Almost sure convergence of recursively defined Random Variables

Question

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

Answer 1

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.