$mathbb E[S_n'^4]le Ccdot n^2$











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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?










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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?










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      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      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?










      share|cite|improve this question













      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






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      asked Nov 21 at 16:40









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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 $mathbb{E} [(X_i')^2 (X_j')^2]$.






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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 $mathbb{E} [(X_i')^2 (X_j')^2]$.






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              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 $mathbb{E} [(X_i')^2 (X_j')^2]$.






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                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 $mathbb{E} [(X_i')^2 (X_j')^2]$.






                share|cite|improve this answer












                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 $mathbb{E} [(X_i')^2 (X_j')^2]$.







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                answered Nov 21 at 17:02









                Daniel

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