Let $(X_n)$ be a sequence of i.i.d $\mathcal N(0,1)$ random variables. Define $S_0=0$ and $S_n=\sum_{k=1}^n X_k$ for $n\geq 1$. Find the limiting distribution of $$\frac1n \sum_{k=1}^{n}|S_{k-1}|(X_k^2 - 1)$$

This problem is from Shiryaev's Problems in Probability, in the chapter on the Central Limit Theorem. It was asked on this site in 2014, but remains unanswered. I posted it yesterday on Cross Validated, and I think it's worth to cross-post it here as well.

Since $S_{k-1}$ and $X_k$ are independent, $E(|S_{k-1}|(X_k^2 - 1))=0$ and $$V(|S_{k-1}|(X_k^2 - 1)) = E(S_{k-1}^2(X_k^2 - 1)^2)= E(S_{k-1}^2)E((X_k^2 - 1)^2) =2(k-1)$$

Note that the $|S_{k-1}|(X_k^2 - 1)$ are clearly not independent. However, as observed by Clement C. in the comments, they are uncorrelated since for $j>k$ $$\begin{aligned}Cov(|S_{k-1}|(X_k^2 - 1), |S_{j-1}|(X_j^2 - 1)) &= E(|S_{k-1}|(X_k^2 - 1)|S_{j-1}|)E(X_j^2 - 1)\\ &=0 \end{aligned}$$

Hence $\displaystyle V(\frac1n \sum_{k=1}^{n}|S_{k-1}|(X_k^2 - 1)) = \frac 1{n^2}\sum_{k=1}^{n} 2(k-1) = \frac{n-1}n$ and the variance converges to $1$.

I have run simulations to get a feel of the answer

import numpy as np
import scipy as sc
import scipy.stats as stats
import matplotlib.pyplot as plt

n = 30000 #summation index
m = 10000 #number of samples

X = np.random.normal(size=(m,n))
sums = np.cumsum(X, axis=1)
sums = np.delete(sums, -1, 1)
prods = np.delete(X**2-1, 0, 1)*np.abs(sums)
samples = 1/n*np.sum(prods, axis=1)

plt.hist(samples, bins=100, density=True)

Below is a histogram of $10.000$ samples ($n=30.000$). The variance from the generated samples is $0.9891$ (this complies with the computations above). If the limiting distribution was $\mathcal N(0,\sigma^2)$, then $\sigma=1$. However the histogram peaks at around $0.6$, while the max of the density of $\mathcal N(0,1)$ is $\frac 1{\sqrt{2 \pi}}\approx 0.4$. Thus simulations suggest that the limiting distribution is not Gaussian.

It might help to write $|S_{k-1}| = (2\cdot 1_{S_{k-1}\geq 0} -1)S_{k-1}$.

It might also be helpful to note that if $Z_n=\frac1n \sum_{k=1}^{n}|S_{k-1}|(X_k^2 - 1)$, conditioning on $(X_1,\ldots,X_{n-1})$ yields $$E(e^{itnZ_n}) = E\left(e^{it(n-1)Z_{n-1}} \frac{e^{-it|S_{n-1}|}}{\sqrt{1-2it|S_{n-1}|}}\right)$$

enter image description here

  • 1
    $\begingroup$ A quick check from my phone (sorry for typos): aren't the covariances zero? It looks like you get an explicit variance of $(n-1)/n$, for what it's worth. $\endgroup$
    – Clement C.
    Aug 24, 2019 at 19:48
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    $\begingroup$ My simulation also suggests that the limiting distribution may not be gaussian. Let us look at a similar problem that produces non-gaussian distribution. Let us replace $|S_{k-1}|$ by $S_{k-1}$ and $X_k^2 - 1$ by $X_k$. Then the resulting sum $$ \frac{1}{n} \sum_{k=1}^{n} S_{k-1} X_k $$ converges in distribution to the Ito integral $$ \int_{0}^{1} W_s \, \mathrm{d}W_s = \frac{W_1^2 - 1}{2}, $$ where $(W_t)_{t\geq 0}$ is a Wiener process. Since $W_1 \sim \mathcal{N}(0, 1)$, this limiting distribution cannot be gaussian. Now I suspect that OP's problem also suffers the same issue. $\endgroup$ Aug 24, 2019 at 23:40
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    $\begingroup$ @ClementC., My guess is that he found an error in his answer. For instance, the inequality $$\mathbb{P}\left(\sqrt{i-1}\left|X_1\right|\left|X_2^2-1\right|>n\varepsilon\right)\leq\mathbb{P}\left(\left|X_1\right|\left|X_2^2-1\right|>n\varepsilon\right)$$ in his verification of the 1st condition is false when $i \geq 2$. Still, I think the proof can be salvaged to deduce the same conclusion. $\endgroup$ Aug 31, 2019 at 5:21
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    $\begingroup$ @ClementC. I also noticed another error in Davide Giraudo's answer, and it is more fundamental than the previous one. Indeed, if we write $I_n=\sum_{i=1}^{n}S_{i-1}^2/n^2$, then $I_m$ and $I_n$ are asymptotically independent for $n\gg m$. So $I_n$ can only converge in distribution, not in probability, and the martingale difference CLT described in Theorem 3.2 of Hall & Heyde cannot be applied directly. $\endgroup$ Sep 2, 2019 at 11:25
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    $\begingroup$ @SangchulLee But in this case we have the same limiting process. Moreover, there exists central limit theorems for martingales where we only have convergence in distribution of the sum of conditional variances. See for instance Theorem 2 in arxiv.org/pdf/1803.09100.pdf $\endgroup$ Sep 2, 2019 at 22:08

2 Answers 2


Let $(X_n)_{n\geq 1}$ be a sequence of i.i.d. standard normal variables. Let $(S_n)_{n\geq 0}$ and $(T_n)_{n\geq 0}$ be given by

$$ S_n = \sum_{i=1}^{n} X_i \qquad\text{and}\qquad T_n = \sum_{i=1}^{n} (X_i^2 - 1). $$

We will also fix a partition $\Pi = \{0 = t_0 < t_1 < \cdots < t_k = 1\}$ of $[0, 1]$. Then define

$$ \begin{gathered} Y_n = \frac{1}{n}\sum_{i=1}^{n} | S_{i-1} | (X_i^2-1), \\ Y_{\Pi,n} = \frac{1}{n} \sum_{j=1}^{k} |S_{\lfloor nt_{j-1}\rfloor}| (T_{\lfloor nt_j\rfloor} - T_{\lfloor nt_{j-1} \rfloor}). \end{gathered}$$

Ingredient 1. If $\varphi_{X}(\xi) = \mathbb{E}[\exp(i\xi X)]$ denotes the characteristic function of the random variable $X$, then the inequality $|e^{ix} - e^{iy}| \leq |x - y|$ followed by Jensen's inequality gives

\begin{align*} \big| \varphi_{Y_n}(\xi) - \varphi_{Y_{\Pi,n}}(\xi) \big|^2 &\leq \xi^2 \mathbb{E}\big[ (Y_n - Y_{\Pi,n})^2 \big] \\ &= \frac{\xi^2}{n^2}\sum_{j=1}^{k} \sum_{i \in (nt_{j-1}, nt_j]} 2 \mathbb{E} \big[ \big( | S_{\lfloor n t_{j-1} \rfloor} | - | S_{i-1} | \big)^2 \big]. \end{align*}

From the reverse triangle inequality, the inner expectation is bounded by

\begin{align*} 2 \mathbb{E} \big[ \big( | S_{\lfloor n t_{j-1} \rfloor} | - | S_{i-1} | \big)^2 \big] \leq 2 \mathbb{E} \big[ \big( S_{i-1} - S_{\lfloor n t_{j-1} \rfloor} \big)^2 \big] = 2(i-1-\lfloor nt_{j-1} \rfloor), \end{align*}

and summing this bound over all $i \in (nt_{j-1}, nt_j]$ yields

$$ \big| \varphi_{Y_n}(\xi) - \varphi_{Y_{\Pi,n}}(\xi) \big|^2 \leq \frac{\xi^2}{n^2} \sum_{j=1}^{k} (\lfloor n t_j \rfloor - \lfloor n t_{j-1} \rfloor)^2 \xrightarrow[n\to\infty]{} \xi^2 \sum_{j=1}^{k} (t_j - t_{j-1})^2. \tag{1} $$

Ingredient 2. From the Multivariate CLT, we know that

$$ \Bigg( \frac{S_{\lfloor nt_j\rfloor} - S_{\lfloor nt_{j-1}\rfloor}}{\sqrt{n}}, \frac{T_{\lfloor nt_j\rfloor} - T_{\lfloor nt_{j-1}\rfloor}}{\sqrt{n}} : 1 \leq j \leq k \Bigg) \xrightarrow[n\to\infty]{\text{law}} ( W_{t_j} - W_{t_{j-1}}, N_j : 1 \leq j \leq k ), $$

where $W$ is the standard Brownian motion, $N_j \sim \mathcal{N}(0, 2(t_j - t_{j-1}))$ for each $1 \leq j \leq k$, and all of $W, N_1, \cdots, N_k$ are independent. By the continuous mapping theorem, this shows that

$$ Y_{\Pi,n} \xrightarrow[n\to\infty]{\text{law}} \sum_{j=1}^{k} W_{t_{j-1}} N_j. $$

Moreover, conditioned on $W$, the right-hand side has normal distribution with mean zero and variance $2\sum_{j=1}^{k} W_{t_{j-1}}^2 (t_j - t_{j-1}) $, and so,

$$ \lim_{n\to\infty} \varphi_{Y_{\Pi,n}}(\xi) = \mathbb{E}\left[ \exp\bigg( -\xi^2 \sum_{j=1}^{k} W_{t_{j-1}}^2 (t_j - t_{j-1}) \bigg) \right]. \tag{2} $$

Ingredient 3. Again let $W$ be the standard Brownian motion. Since the sample path $t \mapsto W_t$ is a.s.-continuous, we know that

$$ \sum_{j=1}^{k} W_{t_{j-1}}^2 (t_j - t_{j-1}) \longrightarrow \int_{0}^{1} W_t^2 \, \mathrm{d}t $$

almost surely along any sequence of partitions $(\Pi_k)_{k\geq 1}$ such that $\|\Pi_k\| \to 0$. So, by the bounded convergence theorem,

$$ \mathbb{E}\left[ \exp\bigg( -\xi^2 \sum_{j=1}^{k} W_{t_{j-1}}^2 (t_j - t_{j-1}) \bigg) \right] \longrightarrow \mathbb{E}\left[ \exp\bigg( -\xi^2 \int_{0}^{1} W_t^2 \, \mathrm{d}t \bigg) \right] \tag{3} $$

as $k\to\infty$ along $(\Pi_k)_{k\geq 1}$.

Conclusion. Combining $\text{(1)–(3)}$ and letting $\|\Pi\| \to 0$ proves that

$$ \lim_{n\to\infty} \varphi_{Y_n}(\xi) = \mathbb{E}\left[ \exp\bigg( -\xi^2 \int_{0}^{1} W^2_t \, \mathrm{d}t \bigg) \right], $$

and therefore $Y_n$ converges in distribution to $\mathcal{N}\big( 0, 2\int_{0}^{1} W_t^2 \, \mathrm{d}t \big)$ as desired.

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    $\begingroup$ In the application of the multivariate CLT, how do you guarantee that in the limiting law $W$ is independent from the $N_j$'s? Is it obvious (since the $S_j$'s and $T_j$'s are not)? $\endgroup$
    – Clement C.
    Sep 2, 2019 at 18:40
  • $\begingroup$ @ClementC., That is the good point which I did not stressed out but is in fact important. A quick answer: it is because $S_m$'s and $T_n$'s are uncorrelated. This is very specific to this problem, and in general we do not (and should not) expect independence of the limiting $W$ and $N_j$'s, such as in the case of $$\frac{1}{n}\sum_{k=1}^{n}S_{k-1}X_k \to \int_{0}^{1}W_s\,\mathrm{d}W_s.$$ $\endgroup$ Sep 2, 2019 at 21:43
  • $\begingroup$ A longer answer: The $2k$ random variables in $\mathcal{S}=\{W_{t_j}-W_{t_{j-1}},N_j:1\leq j\leq k\}$ are jointly normal, meaning that any linear combinations are again normal (or equivalently, they form a $2k$-dimensional multivariate normal distribution). Also we easily find that any two distinct RVs from $\mathcal{S}$ are pairwise uncorrelated. Being jointly normal, this implies that RVs in $\mathcal{S}$'s are mutually independent. So we are critically exploiting the property of joint normality here. $\endgroup$ Sep 2, 2019 at 22:06
  • $\begingroup$ I see. Yes, that's an important point... $\endgroup$
    – Clement C.
    Sep 2, 2019 at 22:13
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    $\begingroup$ @GabrielRomon, Ah, yes, I forgot to write the factor $|\xi|^2$. Thank you for pointing this out, I fixed it now. Thankfully, this does not affect the validity of the proof as we are essentially fixing $\xi\in\mathbb{R}$ to the very end. $\endgroup$ Sep 3, 2019 at 9:34

An idea is to use the following result in Martingale limit theory and its applications by Hall and Heyde (Theorem 3.2):

Let $(X_{n,i},\mathcal F_{n,i})_{1\leqslant i\leqslant k_n,n\geqslant 1 }$ by a martingale differences array where $X_{n,i}\in L^2$ and $\mathcal F_{n,i-1}\subset \mathcal F_{n,i}$ for all $n$ and $i$. Suppose that there exists a random variable $\eta^2$ which is a.s. finite and such that

  1. $\max_{1\leq i\leq k_n} \left\lvert X_{n,i}\right\rvert\to 0$ in probability;
  2. $\sup_{n\geqslant 1}\mathbb E\left[\max_{1\leq i\leq k_n}X_{n,i}^2\right]$ is finite.
  3. $\sum_{i=1}^{k_n}X_{n,i}^2\to \eta^2$ in probability;

Then $\sum_{i=1}^{k_n}X_{n,i}\to Z$ in distribution, where $Z=\eta N$, with $N$ standard normal and independent of $\eta$.

However, unfortunately, I am not sum whether this works because here. The sum of the conditional variances converges in law and not in probability.

We will use this result with $\mathcal F_{n,i}=\sigma(X_j,1\leq j\leq n)$, $k_n=n$ and $X_{n,i}=\frac 1n\left\lvert S_{i-1}\right\rvert (X_i^2-1)$.

  1. For a fixed $\varepsilon$, $$\mathbb P\left(\max_{1\leq i\leq n} \frac 1n\left\lvert S_{i-1}\right\rvert \left\lvert X_i^2-1\right\rvert>\varepsilon \right) \leqslant \sum_{i=2}^n \mathbb P\left( \left\lvert S_{i-1}\right\rvert \left\lvert X_i^2-1\right\rvert>n\varepsilon \right).$$ Using the independence between $S_{i-1}$ and $X_i$ and the fact that $S_{i-1}$ has a normal distribution with variance $i-1$, we get that
    $$\mathbb P\left( \left\lvert S_{i-1}\right\rvert \left\lvert X_i^2-1\right\rvert>n\varepsilon \right)=\mathbb P\left( \sqrt{i-1}\left\lvert X_1 \right\rvert \left\lvert X_2^2-1\right\rvert>n\varepsilon \right)\leq \mathbb P\left( \left\lvert X_1 \right\rvert \left\lvert X_2^2-1\right\rvert>n^{1/2}\varepsilon \right)$$ and we conclude that 1. holds by the integrability of $\left\lvert X_1 \right\rvert \left\lvert X_2^2-1\right\rvert$.

  2. It follows from the fact that $$\mathbb E[X_{n,i}^2]=\frac 1{n^2}(i-1)\mathbb E\left[(X_1^2-1)^2\right]$$ hence we even have finiteness of $\sup_{n\geqslant 1}\mathbb E\left[\sum_{1\leq i\leq k_n}X_{n,i}^2\right].$

  3. For the third condition, it would be better to deal with conditional variances. Let $$ \delta_n:= \frac 1{n^2}\sum_{i=2}^n\left( S_{i-1}^2 (X_i^2-1)^2-\mathbb E\left[S_{i-1}^2 (X_i^2-1)^2\mid \mathcal F_{n,i-1}\right]\right). $$ Then $\delta_n$ is a sum of martingale differences and we can check that $\mathbb E\left[\delta_n^2\right]\to 0$. Therefore, we have to look at the limit in probability of $$ A_n:=\frac 1{n^2}\sum_{i=1}^nS_{i}^2= \frac 1n \sum_{i=1}^n W_{i/n}^2, $$ where the equality takes place in distribution and $(W_t)_{t\in [0,1]}$ is a standard Brownian motion. The latter quantity converges in probability to $\int_0^1 W(t)^2dt=:\eta^2$. But there is no convergence in probability. Indeed, $$A_{2n}-A_n= \frac 1{n^2}\sum_{i=1}^nS_i^2-\frac{3}{4n^2}\sum_{i=n+1}^{2n}S_i^2\overset{\mbox{law}}{=}\frac 1n\sum_{i=1}^nW_{i/(2n)}^2-\frac{3}{4n}\sum_{i=n+1}^{2n}W_{i/(2n)}^2.$$

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    $\begingroup$ That's... non-trivial, I wonder how this problem made it in that chapter of the problem book. Out of curiosity, that does lead to unit variance? (and what is the corresponding max of the pdf?) $\endgroup$
    – Clement C.
    Aug 26, 2019 at 16:02
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    $\begingroup$ What do you mean by unit variance? $\endgroup$ Aug 26, 2019 at 16:57
  • $\begingroup$ Just, as a sanity check, that the limiting distribution does have variance 1 (since that is the limit that's expected). $\endgroup$
    – Clement C.
    Aug 26, 2019 at 17:07
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    $\begingroup$ Yes. I realized that $\mathbb E\left[(X_1^2-1)^2\right]=3-2+1=2$ and the expectation of $\eta^2$ is $\int_0^1\mathbb E[W(t)^2]dt= \int_0^1t=1/2$. $\endgroup$ Aug 26, 2019 at 18:41

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