Consider the following martingale: $X\_1 \sim \mathcal{N}(0, 1)$, and for any $n > 1$, $X\_n \sim \mathcal{N}(X\_{n-1}, X\_{n-1}^2)$ (notice, this is a conditional distribution given $X\_{n-1}$).


I am looking for concentration bounds for $X\_n$, which I suspect exist, based on numerical simulation.


**Additional information:**


It is easy to see that $\operatorname{Var}(X\_n) = 2^{n-1}$, by induction over $n$.


The $n=1$ case is trivial. Since $X\_n$ is unbiased, for any $n > 1$, $$\underset{X\_n}{\mathbb{E}} \left[X\_n^2 \mid X\_{n-1} \right] = \underset{X\_n}{\mathbb{E}} \left[(X\_{n-1} + (X\_n - X\_{n-1}))^2 \,|\, X\_{n-1} \right] = X\_{n-1}^2 + \underset{X\_n}{\mathbb{E}} \left[(X\_n - X\_{n-1})^2 \mid X\_{n-1} \right] = 2 X\_{n-1}^2.$$
Using this fact we get, $$\operatorname{Var}(X\_n) = \underset{X\_1, \ldots, X\_n}{\mathbb{E}} \left[X\_n^2 \right] = \underset{X\_1, \ldots, X\_{n-1}}{\mathbb{E}} \left[ \underset{X\_n}{\mathbb{E}} \left[X\_n^2 \mid X\_{n-1} \right] \right] = \underset{X\_1, \ldots, X\_{n-1}}{\mathbb{E}} \left[2 X\_{n-1}^2 \right] = 2 \operatorname{Var}(X\_{n-1}).$$


At first glance, this might lead to the assumption that the distribution becomes less concentrated as $n$ grows, but running some simulations, it seems like the distribution actually gets more concentrated around $0$, and the growth of the variance results from the tails becoming heavier.


It is possible to state this question in a more general form, by defining $X\_n \sim \mathcal{N}(X\_{n-1}, f(X\_{n-1}))$ for any non negative function $f$. A similar question was presented in [this thread](https://mathoverflow.net/questions/227556/adaptive-version-of-the-azuma-hoeffding-inequality), but the answers do not apply to this example.


This question has close ties to the method of mixtures (see e.g., Theorem 2.7 in [De la Pena et al.](https://arxiv.org/pdf/0709.2233.pdf)), but as far as I understand, none of their results can be used to answer the question, the way it was formalized here.



Concentration bounds for martingales with adaptive Gaussian steps
Observe that $X\_n=X\_{n-1}(1+Z\_n)$ where $\{Z\_k\}\_{k \ge 1}$ are i.i.d. standard normal. Hence to analyze the asymptotic distribution of $|X\_n|$, pass to logarithms, to get $$\log(|X\_n|)= \log(|X\_1|) +\sum\_{k=2}^n \log(|1+Z\_k|) \tag{$\*$} \,.$$


According to Wolfram alpha,
\begin{align}
& \mu:=E\bigl[ \log(|1+Z\_k|) \bigr] \\[8pt] = {} & \frac1{\sqrt{2\pi}}\int\_{-\infty}^\infty \log(|1+z|)e^{-z^2/2} \, dz\in [-0.2085,-0.2084] \,.
\end{align}


Thus the law of large numbers yields that for large $n$, with high probability,
$$ \log(|X\_n|) <-n/5 \quad \text{whence } |X\_n|<e^{-n/5} \,.$$
This confirms the prediction of the O.P.
More precise information can be extracted from $(\*)$ via the asymptotic distribution theory for sums of i.i.d. variables [1].


[1] Petrov, Valentin V. "Sums of independent random variables." De Gruyter, 2022.


