# Limiting distribution of the sum of squared independent normals

Let $t\in[0,1]$ and suppose $Z_t$ is a standard normal random variable, such that $Z_t$ and $Z_s$ are independent if $s\neq t$. If we pick $0\leq t_1\leq \ldots\leq t_k\leq 1$ then

$$X = \sum_{j=1}^k Z_{t_j}^2$$

is a chi-squared distribution with $k-1$ degrees of freedom. What happens in the limit? What is the distribution of

$$Y = \int_{0}^1 Z_t^2 dt?$$

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Dear alext87, you might be accepting answers too shortly after having proposed your questions, with the obvious drawbacks associated to this modus operandi... – Did Aug 25 '11 at 15:52

Consider Riemann sum $Y_n = \frac{1}{n} \sum_{i=1}^n Z_i^2$, the $Y_n$ would correspond to $\frac{1}{n} \chi^2_{n}$. Moments of this variable are

$$m_r(Y_n) = \frac{1}{n^r} \prod_{k=1}^r (n+2k) = \prod_{k=1}^r ( 1+ \frac{2k}{n})$$

Now $\lim_{n \to \infty} m_r(Y_n) = 1$. Hence the characteristic function of the new distribution is $\exp(i t)$, i.e. limiting distribution is degenerate, as proved by Did in comments.

Added: As observed by Did, the integral $Y = \int_0^1 Z_t^2 \mathrm{d} t$ does not exist.

In order to understand this better, I consider a different mesh, say with $\Delta t_k = \frac{2 k}{n (n+1)}$, so that $\sum_{k=1}^n \Delta t_k = 1$. Then the Riemann sum is

$$Y^{(1)}_n =\sum_{k=1}^n Z_k^2 \frac{2k}{n (n+1)}$$ The characteristic function of $Y^{(1)}_n$ is

$$\phi_{Y^{(1)}_n}(t) = \prod_{k=1}^n \phi_{\chi^2_1} \left( \frac{2 t k}{n (n+1)} \right) = \left( \left( \frac{-4 i t}{n (n+1)} \right)^n \frac{ \Gamma( 1 + n + \frac{ i n(n+1)}{4 t} )}{\Gamma(1 + \frac{ i n (n+1)}{4 t} )} \right)^{-\frac{1}{2}}$$

Sparing you tedious details, the large $n$ limit is

$$\phi_{Y^{(1)}_n}(t) \approx \exp\left( i t - \frac{4}{3} \frac{t^2}{n} + o\left(\frac{1}{n}\right) \right)$$ and seems to converge to degenerate distribution as well.

Question: It would be very interesting to construct partitions of the unit interval which would lead to a different limit.

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Sasha, I am surprised to see you leave unchanged your answer, given the argument in mine. – Did Aug 30 '11 at 11:22
@Didier Please see my updated answer. I would appreciate your comment on it. Thank you – Sasha Aug 30 '11 at 12:59
Sasha, see my Edit for an answer to the question at the end of your post. – Did Aug 30 '11 at 14:06

For every $t$, let $U_t=(Z_t)^2$. Almost surely, the integral which defines $Y$ has infinite upper Riemann sums and zero lower Riemann sums. To see this, note that, on every interval of the subdivision, infinitely many independent random variables $(U_t)$ are involved, whose distributions have common support $S=[0,+\infty)$, hence their supremum is the supremum of $S$ and their infimum is the infimum of $S$.

As a consequence, almost surely the function $t \mapsto U_t$ is not Riemann(-Darboux) integrable and $Y$ does not exist. (That is, unless one precises another way to define the integral.)

As explained above, almost surely and for every partition $\{t_k\}_k$ of $[0,1]$, the upper Riemann sum is infinite and the lower Riemann sum is zero, that is, $$\sum\limits_k(t_k-t_{k-1})\sup\{U_s\mid s\in[t_{k-1},t_k]\}=+\infty,$$ and $$\sum\limits_k(t_k-t_{k-1})\inf\{U_s\mid s\in[t_{k-1},t_k]\}=0.$$ However, the Riemann sums $Y(P)$ associated to tagged partitions $P$ do converge uniformly to $1$ in $L^2$ when the mesh of $P$ goes to zero.
To see this, for any tagged partition $P=\{(s_k,t_k)\}_k$ of $[0,1]$, write $Y(P)$ for the Riemann sum associated to $P$. That is, $s_k$ is in $[t_{k-1},t_k]$ for every $k$ and $$Y(P)=\sum\limits_k(t_k-t_{k-1})U_{s_k}.$$ Then, $E(U_s)=1$ and $\text{Var}(U_s)=3$ for every $s$, hence $E(Y(P))=1$ and, since the random variables $(U_{s_k})$ are independent, $$\text{Var}(Y(P))=3\sum\limits_k(t_k-t_{k-1})^2\le3\text{mesh}(P).$$ This proves that $Y(P)\to1$ in $L^2$ when $\text{mesh}(P)\to0$, in the sense that $$\sup\{\|Y(P)-1\|_2\mid\text{mesh}(P)\le m\}\to0$$ when $m\to0$. In particular, for every sequence $(P_n)$ of tagged partitions, $Y(P_n)\to1$ (in $L^2$ and in particular, in probability) as soon as $\text{mesh}(P_n)\to0$.
(+1) Thank you very much for writing up an answer to my question. Links to corresponding definitions proved useful. I do suspect, however, that OP had in mind $L^2$ limit of the Riemann sums when mesh get finer. – Sasha Aug 30 '11 at 15:02