Convex combination of a.s. convergent sequences

Question

Assume we have two sequences: $\{X_{n}\}$ and $\{Y_{n}\}$, such that $X_{n}\overset{a.s.}{\to} a$ and $X_{n}\overset{a.s.}{\to} a$, with $a$ some constant. Next, let $$ Z_{n} = \gamma_{n}X_{n} + (1 - \gamma_{n})Y_{n}, $$ where $\{\gamma_{n}\}$ is some sequence (random or not), such that $0 \leq \gamma_{n}\leq 1$, for all $n$.

Does there exist some sequence $\{\gamma_{n}\}$ such that

$$ Z_{n}\overset{a.s.}{\not\to} a \,? $$

What if we weaken the condition from a.s. convergence to convergence in probability? My hypothesis is that in this case the sequence exist.

Answer 1

Let $X'_n=X_n-a$ and $Y'_n=Y_n-a$. Then $$ Z_n=\gamma_n(X'_n+a)+(1-\gamma_n)(Y'_n+a)= a+\gamma_n X'_n +(1-\gamma_n) Y'_n. $$ Since the convergence of $X_n\to a$ and $Y_n\to a$ imply that $X'_n\to 0$ and $Y'_n\to 0$, it follows from $0\leqslant \lvert \gamma_n X'_n\rvert\leqslant \lvert X'_n\rvert$ and $0\leqslant \lvert \gamma_n Y'_n\rvert\leqslant \lvert Y'_n\rvert$ that $Z_n\to a$ almost surely.

If we assume that $X_n\to a$ and $Y_n\to a$ in probability, the same reasoning gives the convergence in probability of $(Z_n)$ to $a$.

Answer 2

Observe that for any $n$, \begin{align*} 0\leq|Z_n-a| &= |\gamma_n (X_n-a)+(1-\gamma_n )(Y_n-a)|\\ &\leq \gamma_n |X_n-a|+(1-\gamma_n )|Y_n-a|\\ &\leq |X_n-a| + |Y_n-a|\\ &\overset{a.s.}{\to} 0 \end{align*} This means that $|Z_n-a|\overset{a.s.}{\to} 0$ which implies that $Z_n\overset{a.s.}{\to} a$.

Answer 3

Note that $Z_n = \gamma_n (X_n - Y_n) + Y_n$, so it is sufficient to show $\gamma_n(X_n - Y_n) \to 0$ almost surely, which is clear, since $0 \le |\gamma_n(X_n-Y_n)| \le |X_n-Y_n|$.

If you assume that $X_n,Y_n \to a$ in probability, you can use characterisation:

$W_n \to W $ in probability iff for every subsequence $(n_k)$ there exists sub-subsequence $(n_{k_m})$ such that $W_{n_{k_m}} \to W$ almost surely.

Using it, the same reasoning gives you $\gamma_n (X_n - Y_n) \to 0$ in probability. (Taking any subsequence $(n_k)$ you have subsubsequnece $(n_{k_m})$ such that $X_{n_{k_m}} \to a$ almost surely, $Y_{n_{k_m}} \to a$ in probability, and for $Y$ we can choose sub-sub-subsequence $r_l = (n_{k_{m_l}})$ such that both $X_{r_l},Y_{r_l}$ converge to $a$ almost surely, and you're done)