If $X_nto X$ in $L^p$, is $E(X_n^p)to E(X^p)$ true?












1















Let $p$ denote a positive integer. If $X_nto X$ in $L^p$, is $E(X_n^p)to E(X^p)$ true?




If each $X_n$ and $X$ are nonnegative, then it follows directly from Minkowski's inequality.



Then how about the general case? I separated each $X_n$ into its positive and negative parts as $X_n=X_n^+-X_n^-$, but it might not be the case that $X_n^+to X^+$ and $X_n^-to X^-$ in $L^p$.






probability-theory lp-spaces







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  • math.stackexchange.com/a/1250473
    – Did
    Dec 2 at 12:04










  • @Did It seems to me like the post you've linked deals with a kind of converse to what is asked here?
    – Rhys Steele
    Dec 2 at 12:22










  • True, I was too hasty, sorry about that. Leaving the link nonetheless, as an element of context.
    – Did
    Dec 2 at 12:24
















1















Let $p$ denote a positive integer. If $X_nto X$ in $L^p$, is $E(X_n^p)to E(X^p)$ true?




If each $X_n$ and $X$ are nonnegative, then it follows directly from Minkowski's inequality.



Then how about the general case? I separated each $X_n$ into its positive and negative parts as $X_n=X_n^+-X_n^-$, but it might not be the case that $X_n^+to X^+$ and $X_n^-to X^-$ in $L^p$.






probability-theory lp-spaces







share|cite|improve this question
























  • math.stackexchange.com/a/1250473
    – Did
    Dec 2 at 12:04










  • @Did It seems to me like the post you've linked deals with a kind of converse to what is asked here?
    – Rhys Steele
    Dec 2 at 12:22










  • True, I was too hasty, sorry about that. Leaving the link nonetheless, as an element of context.
    – Did
    Dec 2 at 12:24














1












1








1








Let $p$ denote a positive integer. If $X_nto X$ in $L^p$, is $E(X_n^p)to E(X^p)$ true?




If each $X_n$ and $X$ are nonnegative, then it follows directly from Minkowski's inequality.



Then how about the general case? I separated each $X_n$ into its positive and negative parts as $X_n=X_n^+-X_n^-$, but it might not be the case that $X_n^+to X^+$ and $X_n^-to X^-$ in $L^p$.






probability-theory lp-spaces







share|cite|improve this question
















Let $p$ denote a positive integer. If $X_nto X$ in $L^p$, is $E(X_n^p)to E(X^p)$ true?




If each $X_n$ and $X$ are nonnegative, then it follows directly from Minkowski's inequality.



Then how about the general case? I separated each $X_n$ into its positive and negative parts as $X_n=X_n^+-X_n^-$, but it might not be the case that $X_n^+to X^+$ and $X_n^-to X^-$ in $L^p$.






probability-theory lp-spaces




probability-theory lp-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 at 12:42









Did

246k23220454




246k23220454










asked Dec 2 at 11:37









bellcircle

1,327411




1,327411












  • math.stackexchange.com/a/1250473
    – Did
    Dec 2 at 12:04










  • @Did It seems to me like the post you've linked deals with a kind of converse to what is asked here?
    – Rhys Steele
    Dec 2 at 12:22










  • True, I was too hasty, sorry about that. Leaving the link nonetheless, as an element of context.
    – Did
    Dec 2 at 12:24


















  • math.stackexchange.com/a/1250473
    – Did
    Dec 2 at 12:04










  • @Did It seems to me like the post you've linked deals with a kind of converse to what is asked here?
    – Rhys Steele
    Dec 2 at 12:22










  • True, I was too hasty, sorry about that. Leaving the link nonetheless, as an element of context.
    – Did
    Dec 2 at 12:24
















math.stackexchange.com/a/1250473
– Did
Dec 2 at 12:04




math.stackexchange.com/a/1250473
– Did
Dec 2 at 12:04












@Did It seems to me like the post you've linked deals with a kind of converse to what is asked here?
– Rhys Steele
Dec 2 at 12:22




@Did It seems to me like the post you've linked deals with a kind of converse to what is asked here?
– Rhys Steele
Dec 2 at 12:22












True, I was too hasty, sorry about that. Leaving the link nonetheless, as an element of context.
– Did
Dec 2 at 12:24




True, I was too hasty, sorry about that. Leaving the link nonetheless, as an element of context.
– Did
Dec 2 at 12:24










2 Answers
2






active

oldest

votes


















3














For a simple, self-contained, proof, one can start from the binomial identity $$X_n^p-X^p=sum_{k=1}^p{pchoose k}(X_n-X)^kX^{p-k}$$ which yields $$|E(X_n^p)-E(X^p)|leqslantsum_{k=1}^p{pchoose k}E(|X_n-X|^k|X|^{p-k})$$ Now, for each $1leqslant kleqslant p$, by convexity, $$E(|X_n-X|^k|X|^{p-k})leqslant E(|X_n-X|^p)^{k/p}E(|X|^p)^{1-k/p}$$ hence each term in the sum on the RHS goes to $0$ as soon as $X_nto X$ in $L^p$.



The method yields an effective upper bound of $|E(X_n^p)-E(X^p)|$ in terms of $|X_n-X|_p$ and $|X|_p$, which we leave to the reader.






share|cite|improve this answer





























    0














    We have
    $$|mathbb{E}[X_n^p - X^p]| leq p mathbb{E}[(|X_n| + |X|)^{p-1} |X_n - X|]$$
    since for any $x,y in mathbb{R}$, $|x^p - y^p| leq p (|x| + |y|)^p |x-y|$ (for example, apply the mean value theorem to $t mapsto t^p$).



    So by Holder's inequality,
    begin{align*}
    |mathbb{E}[X_n^p - X^p]| leq& p |X_n - X|_p mathbb{E}[(|X_n| + |X|)^p]^{frac{p-1}{p}} \
    leq & 2^{p-1} p |X_n - X|_p (|X_n|_p^{p} + |X|_p^p)^{frac{p-1}{p}} to 0
    end{align*}
    since $|X_n|_p$ is a bounded sequence and $|X_n - X|_p to 0$.






    share|cite|improve this answer























    • $L^p$ norm is $mathbb{E}|X^p|^{1/p}$, not merely $(mathbb{E}X^p)^{1/p}$.
      – bellcircle
      Dec 2 at 11:56











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    For a simple, self-contained, proof, one can start from the binomial identity $$X_n^p-X^p=sum_{k=1}^p{pchoose k}(X_n-X)^kX^{p-k}$$ which yields $$|E(X_n^p)-E(X^p)|leqslantsum_{k=1}^p{pchoose k}E(|X_n-X|^k|X|^{p-k})$$ Now, for each $1leqslant kleqslant p$, by convexity, $$E(|X_n-X|^k|X|^{p-k})leqslant E(|X_n-X|^p)^{k/p}E(|X|^p)^{1-k/p}$$ hence each term in the sum on the RHS goes to $0$ as soon as $X_nto X$ in $L^p$.



    The method yields an effective upper bound of $|E(X_n^p)-E(X^p)|$ in terms of $|X_n-X|_p$ and $|X|_p$, which we leave to the reader.






    share|cite|improve this answer


























      3














      For a simple, self-contained, proof, one can start from the binomial identity $$X_n^p-X^p=sum_{k=1}^p{pchoose k}(X_n-X)^kX^{p-k}$$ which yields $$|E(X_n^p)-E(X^p)|leqslantsum_{k=1}^p{pchoose k}E(|X_n-X|^k|X|^{p-k})$$ Now, for each $1leqslant kleqslant p$, by convexity, $$E(|X_n-X|^k|X|^{p-k})leqslant E(|X_n-X|^p)^{k/p}E(|X|^p)^{1-k/p}$$ hence each term in the sum on the RHS goes to $0$ as soon as $X_nto X$ in $L^p$.



      The method yields an effective upper bound of $|E(X_n^p)-E(X^p)|$ in terms of $|X_n-X|_p$ and $|X|_p$, which we leave to the reader.






      share|cite|improve this answer
























        3












        3








        3






        For a simple, self-contained, proof, one can start from the binomial identity $$X_n^p-X^p=sum_{k=1}^p{pchoose k}(X_n-X)^kX^{p-k}$$ which yields $$|E(X_n^p)-E(X^p)|leqslantsum_{k=1}^p{pchoose k}E(|X_n-X|^k|X|^{p-k})$$ Now, for each $1leqslant kleqslant p$, by convexity, $$E(|X_n-X|^k|X|^{p-k})leqslant E(|X_n-X|^p)^{k/p}E(|X|^p)^{1-k/p}$$ hence each term in the sum on the RHS goes to $0$ as soon as $X_nto X$ in $L^p$.



        The method yields an effective upper bound of $|E(X_n^p)-E(X^p)|$ in terms of $|X_n-X|_p$ and $|X|_p$, which we leave to the reader.






        share|cite|improve this answer












        For a simple, self-contained, proof, one can start from the binomial identity $$X_n^p-X^p=sum_{k=1}^p{pchoose k}(X_n-X)^kX^{p-k}$$ which yields $$|E(X_n^p)-E(X^p)|leqslantsum_{k=1}^p{pchoose k}E(|X_n-X|^k|X|^{p-k})$$ Now, for each $1leqslant kleqslant p$, by convexity, $$E(|X_n-X|^k|X|^{p-k})leqslant E(|X_n-X|^p)^{k/p}E(|X|^p)^{1-k/p}$$ hence each term in the sum on the RHS goes to $0$ as soon as $X_nto X$ in $L^p$.



        The method yields an effective upper bound of $|E(X_n^p)-E(X^p)|$ in terms of $|X_n-X|_p$ and $|X|_p$, which we leave to the reader.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 at 12:36









        Did

        246k23220454




        246k23220454























            0














            We have
            $$|mathbb{E}[X_n^p - X^p]| leq p mathbb{E}[(|X_n| + |X|)^{p-1} |X_n - X|]$$
            since for any $x,y in mathbb{R}$, $|x^p - y^p| leq p (|x| + |y|)^p |x-y|$ (for example, apply the mean value theorem to $t mapsto t^p$).



            So by Holder's inequality,
            begin{align*}
            |mathbb{E}[X_n^p - X^p]| leq& p |X_n - X|_p mathbb{E}[(|X_n| + |X|)^p]^{frac{p-1}{p}} \
            leq & 2^{p-1} p |X_n - X|_p (|X_n|_p^{p} + |X|_p^p)^{frac{p-1}{p}} to 0
            end{align*}
            since $|X_n|_p$ is a bounded sequence and $|X_n - X|_p to 0$.






            share|cite|improve this answer























            • $L^p$ norm is $mathbb{E}|X^p|^{1/p}$, not merely $(mathbb{E}X^p)^{1/p}$.
              – bellcircle
              Dec 2 at 11:56
















            0














            We have
            $$|mathbb{E}[X_n^p - X^p]| leq p mathbb{E}[(|X_n| + |X|)^{p-1} |X_n - X|]$$
            since for any $x,y in mathbb{R}$, $|x^p - y^p| leq p (|x| + |y|)^p |x-y|$ (for example, apply the mean value theorem to $t mapsto t^p$).



            So by Holder's inequality,
            begin{align*}
            |mathbb{E}[X_n^p - X^p]| leq& p |X_n - X|_p mathbb{E}[(|X_n| + |X|)^p]^{frac{p-1}{p}} \
            leq & 2^{p-1} p |X_n - X|_p (|X_n|_p^{p} + |X|_p^p)^{frac{p-1}{p}} to 0
            end{align*}
            since $|X_n|_p$ is a bounded sequence and $|X_n - X|_p to 0$.






            share|cite|improve this answer























            • $L^p$ norm is $mathbb{E}|X^p|^{1/p}$, not merely $(mathbb{E}X^p)^{1/p}$.
              – bellcircle
              Dec 2 at 11:56














            0












            0








            0






            We have
            $$|mathbb{E}[X_n^p - X^p]| leq p mathbb{E}[(|X_n| + |X|)^{p-1} |X_n - X|]$$
            since for any $x,y in mathbb{R}$, $|x^p - y^p| leq p (|x| + |y|)^p |x-y|$ (for example, apply the mean value theorem to $t mapsto t^p$).



            So by Holder's inequality,
            begin{align*}
            |mathbb{E}[X_n^p - X^p]| leq& p |X_n - X|_p mathbb{E}[(|X_n| + |X|)^p]^{frac{p-1}{p}} \
            leq & 2^{p-1} p |X_n - X|_p (|X_n|_p^{p} + |X|_p^p)^{frac{p-1}{p}} to 0
            end{align*}
            since $|X_n|_p$ is a bounded sequence and $|X_n - X|_p to 0$.






            share|cite|improve this answer














            We have
            $$|mathbb{E}[X_n^p - X^p]| leq p mathbb{E}[(|X_n| + |X|)^{p-1} |X_n - X|]$$
            since for any $x,y in mathbb{R}$, $|x^p - y^p| leq p (|x| + |y|)^p |x-y|$ (for example, apply the mean value theorem to $t mapsto t^p$).



            So by Holder's inequality,
            begin{align*}
            |mathbb{E}[X_n^p - X^p]| leq& p |X_n - X|_p mathbb{E}[(|X_n| + |X|)^p]^{frac{p-1}{p}} \
            leq & 2^{p-1} p |X_n - X|_p (|X_n|_p^{p} + |X|_p^p)^{frac{p-1}{p}} to 0
            end{align*}
            since $|X_n|_p$ is a bounded sequence and $|X_n - X|_p to 0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 2 at 12:19

























            answered Dec 2 at 11:50









            Rhys Steele

            6,0431829




            6,0431829












            • $L^p$ norm is $mathbb{E}|X^p|^{1/p}$, not merely $(mathbb{E}X^p)^{1/p}$.
              – bellcircle
              Dec 2 at 11:56


















            • $L^p$ norm is $mathbb{E}|X^p|^{1/p}$, not merely $(mathbb{E}X^p)^{1/p}$.
              – bellcircle
              Dec 2 at 11:56
















            $L^p$ norm is $mathbb{E}|X^p|^{1/p}$, not merely $(mathbb{E}X^p)^{1/p}$.
            – bellcircle
            Dec 2 at 11:56




            $L^p$ norm is $mathbb{E}|X^p|^{1/p}$, not merely $(mathbb{E}X^p)^{1/p}$.
            – bellcircle
            Dec 2 at 11:56


















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