Suppose we have a sequence of positive random variables X 1 </msub> ,

qtbabe9876a9

qtbabe9876a9

Answered question

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

Beginner2022-05-30Added 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

Beginner2022-05-31Added 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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