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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

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

Thanks in advance.

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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
up vote 2 down vote accepted

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.)

Edit (Added to answer a question raised by @Sasha.)

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 $\text{mesh}(P_n)\to0$.

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(+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
Sasha (thanks). Mindreader, are you? :-) If you are right, it is good that my post answers this (not explicitely asked) question as well. – Did Aug 30 '11 at 15:05
Not a mindreader, but I confess I was not aware of intricacies you exposed, and was in fact interpreting OP's question as request to find the limiting distribution that Riemann sums convergence in probability to as mesh gets infinitely thin, analogously to limit of quadratic variation of the Brownian motion. – Sasha Aug 30 '11 at 15:13
A lesson to be learned here is that no good can come of uncountable iid families. – Nate Eldredge Sep 2 '11 at 11:28
@Nate, indeed. And you noticed that the question of the existence of such families (for what sigma-algebra?) was carefully avoided... – Did Sep 2 '11 at 13:18

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