Is the following process bounded (iterative normal sampling)

Question

We define the following stochastic process:

• $X_0=1$
• $\forall i\geq i:X_i\sim\mathcal N(0,X_{i-1}^2)$

That is, we first sample $X_1$ from the normal distribution with variance $1$, then in the $i$-th iteration, $X_i$ is sampled from a normal distribution with mean $0$ and variance $X_{i-1}^2$.

Does the process $X$ converge (let's say $X$ converge if there exists $c, N_0\in \mathbb N$ such that $\forall n\geq N_0: Pr[|X_n|\leq c]>0$) ?

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If it converges, does it converge to $0$? (can we pick $N_0$ such that $c$ is arbitrarily small?).

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If we replace the normal distribution by any other (full-support) distribution with mean 0 and variance $X_i$, does the answer depend on the actual distribution?

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EDIT: A ran a short python code to check this and it seems $X$ converges to 0 every time.

Rerunning the code it seems that (everytime):

• If $\forall i\geq i:X_i\sim\mathcal N(0,1.5\cdot X_{i-1}^2)$ then $X\to 0$.
• If $\forall i\geq i:X_i\sim\mathcal N(0,2\cdot X_{i-1}^2)$ then $X$ diverges.

Can we compute the magic number $\tau$ (which is probably in $(1.5,2)$) such that if $\forall i\geq i:X_i\sim\mathcal N(0,\tau\cdot X_{i-1}^2)$ then $X$ stays bounded but doesn't converge to 0?

Answer 1

Note that, for every $i$, $X_i$ can be realized as $X_i=Z_iZ_{i-1}\cdots Z_1$ where $Z$ and $(Z_i)$ are i.i.d. standard notmal, hence $|X_i|=\exp(Y_i)$ where $Y_i$ is the sum of $i$ independent random variables distributed like $\log |Z|$. By concavity (twice), $E(\log|Z|)\lt\log E(|Z|)\lt\frac12\log E(Z^2)=0$ hence, by the law of large numbers, $\frac1iY_i\to\log E(|Z|)\lt0$ almost surely, hence $Y_i\to-\infty$ almost surely, which implies that $X_i\to0$ almost surely. This only uses that $E(Z^2)=1$.

Edit: To deal with recursions $X_i\sim\mathcal N(0,t\cdot X_{i-1}^2)$, use the representation $X_i=t^{i/2}Z_iZ_{i-1}\cdots Z_1$ hence $|X_i|=\exp(Y_i)$ where $Y_i$ is the sum of $i$ independent random variables distributed like $\frac12\log t+\log |Z|$ and the same reasoning shows that $|X_i|\to0$ for every $t\lt t_*$ and $|X_i|\to\infty$ for every $t\gt t_*$, where $$t_*=\exp(-2E(\log |Z|))=2\mathrm e^\gamma\approx3.562.$$ For recursions $X_i\sim\mathcal N(0,\tau^2\cdot X_{i-1}^2)$, the phase transition occurs at $\tau_*=\sqrt{2\mathrm e^\gamma}\approx1.887$.

At the critical value $\tau_*$ or $t_*$, CLT effects show that $(X_i)$ diverges since $$\liminf|X_i|=0,\qquad\limsup|X_i|=+\infty.$$