# Limiting distribution of sum of normals

How would I go about solving this problem below? I am not exactly sure where to start. I know that I need to make use of the Lebesgue Dominated Convergence theorem as well. Thanks for the help.

Let $X_1, X_2, \ldots, X_n$ be a random sample of size $n$ from a distribution that is $N(\mu, \sigma^2)$ where $\sigma^2 > 0$. Show that $Z_n = \sum\limits_{i=1}^n X_i$ does not have a limiting distribution.

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I like this problem. Many people who have taken only a basic course in probability retain the impression that the Central Limit Theorem says that "if you add up a bunch of random numbers, you get a normal distribution (in the limit)." Finite variance or the need for scaling etc are long forgotten. This problem points out that even the sum of iid normal random variables does not approach a normal distribution, let alone the sum of a "bunch of random numbers." – Dilip Sarwate Oct 17 '11 at 11:24
Do you know about characteristic functions? – Nate Eldredge Oct 17 '11 at 12:42
I know about mgfs – MathMan Oct 17 '11 at 14:44

## 1 Answer

Do not see Lebesgue dominated convergence in here.

Rather, you could try to compute the distribution of $Z_n$ for each $n$ (hint: this is gaussian) and then show that $\mathrm P(|Z_n|\leqslant z)\to0$ for every fixed $z$ (hint: write each $Z_n$ as $a_nY_n+b_n$ with $Y_n$ standard gaussian and try to include each event $[|Z_n|\leqslant z]$ in an event $[|Y_n|\leqslant y_n]$ with $y_n\to0$).

If one of these steps causes you some trouble and once you will have shown what you tried, please ask for further indications.

Edit After some exchanges in the comments, we know that $Z_n=\sqrt{n}\sigma Y_n+n\mu$ hence $|Z_n|\leqslant z$ if and only if $|Y_n+\sqrt{n}\mu/\sigma|\leqslant z/(\sqrt{n}\sigma)$. For the condition that $|Z_n|\leqslant z$ to hold, $Y_n$ must belong to the interval centered at $-\sqrt{n}\mu/\sigma$ of length $2z/(\sqrt{n}\sigma)$. Since the density of $Y_n$ is uniformly bounded by $1/\sqrt{2\pi}$, this shows that $$\mathrm P(|Z_n|\leqslant z)\leqslant2z/(\sqrt{2\pi n}\sigma)\to0.$$ Alternatively, one could have noted from the start that the density of $Z_n$ is uniformly bounded by $1/(\sqrt{2\pi n}\sigma)$ and that one asks $Z_n$ to be in an interval of length $2z$.

In any case, if $Z_n\to \zeta$ in distribution and if $\zeta$ is not almost surely infinite, there exists $z$ such that $\mathrm P(|\zeta|\leqslant z)\geqslant\frac12$, say, hence $\mathrm P(|Z_n|\leqslant z)$ cannot converge to zero. This shows $Z_n$ cannot converge in distribution to a random variable $\zeta$ such that $\mathrm P(|\zeta|=\infty)\ne1$.

Note As this proof shows, my first indication (which was to include $[|Z_n|\leqslant z]$ in $[|Y_n|\leqslant y_n]$ with $y_n\to0$) was wrong since the suitable event involving $Y_n$ is NOT of the form $[|Y_n|\leqslant y_n]$ with $y_n\to0$ but $[|Y_n+t_n|\leqslant y_n]$ with $y_n\to0$ and $|t_n|\to\infty$. I leave everything as it was first written.

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No. First, each $X_n$ is distributed like a standard gaussian $X$ hence $P(|X_n|\leqslant x_n)=P(|X|\leqslant x_n)$ for every $n$. Second, the distribution of $X$ has no atom at $0$ hence, if $x_n\to0$, $P(|X|\leqslant x_n)\to0$. Hence all this reduces to show that $x_n\to0$. The first step is to compute $a_n$ and $b_n$. – Did Oct 17 '11 at 6:27
Multivariate refers to random variables with values in $\mathbb R^n$ with $n>1$. Your $Z$ lives in $\mathbb R$, so no. // Random sample usually means independent random variables $X_i$. If the $X_i$ are not independent, the distribution of $Z$ can be VERY complicated hence their independence is a hypothesis you are supposed to make, I think. – Did Oct 17 '11 at 7:39
Just realized I used $X_n$ in my answer, forgetting that you used this notation for something else in the question, so my hint might have been difficult to decipher. Sorry about that. Try the new version. Anyway, yes $Z_n$ is normal with mean $n\mu$ and variance $n\sigma^2$ and you want $Z_n=a_nY_n+b_n$ with $Y_n$ normal with mean $0$ and variance $1$. So $a_n=\ldots$ and $b_n=\ldots$ – Did Oct 18 '11 at 4:48
Well since all parameters of Z are being multiplied by n, I would guess that a = n and b = 0. If that is correct, then do I need to express Z as an event that involves Y? Then do I need to find the limiting distribution of that? And I assume that there will be no limiting distribution given what the question has asked... – MathMan Oct 18 '11 at 4:58
$a_n=n$ and $b_n=0$ are both incorrect, since, if $Y_n$ is normal with mean $0$ and variance $1$, then $nY_n+0$ is normal with mean $0$ and variance $n^2$. You want $a_nY_n+b_n$ to be normal with mean $n\mu$ and variance $n\sigma^2$. As I said in my answer, this is the first step. The second step is to include $[|Z_n|\leqslant z]$ into $[|Y_n|\leqslant y_n]$ for a suitable $y_n$. And the third and final step is to show that $y_n\to0$ and to draw some consequences. But, first things first: what are $a_n=\ldots$ and $b_n=\ldots$? – Did Oct 18 '11 at 5:31