '18 at 16:40
newbie
318212
asked Nov 21 '18 at 16:40
newbie
318212
asked Nov 21 '18 at 16:40
newbie
318212
asked Nov 21 '18 at 16:40
newbie
318212
318212
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add a comment |
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Expand $mathbb{E} [(S_n')^4] = mathbb{E} [(X_1' + ldots + X_n')^4]$. Notice that terms $mathbb{E}[X_i' (X_j')^3]$ with $i neq j$ will vanish using independence and $mathbb{E}[X'_i] = 0$. For terms $mathbb{E} [(X_i')^2 (X_j')^2]$ with $i neq j$, notice that they are also independent and that by Cauchy-Schwarz $mathbb{E} [(X_i')^4] geq mathbb{E}[(X_i')^2]^2$. So, in the end all non-vanishing terms can be bounded by the same constant $C = mathbb{E} [(X_i')^4]$ and there are at $n$ terms $mathbb{E} [(X_i')^4]$ and $binom{n}{2} cdot binom{4}{2}$ terms $mathbb{E} [(X_i')^2 (X_j')^2]$.
answered Nov 21 '18 at 17:02
Daniel
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Expand $mathbb{E} [(S_n')^4] = mathbb{E} [(X_1' + ldots + X_n')^4]$. Notice that terms $mathbb{E}[X_i' (X_j')^3]$ with $i neq j$ will vanish using independence and $mathbb{E}[X'_i] = 0$. For terms $mathbb{E} [(X_i')^2 (X_j')^2]$ with $i neq j$, notice that they are also independent and that by Cauchy-Schwarz $mathbb{E} [(X_i')^4] geq mathbb{E}[(X_i')^2]^2$. So, in the end all non-vanishing terms can be bounded by the same constant $C = mathbb{E} [(X_i')^4]$ and there are at $n$ terms $mathbb{E} [(X_i')^4]$ and $binom{n}{2} cdot binom{4}{2}$ terms $mathbb{E} [(X_i')^2 (X_j')^2]$.
answered Nov 21 '18 at 17:02
Daniel
1,516210
add a comment |
Expand $mathbb{E} [(S_n')^4] = mathbb{E} [(X_1' + ldots + X_n')^4]$. Notice that terms $mathbb{E}[X_i' (X_j')^3]$ with $i neq j$ will vanish using independence and $mathbb{E}[X'_i] = 0$. For terms $mathbb{E} [(X_i')^2 (X_j')^2]$ with $i neq j$, notice that they are also independent and that by Cauchy-Schwarz $mathbb{E} [(X_i')^4] geq mathbb{E}[(X_i')^2]^2$. So, in the end all non-vanishing terms can be bounded by the same constant $C = mathbb{E} [(X_i')^4]$ and there are at $n$ terms $mathbb{E} [(X_i')^4]$ and $binom{n}{2} cdot binom{4}{2}$ terms $mathbb{E} [(X_i')^2 (X_j')^2]$.
answered Nov 21 '18 at 17:02
Daniel
1,516210
add a comment |
Expand $mathbb{E} [(S_n')^4] = mathbb{E} [(X_1' + ldots + X_n')^4]$. Notice that terms $mathbb{E}[X_i' (X_j')^3]$ with $i neq j$ will vanish using independence and $mathbb{E}[X'_i] = 0$. For terms $mathbb{E} [(X_i')^2 (X_j')^2]$ with $i neq j$, notice that they are also independent and that by Cauchy-Schwarz $mathbb{E} [(X_i')^4] geq mathbb{E}[(X_i')^2]^2$. So, in the end all non-vanishing terms can be bounded by the same constant $C = mathbb{E} [(X_i')^4]$ and there are at $n$ terms $mathbb{E} [(X_i')^4]$ and $binom{n}{2} cdot binom{4}{2}$ terms $mathbb{E} [(X_i')^2 (X_j')^2]$.
answered Nov 21 '18 at 17:02
Daniel
1,516210
Expand $mathbb{E} [(S_n')^4] = mathbb{E} [(X_1' + ldots + X_n')^4]$. Notice that terms $mathbb{E}[X_i' (X_j')^3]$ with $i neq j$ will vanish using independence and $mathbb{E}[X'_i] = 0$. For terms $mathbb{E} [(X_i')^2 (X_j')^2]$ with $i neq j$, notice that they are also independent and that by Cauchy-Schwarz $mathbb{E} [(X_i')^4] geq mathbb{E}[(X_i')^2]^2$. So, in the end all non-vanishing terms can be bounded by the same constant $C = mathbb{E} [(X_i')^4]$ and there are at $n$ terms $mathbb{E} [(X_i')^4]$ and $binom{n}{2} cdot binom{4}{2}$ terms $mathbb{E} [(X_i')^2 (X_j')^2]$.
answered Nov 21 '18 at 17:02
Daniel
1,516210
answered Nov 21 '18 at 17:02
Daniel
1,516210
answered Nov 21 '18 at 17:02
Daniel
1,516210
answered Nov 21 '18 at 17:02
Daniel
1,516210
1,516210
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