If $X_n to X_n$ in r-th mean implies $E|X_n^r|to E|X^r|$











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Im reading this article in Wikipedia : https://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_mean



In the section Convergence in mean say that if $rgeq 1$ real number and ${X_n}$ a sequence of v.a. souch that $X_n to X$ in r-th mean, this implies $E|X_n^r|to E|X^r|$.



I don't have clear why this result is true and in the wiki there is no explanation.






probability-theory







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    It's a consequence of the reverse triangle inequality for the $L^r$-norm.
    – saz
    2 days ago















up vote
0
down vote

favorite












Im reading this article in Wikipedia : https://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_mean



In the section Convergence in mean say that if $rgeq 1$ real number and ${X_n}$ a sequence of v.a. souch that $X_n to X$ in r-th mean, this implies $E|X_n^r|to E|X^r|$.



I don't have clear why this result is true and in the wiki there is no explanation.






probability-theory







share|cite|improve this question


















  • 1




    It's a consequence of the reverse triangle inequality for the $L^r$-norm.
    – saz
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Im reading this article in Wikipedia : https://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_mean



In the section Convergence in mean say that if $rgeq 1$ real number and ${X_n}$ a sequence of v.a. souch that $X_n to X$ in r-th mean, this implies $E|X_n^r|to E|X^r|$.



I don't have clear why this result is true and in the wiki there is no explanation.






probability-theory







share|cite|improve this question













Im reading this article in Wikipedia : https://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_mean



In the section Convergence in mean say that if $rgeq 1$ real number and ${X_n}$ a sequence of v.a. souch that $X_n to X$ in r-th mean, this implies $E|X_n^r|to E|X^r|$.



I don't have clear why this result is true and in the wiki there is no explanation.






probability-theory




probability-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









Malena Manzanares

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243








  • 1




    It's a consequence of the reverse triangle inequality for the $L^r$-norm.
    – saz
    2 days ago














  • 1




    It's a consequence of the reverse triangle inequality for the $L^r$-norm.
    – saz
    2 days ago








1




1




It's a consequence of the reverse triangle inequality for the $L^r$-norm.
– saz
2 days ago




It's a consequence of the reverse triangle inequality for the $L^r$-norm.
– saz
2 days ago















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