Let $\mu\_. : \mathbb{R}^+ \rightarrow M\_F(\mathbb{N}) $ a function. We set up :
$$ \mu\_t = \sum a\_i(t) \delta\_i$$
where each $a\_i$ is a positive continuous function from $\mathbb{R}^+$ to $\mathbb{R}^+$. If we have the following hypotheses :


1. Given $\mu\_0$, $<\mu\_0, \chi^2> := \int x^2 \mu\_0(dx) = \sum\_i a\_i(0) i^2$ is finite (also means $\mu\_0$ has finite first moment)
2. $a\_i(t)$ is defined by the following ode :
$$ \frac{d a\_i(t)}{dt} = - \frac{i a\_i(t)}{\sum k a\_k(t)}$$
I want to show that $<\mu\_t, \chi^2>$ would also be finite. And if it is not, what hypotheses can i add on $\mu\_0$ and its moments to make it so?


My intuition is that since $\mu\_t$ loses mass, its would be decreasing, in particular $<\mu\_t, \chi^2>$ would be bounded but I do not know if is enough of an argument...



Second moment of a measure with size biaised variation
Yes: If $a\_i(t)>0$ for some real $T>0$, all $t\in[0,T)$, and all $i$, then the $a\_i$'s will be decreasing on $[0,T)$, so that for all $t\in[0,T)$
we will have
$$0<\sum\_i a\_i(t) i^2\le\sum\_i a\_i(0) i^2.$$




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**Comment:** In fact, if $a\_i(t)>0$ for some real $T>0$, all $t\in[0,T)$, and all $i$, then
$$T\le T\_\*:=\sum\_k k a\_k(0).$$
Indeed, the $a\_i$'s are strictly positive, strictly decreasing, and differentiable on $[0,T)$, and hence your ODE implies
$$\frac{da\_k}{da\_1}=\frac{d a\_k}{dt}\Big/\frac{d a\_1}{dt}
=k\frac{a\_k}{a\_1},$$
so that for $l\_k:=\ln a\_k$ we have
$$\frac{dl\_k}{dl\_1}=k,$$
whence $l\_k(t)=c\_k+kl\_1(t)$ for some real constants $c\_k$. So, for $t\in[0,T)$,
$$\frac{a\_k(t)}{a\_k(0)}=\Big(\frac{a\_1(t)}{a\_1(0)}\Big)^k$$
and hence
$$\frac{d a\_1}{dt}=-\frac{a\_1(t)}{\sum\_k k a\_k(0)(a\_1(t)/a\_1(0))^k}
=-\frac{a\_1(0)}{\sum\_k k a\_k(0)(a\_1(t)/a\_1(0))^{k-1}}
\le-\frac{a\_1(0)}{\sum\_k k a\_k(0)},$$
which implies
$$a\_1(t)
\le a\_1(0)-\frac{a\_1(0)}{\sum\_k k a\_k(0)}\,t\le0$$
if $t\ge T\_\*$. Since $a\_1(t)>0$ for $t\in[0,T)$, we conclude that $T\le T\_\*$, as claimed.


