Jensen's inequality applied to Liapunov's CLT condition

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The Liapunov's sufficient condition to the Central Limit Theorem says that if $exists delta$ for whom the following expression converges to $0$ then you can apply the CLT to the sequence's standardized partial sums:



$$ frac1left(sqrtoperatornameVarS_nright)^2+deltasum_k=1^nmathbbE(|X_k-mu_k|^2+delta) to 0 implies fracS_n-mathbbE(S_n)sqrtoperatornameVarS_n xrightarrowD mathcalN(0,1)
$$




My question is if when we choose $delta =2$, can i apply Jensen's inequality, $mathbbE(phi(X)) geq phi(mathbbE(X))$ when $phi$ is a convex function, as in:



beginalign
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE(|X_k-mu_k|^4) &=\
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE((|X_k-mu_k|^2)^2) &geq\
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE(|X_k-mu_k|^2)^2 & \
endalign



It's a fairly simple doubt. Thanks in advance for the help.




probability inequality jensen-inequality




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  • 2




    Seems like a valid application of Jensen's inequality.
    – Alex R.
    Jul 31 at 21:14














up vote
2
down vote

favorite













The Liapunov's sufficient condition to the Central Limit Theorem says that if $exists delta$ for whom the following expression converges to $0$ then you can apply the CLT to the sequence's standardized partial sums:



$$ frac1left(sqrtoperatornameVarS_nright)^2+deltasum_k=1^nmathbbE(|X_k-mu_k|^2+delta) to 0 implies fracS_n-mathbbE(S_n)sqrtoperatornameVarS_n xrightarrowD mathcalN(0,1)
$$




My question is if when we choose $delta =2$, can i apply Jensen's inequality, $mathbbE(phi(X)) geq phi(mathbbE(X))$ when $phi$ is a convex function, as in:



beginalign
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE(|X_k-mu_k|^4) &=\
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE((|X_k-mu_k|^2)^2) &geq\
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE(|X_k-mu_k|^2)^2 & \
endalign



It's a fairly simple doubt. Thanks in advance for the help.




probability inequality jensen-inequality




share|cite|improve this question

















  • 2




    Seems like a valid application of Jensen's inequality.
    – Alex R.
    Jul 31 at 21:14












up vote
2
down vote

favorite









up vote
2
down vote

favorite












The Liapunov's sufficient condition to the Central Limit Theorem says that if $exists delta$ for whom the following expression converges to $0$ then you can apply the CLT to the sequence's standardized partial sums:



$$ frac1left(sqrtoperatornameVarS_nright)^2+deltasum_k=1^nmathbbE(|X_k-mu_k|^2+delta) to 0 implies fracS_n-mathbbE(S_n)sqrtoperatornameVarS_n xrightarrowD mathcalN(0,1)
$$




My question is if when we choose $delta =2$, can i apply Jensen's inequality, $mathbbE(phi(X)) geq phi(mathbbE(X))$ when $phi$ is a convex function, as in:



beginalign
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE(|X_k-mu_k|^4) &=\
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE((|X_k-mu_k|^2)^2) &geq\
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE(|X_k-mu_k|^2)^2 & \
endalign



It's a fairly simple doubt. Thanks in advance for the help.




probability inequality jensen-inequality




share|cite|improve this question














The Liapunov's sufficient condition to the Central Limit Theorem says that if $exists delta$ for whom the following expression converges to $0$ then you can apply the CLT to the sequence's standardized partial sums:



$$ frac1left(sqrtoperatornameVarS_nright)^2+deltasum_k=1^nmathbbE(|X_k-mu_k|^2+delta) to 0 implies fracS_n-mathbbE(S_n)sqrtoperatornameVarS_n xrightarrowD mathcalN(0,1)
$$




My question is if when we choose $delta =2$, can i apply Jensen's inequality, $mathbbE(phi(X)) geq phi(mathbbE(X))$ when $phi$ is a convex function, as in:



beginalign
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE(|X_k-mu_k|^4) &=\
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE((|X_k-mu_k|^2)^2) &geq\
frac1left(sqrtoperatornameVarS_nright)^4sum_k=1^nmathbbE(|X_k-mu_k|^2)^2 & \
endalign



It's a fairly simple doubt. Thanks in advance for the help.





probability inequality jensen-inequality





share|cite|improve this question












share|cite|improve this question




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edited Jul 31 at 20:57
























asked Jul 31 at 18:22









Sergio Andrade

14312




14312







  • 2




    Seems like a valid application of Jensen's inequality.
    – Alex R.
    Jul 31 at 21:14












  • 2




    Seems like a valid application of Jensen's inequality.
    – Alex R.
    Jul 31 at 21:14







2




2




Seems like a valid application of Jensen's inequality.
– Alex R.
Jul 31 at 21:14




Seems like a valid application of Jensen's inequality.
– Alex R.
Jul 31 at 21:14















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