# Non-normality of limit of random variables

I have encounter the following difficulty in the study of limits of random variables. Assume that $\{X_n\}_{n\geq 1}$ is a sequence of real-valued random variables such that

$$\mathbb{E}[X_n]=\lambda_n,\,\, \mathbb{V}[X_n]=\sigma_n^2,$$

where, as usual, $\mathbb{E}$ and $\mathbb{V}$ are the expectation and variance. The condition that is satisfied is that $\sigma_n=o(\lambda_n)$.

Question: can we assure that the (conveniently normalized) sequence of random variables does NOT converge to a normal distribution $N(0,1)$?

As far as my intuition says a simple argument (Markov, Chebyshev,...) would be enough, but I am not able to get it.

• They certainly can converge to $N(0,1)$ after being "conveniently normalized" (love that phrase!), for example you can have $X_n \sim N(\lambda_n,\sigma_n^2)$. So if by "can we assure that ... does NOT converge ..." you meant "is it always the case that ... does NOT converge ...", the answer is No. – Dan Romik Oct 9 '15 at 7:21
• also translated: $(X_n-\lambda_n)/\sigma_n$ – gaussian-matter Oct 9 '15 at 7:34
• If you are allowing translations, then the condition $\sigma_n = o(E[X_n])$ means almost nothing, so it is strange that you include it. – Douglas Zare Oct 9 '15 at 7:47
• As pointed out, at this level of generality you can't ensure anything. You have to know something more about the sequence. Perhaps you should state precisely the claim made in the paper, and people can help figure out why it is true. – Nate Eldredge Oct 9 '15 at 14:26
• The paper is Rucinski: When are small subgraphs of a random graph normally distributed? (year, 1988): the main problem is that the author cites Equation (4) in the paper, which is missing. The context is the following: the random model studied is the random graph $G(n,p)$ with $n^2(1-p)$ tending to 0. Then, for any fixed graph $H$, the random variable $X_n$ counting copies of $H$ is NOT normally distributed in this regime. The argument is missing, and apparently taking the expectation and variance of $X_n$ and applying Markov should give that $X_n$ tends to 0 in distribution... – gaussian-matter Oct 10 '15 at 16:09

From $n^2(1-p_n) \to 0$ it follows that $\mathbb{E}[\sum_{u, v \in V(G_n)} 1_{\{uv \notin E(G_n)\}}] \to 0$, so we have $P(\exists u, v \in V(G_n)\ \text{s.t.}\ uv \notin E(G_n)) \to 0$ by the first moment method. It is now immediate that $X_n$ will go to $\infty$ if $H$ is a finite complete graph and will converge in probability to $0$ otherwise.

• For me it is clear that the expectation of $X_n$ will go to infinity for finite complete graphs and indeed for every subgraph $H$. But I do not see how you conclude the second statement. – gaussian-matter Oct 12 '15 at 8:00
• For any finite graph $H$ (other than the complete graphs) $\{X_n > 0\} \subseteq \{\exists u, v \in V(G_n)\ \text{s.t.}\ uv \notin E(G_n)\}$ so it follows that we must have $P(X_n >0) \to 0$. Since $X_n$ is non-negative w.p. 1 for all $n$, it follows that $X_n \to 0$ in probability. – Jason Lenderman Oct 12 '15 at 8:24
• I think that this argument works when dealing with induced subgraphs, but not subgraphs in general. – gaussian-matter Oct 12 '15 at 9:48
• That's true. But if you allow any subgraph then won't $X_n$ always go to infinity? I haven't looked at the paper, but I'm fairly certain it must be talking about induced subgraphs. – Jason Lenderman Oct 12 '15 at 14:46

The characteristic function framework allows you to think a ton about sequences that won't go to a normal distribution in the limit. If I recall right, each distribution is in one to one correspondance with its characteristic function. So look at,

$$f_n(t) = \mathbb{E}[e^{itX_n}]$$

and lets say for example $f_n(t)\in C^2$ $$\log f_n(t) = \mu_n t -\frac{1}{2}\sigma_n^2 t^2 + O(\kappa_n^{(3)} t^3).$$

Then the normalized random variable's characteristic function converges weakly to normal in the limit if and only if $$\lim_{n\to \infty } \log f_n(\frac{t}{\sigma_n}) - \frac{\mu_n}{\sigma_n} t = -\frac{1}{2}t^2$$

so I would suggest playing round with functions that don't have this property until you find one you like.