Let $(X_{n})_{n \in \mathbb{N}}$ be independent random variables with

\begin{equation} \mathbb{P}(X_{n} = -n^{3/2}) = \mathbb{P}(X_{n} = n^{3/2}) = \frac{1}{2n},~ \mathbb{P}(X_{n} = 0) = 1 - \frac{1}{n}. \end{equation}

Let $S_{n} = \sum_{k=1}^{n}X_{k}$. It is to show that

\begin{equation} \frac{S_{n}}{\sqrt{var(S_{n})}} \xrightarrow d S, \end{equation} for $n \to\infty$ where $S$ is an non standard random variable. Now the professor give us a solution but it is about 2 pages long. I understand the proof. Is there a shorter proof?

  • 3
    $\begingroup$ What is a "non standard" random variable...? $\endgroup$
    – saz
    May 18, 2018 at 8:45
  • 1
    $\begingroup$ What was your professor's proof? $\endgroup$
    – Math1000
    May 18, 2018 at 9:20
  • 1
    $\begingroup$ By not standard I mean a random variable that has no normal distribution with mean 0 and variance 1 $\endgroup$
    – user562844
    May 18, 2018 at 15:40

1 Answer 1


Not sure what you mean by "non-standard" random variable (perhaps not having a "name" ?!), but one may approach the problem of convergence in distribution via characteristic functions, which seems pretty straightforward in this case.

Let $Y_n = \frac{S_n}{\sqrt{var(S_n)}}$, and consider its characteristic function $$ \varphi_{n}(\lambda) = \mathbb{E} e^{i\lambda Y_n}, \qquad \lambda \in \mathbb{R}. $$ If we prove that $\varphi_n(\lambda)$ converges pointwise to some function $\varphi(\lambda)$ which is continuous at the origin, then $Y_n$ would converge in distribution to some random variable having $\varphi $ as its characteristic function (this is standard, see Levy's theorem).

What we do next, is studying convergence of $\varphi_n$, and show in particular that it does NOT converge to the characteristic function of an $N(0,1)$ random variable.

Using independence of $X_k$ we have $$ var S_n = \sum_{k=1}^n var X_k = \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6} =: a_n. $$

Again, in view of independence we get $$ \varphi_n (\lambda) = \prod_{k=1}^n \mathbb{E} e^{i\lambda X_k (a_n)^{-1/2} } = \prod_{k=1}^n\left[ \frac{\exp(i\lambda k^{3/2} a_n^{-1/2} ) + \exp(-i\lambda k^{3/2} a_n^{-1/2} )}{2k} + 1- \frac 1k \right]= \\ \prod_{k=1}^n \left[1- \frac 1k + \frac 1k \cos \frac{\lambda k^{3/2}}{a_n^{1/2}} \right]. $$

Taking $\log$ of $\varphi_n$ we get $$ (1) \qquad \log \varphi_n (\lambda) = \sum_{k=1}^n \log \left( 1 - \frac 1k \left( 1 - \cos \frac{\lambda k^{3/2}}{a_n^{1/2}} \right) \right). $$ The convergence of the series in $(1)$ implies convergence of $\varphi_n(\lambda)$. Observe that we have series of non-positive terms and in order to show convergence, it is enough to establish a bound from below on $\log \varphi_n(\lambda)$, which we do next.

Fix a constant $a>0$ and consider $|\lambda| \leq a$. This bound will be used to get uniform estimates for $(1)$ with respect to $\lambda$. Let also $0<\varepsilon <1 $ be fixed, and assume $n\in \mathbb{N}$ is large enough that we have $$ (2) \qquad 0\leq \frac 1k \left( 1 - \cos \frac{\lambda k^{3/2}}{a_n^{1/2}} \right) \leq \varepsilon, \ k=1,2..., n. $$ Here $(2)$ is a consequence of the fact that $a_n \to \infty$, so we get $\cos$ converging to $1$. This handles the case for smaller $k$, and for larger $k$ the factor $\frac 1k$ takes over.

We next get rid of the $\log$ in $(1)$. In view of Taylor's expansion for $\log(1-x) $ around $0$, we have $\log (1 - x) = -x + O(x^2)$, as $x \to 0$. In view of $(2)$, we can employ this expansion in $(1)$. Now using the inequality $1-\cos x \leq \frac 12 x^2$ for all $x\in \mathbb{R}$, we have $$ \frac{1}{k^2} \left( 1 - \cos \frac{\lambda k^{3/2}}{a_n^{1/2}} \right)^2 \leq \frac{1}{k^2} \left( \frac 12 \frac{\lambda^2 k^3}{a_n} \right)^2 \leq \frac{\lambda^4}{4} \frac{k^4}{a_n^2}. $$ Since $\sum_{k=1}^n k^4 \leq C n^5$, and $a_n^2 \sim n^6$, summing the last inequality over $k=1,...,n$, and getting back to $(1)$ we obtain $$ (3) \qquad \log \varphi_n(\lambda) = - \sum_{k=1}^n \frac{1}{k} \left( 1 - \cos \frac{\lambda k^{3/2}}{a_n^{1/2}} \right) + O(\frac{1}{n}). $$

After this reduction, the goal is to establish a lower bound on the sum in $(3)$, which will enable us to compare $\varphi_n(\lambda)$ with $e^{-\frac 12 \lambda^2}$ - the characteristic function of $N(0,1)$ random variable. This is now the final step of the proof.

Based on the Taylor expansion of $\cos x$ we have the inequality $$ 1- \cos x \leq \frac{x^2}{2} - \frac{x^4}{4!} + \frac{x^6}{6!}, \text{ for } x \in \mathbb{R}. $$ This inequality can be deduced, for example, by considering the difference of left and right hand sides and differentiating a few times. Applying the inequality on $(3)$, we have $$ - \frac 1k \left( 1 - \cos \frac{\lambda k^{3/2}}{a_n^{1/2}} \right) \geq - \frac 12 \frac{ \lambda^2 k^2}{a_n} + \frac {1}{24} \frac{ \lambda^4 k^5}{a_n^2} - \frac {1}{720} \frac{ \lambda^6 k^8}{a_n^3}. $$ Summing over $k=1,...,n$ in the last inequality, from $(3)$ we get $$ (4) \qquad \log \varphi_n(\lambda) \geq - \frac{\lambda^2}{2} + \frac{1}{24} \frac{\lambda^4}{a_n^2} \sum_{k=1}^n k^5 - \frac{1}{720} \frac{\lambda^6}{a_n^3} \sum_{k=1}^n k^8 + O(\frac 1n) . $$ Using the fact that $\sum_{k=1}^n k^5 = \frac 16 n^6 + O(n^5)$ and $\sum_{k=1}^n k^8 = \frac 19 n^9 + O(n^8)$ (see here for example), and that $a_n \sim \frac 13 n^3 $, we get a uniform lower bound on $\varphi_n (\lambda)$, which shows the convergence of $\varphi_n (\lambda)$ to some function $\varphi(\lambda)$ . Since bounds are uniform with respect to $|\lambda| \leq a$, we also get that $\varphi$ is continuous at $0$. Passing to the limit in $(4)$ we obtain $$ \log \varphi (\lambda) \geq - \frac 12 \lambda^2 + \frac{1}{24} \frac{3^2}{6}\lambda^4 - \frac{1}{720} \frac{3^3}{9}\lambda^6 = - \frac 12 \lambda^2 + \frac{1}{16} \lambda^4 - \frac{1}{240} \lambda^6 . $$

For smaller $|\lambda|$, the last inequality implies $\varphi(\lambda) > e^{- \frac{1}{2}\lambda^2 }$, and hence $Y_n$ converges in distribution to a random variable $S$ which is not $N(0,1)$.

  • $\begingroup$ but how can we conclude that $\varphi$ is not the characteristic function of a standard normal rndom variable? $\endgroup$
    – wayne
    May 26, 2018 at 19:51
  • $\begingroup$ but that is the important part of the proof $\endgroup$
    – wayne
    May 26, 2018 at 21:38
  • $\begingroup$ @wayne, I filled out all the details, it became a bit longer, but still a straightforward proof. $\endgroup$
    – Hayk
    May 28, 2018 at 8:25

You must log in to answer this question.