Almost sure convergence of recursively defined Random Variables I have a sequence of Random variables that is recursively defined in the following way
$$ Y_t = 1 + |Z_{t-1}|Y_{t-1} $$
Here the $Z_t$ are i.i.d. standard normal random variables.
I need to show almost sure convergence of the sequence $Y_t$. I have managed to show convergence in mean by taking expectations on both sides and finding
$$\lim_{t \rightarrow \infty} E[Y_t] = \frac{\sqrt{\pi}}{\sqrt{\pi}-\sqrt{2}}$$
Can I take the same approach by taking limits to conclude
$$ Y_t \rightarrow \frac{1}{1-|Z|}=Y \quad \text{a.s.}$$
Where $Z$ a standard normal R.V.? Or does this need a more sophisticated argument, i.e. I've tried showing $\sum E|Y_t -Y| <\infty$, but couldn't finish the argument
 A: The sequence never converges almost everywhere.
The proof proceeds in a number of steps.
Step 1: We show by induction that
$$
  Y_{t+1} \geq 1 + |Z_t| + |Z_t \cdots Z_1| \, Y_1
  \qquad \forall \, t \in \mathbb{N}_{\geq 2} .
$$
Indeed, for $t = 2$, we have
\begin{align*}
  Y_{t+1}
  & = 1 + |Z_t| \, Y_t \\
  & = 1 + |Z_2| \, Y_2 \\
  & = 1 + |Z_2| (1 + |Z_1| \, Y_1) \\
  & \geq 1 + |Z_2| + |Z_2 Z_1| \, Y_1 ,
\end{align*}
as claimed.
Next, if the claim holds for some $t \geq 2$, we see
\begin{align*}
  Y_{t+2}
  & = 1 + |Z_{t+1}| \, Y_{t+1} \\
  & \geq 1 + |Z_{t+1}| (1 + |Z_t| + |Z_t\cdots Z_1| \, Y_1) \\
  & \geq 1 + |Z_{t+1}| + |Z_{t+1}\cdots Z_1| \, Y_1 .
\end{align*}
Step 2:
We show that $V_t := |Z_t \cdots Z_1| \, Y_1 \to 0$ almost surely.
Since $Y_1$ is a real-valued random variable, it suffices to show
$|Z_t \cdots Z_1| \to 0$ almost surely.
But it is well-known that $\mathbb{E} |Z_t| = \sqrt{2/\pi} =: \theta < 1$.
By monotone convergence and independence, we thus see
$$
  \mathbb{E} \sum_{t=1}^\infty |Z_t \cdots Z_1|
  = \sum_{t=1}^\infty \mathbb{E} |Z_t \cdots Z_1|
  = \sum_{t=1}^\infty \theta^n
  < \infty,
$$
by convergence of the geometric series.
This shows $\sum_{t=1}^\infty |Z_t \cdots Z_1| < \infty$ almost surely.
Step 3: In this step we complete the proof.
Assume towards a contradiction that $Y_t \to Y$ almost surely.
Then $Y_{t+1} - V_t \to Y$ almost surely.
Since convergence sequences are bounded and since Step 1 shows
$0 \leq |Z_t| \leq 1 + |Z_t| \leq Y_{t+1} - V_t \to Y$,
this implies that $Z^\ast := \sup_t |Z_t| < \infty$ almost surely.
However, since the random variables $Z_t$ are i.i.d. standard normally distributed,
the second Borel-Cantelli Lemma
shows for arbitrary $M \in \mathbb{N}$ that the event $F_M := \{ |Z_t| \geq M \text{ infinitely often} \}$
has probability one.
This easily implies $Z^\ast = \infty$ almost surely,
which is the desired contradiction.
