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Suppose $X_n$ is a random variable such that $X_n=O(b_n)$ almost surely, with $b_n\to 0$ as $n\to \infty$. Let $C$ be a real constant and $S_{j,k}(X_n)=\sum_{i=j}^kC^iX_n^i$ for $0<j<k<\infty$. Then, since $X_n=O(b_n)$ it follows that $S_{j,k}=O(b_n^j)$ almost surely.
I have troubles with the case in which $k$ goes to infinity, that is, $S=\sum_{i=1}^\infty C^iX_n^i$ because $S$ does not necessarily converge. Is it still true that $S=O(b_n)$ almost surely?
Thanks in advance.
Let $(x_n)$ denote a real valued sequence such that $x_n\to0$, and $C$ some constant. Then, for every $n$ large enough, $s_n=\sum\limits_{i=1}^\infty C^ix_n^i$ converges and $|s_n|\leqslant2|C|\,|x_n|$.
(Proof: If $C=0$, $s_n=0$ hence there is nothing to do. If $C\ne0$, there exists some finite $n_0$ such that for every $n\geqslant n_0$, $|C|\,|x_n|\leqslant\frac12$. Then $|s_n|\leqslant\sum\limits_{i=1}^\infty|C|^i|x_n|^i\leqslant|C|\,|x_n|\,\sum\limits_{i=1}^\infty2^{1-i}$ implies that $|s_n|\leqslant2|C|\,|x_n|$. QED.)
Apply this to the random variable $S_n=\sum\limits_{i=1}^\infty C^iX_n^i$. If $X_n=O(b_n)$ and $b_n\to0$, there exists some $N_0$ such that $N_0$ is almost surely finite and, for every $n\geqslant N_0$, $|C|\,|X_n|\leqslant\frac12$. Then, $|S_n|\leqslant2|C|\,|X_n|$ for every $n\geqslant N_0$, in particular $S_n$ converges (at least) for every $n\geqslant N_0$, and $S_n=O(b_n)$ almost surely.
