If $X_n$ iid, $\sum X_n$ conv a.s., then $\sum X_n 1\{|Y_n|\leq n\}$ conv a.s. for some identically distributed $L^1$ RV $Y_n$?

Question

Let $X_n$ be iid RVs, $S_n=X_1+...+X_n$. Suppose $S_n \rightarrow_{n\rightarrow \infty} S$ a.s.(almost surely). Let $Y_n$ be identically distributed and $E|Y_1| \lt \infty$.

Show $S_n':= \sum_{k=1}^n X_k \ 1\{|Y_k| \leq k\} \rightarrow_{n\rightarrow \infty} S'\ \ \text{a.s.}$ for some RV $S'$.

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What I have tried:

By 3 series theorem, for any positive $A$, $\sum P(|X_n|>A) \lt \infty$ $\sum E\bar X_n \in \mathbb R$ , $\sum var \bar X_n \lt \infty$ where $\bar X_n :=X_n 1\{|X_n | \leq A \}$ .

We check the 3 series for $S_n'$;

Fix $A>0$.

1. $\sum P(|X_n \ 1\{|Y_n| \leq n\}|>A) \lt \sum P(|X_n| >A) <\infty$.

2. $\sum Var(\ \ X_n \ 1\{|Y_n| \leq n\} \ \ 1\{|X_n| \ 1\{|Y_n| \leq n\} \leq A\}\ \ )\leq \sum E(\ \ X_n^2 \ 1\{|Y_n| \leq n\} \ \ 1\{|X_n| \ 1\{|Y_n| \leq n\}\leq A\}\ \ ) \leq \sum E(\ \ X_n^2 \ 1\{|Y_n| \leq n \ , |X_n | \ 1\{|Y_n| \leq n\}\leq A\}\ \ ) = \sum E(\ \ X_n^2 \ 1\{|Y_n| \leq n \ , |X_n | \ \leq A\}\ \ ) \leq \sum E(\ \ X_n^2 \ 1\{ |X_n | \ \leq A\}\ \ ) \lt \infty$

since $$\sum E\bar X_n= \sum E\bar X_1 \lt \infty \Rightarrow \forall n,\ E \bar X_n=E \bar X_1 =0 \Rightarrow \sum var \bar X_n=\sum E\bar X_n^2 \lt \infty$$ which is the very last inequality in the previous paragraph.

Now,(this is the part where I get stuck)

3. $\sum E (\ \ X_n \ 1\{|Y_n| \leq n\} \ \ 1\{|X_n| \ 1\{|Y_n| \leq n\}\leq A\}\ \ ) =\sum E(\ \ X_n \ 1\{|Y_n| \leq n, \ \ |X_n| 1\{|Y_n| \leq n\} \leq A\}\ \ ) =\sum E(\ \ X_n \ 1\{|Y_n| \leq n, \ \ |X_n| \leq A\}\ \ )$

I don't see how to proceed on to show this sum exists. I know I need to use the integrability of $Y_n$ somewhere, but can't figure out how.

Any hint is appreciated.

Answer 1

Consider the difference between $\sum E\overline {X_n}$ and $\sum E(X_nI_{|Y_n|\leq n, |X_n| \leq A})$. The difference is dominated by the series $\sum A P\{|Y_n| >n\}$. But $\sum A P\{|Y_1| >n\}=AE|Y_1|<\infty$.