Suppose we have a sequence of positive random variables X 1 , X 2 ,...

qtbabe9876a9

qtbabe9876a9

Answered

2022-05-29

Suppose we have a sequence of positive random variables X 1 , X 2 , . . . , X. I am trying to prove a characterization of almost sure convergence.
It states that X n X almost surely iff for every ϵ > 0, lim n P [ sup k n X k X > 1 + ϵ ] = 0 and lim n P [ sup k n X X k > 1 + ϵ ] = 0.
If I assume almost sure convergence, then the implication is easy but I am not being able to prove the other way round.

Answer & Explanation

concludirgt

concludirgt

Expert

2022-05-30Added 11 answers

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Since all the random variables are positive,
{ ω : X n ( ω ) X ( ω ) } = { ω : X n ( ω ) / X ( ω ) 1 } = { lim sup X n / X > 1 } { lim inf X n / X < 1 } .
Thus,
P ( X n X ) P ( lim sup X n / X > 1 ) + P ( lim inf X n / X < 1 ) .
But
P ( lim sup X n / X > 1 ) = lim m P ( lim sup X n / X 1 + m 1 ) = lim m lim n P ( sup k n X k / X 1 + m 1 ) .
That is, P ( lim sup X n / X > 1 ) = 0 is equivalent to
lim n P ( sup k n X k / X 1 + ϵ ) = 0 ϵ > 0 ,
and, similarly, P ( lim inf X n / X < 1 ) = 0 is equivalent to
lim n P ( inf k n X k / X 1 ϵ ) = 0 ϵ > 0.

The last condition is equivalent to lim n P ( sup k n X / X k 1 + ϵ ) = 0 ϵ > 0.
Trevor Wood

Trevor Wood

Expert

2022-05-31Added 5 answers

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Let
A n ϵ = { sup k n X k X > 1 + ϵ }  and  B n ϵ = { sup k n X X k > 1 + ϵ } .
Then A n + 1 ϵ A n ϵ and B n + 1 ϵ B n ϵ , thus
lim n P ( A n ϵ ) = P ( n = 1 A n ϵ ) = P ( lim sup n X n X > 1 + ϵ )
and
lim n P ( B n ϵ ) = P ( n = 1 B n ϵ ) = P ( lim sup n X X n > 1 + ϵ ) .
Since the above hold for every ϵ > 0 we conclude that
P ( lim sup n X n X > 1 ) = 0  and  P ( lim sup n X X n > 1 ) = 0.
These two are equivalent to the following two
P ( lim sup n X n X 1 ) = 1  and  P ( lim sup n X X n 1 ) = 1.
Now observe that lim sup n X n X = 1 X lim sup n X n and lim sup n X X n = X lim sup n 1 X n = X 1 lim inf n X n .. Therefore,
P ( lim sup n X n X ) = 1  and  P ( lim inf n X n X ) = 1 ,
taking intersections we conclude that X n a.s X

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