I'm trying to prove convergence of a stochastic approximation-like algorithm. I have two questions about prove-techniques when working with randomness.
1. For a non-random sequence $(a_t)_t$ one could prove $a_t \rightarrow a$ by induction as follows: $$\text{Assume } |a_t - a| \leq \epsilon \text{ for some } t \text{ and show that } |a_{t+1} - a| \leq \delta \cdot \epsilon \text{ with } \delta \in (0, 1)$$
Now consider the random case. If one would knew that the induction hypothesis holds with probability $p_t$ and the induction step holds with probability $q_t$, e.g.: $$\mathbb{P}(|a_t - a| \leq \epsilon) \geq p_t, \text{ and if } |a_t - a| \leq \epsilon \text{ is true, then}$$ $$\mathbb{P}(|a_{t+1} - a| \leq \delta \cdot \epsilon \ \big| \ |a_t - a| \leq \epsilon) \geq q_t$$ Could one show that $a_t \rightarrow a$ almost surely? Furthermore, how would the conditions of $p_t$ and $q_t$ look like? I would assume that $\prod_t p_t \cdot q_t > C$ must hold.
2. Alternatively, for a non-random sequence $(a_t)_t$ one could prove $a_t \rightarrow a$ simply by showing that: $$|a_{t+1} - a| \leq \dots \leq \delta |a_t - a| \text{ with } \delta \in (0, 1) \ \forall t$$ Now consider again the random case. If the statement holds with probability $p_t$. Does $a_t \rightarrow a$ follow with probability $\prod_t p_t$ ?
Thanks in advance!
