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, 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.), but as far as I understand, none of their results can be used to answer the question, the way it was formalized here.

  • $\begingroup$ I'd write $X_n \mid X_{n-1} \sim \mathcal{N}(X_{n-1}, X_{n-1}^2)$ rather than $X_n \sim \mathcal{N}(X_{n-1}, X_{n-1}^2).$ $\endgroup$ Dec 11, 2022 at 20:30
  • $\begingroup$ @moshenfeld Is the answer below sufficiently clear? If so, do you know how to accept an answer? $\endgroup$ Dec 15, 2022 at 4:04
  • $\begingroup$ Thank you, @YuvalPeres. Of course, I would hope for a finite sample bound that is better than than what can be achieved using the second moment + Chebyshev or CLT + Berry–Esseen, but the answer is clear and very helpful. And now I learned how to accept an answer as well. $\endgroup$
    – moshenfeld
    Dec 15, 2022 at 16:18

1 Answer 1


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.

  • $\begingroup$ I think you may have misplaced some absolute values in the second displayed equation. $\endgroup$
    – Thomas
    Dec 11, 2022 at 14:11
  • 1
    $\begingroup$ @Thomas Thanks you, corrected now $\endgroup$ Dec 11, 2022 at 14:18

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