Second moment of a measure with size biaised variation 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 :

*

*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)

*$a_i(t)$ 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...
 A: 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
$$0<\sum_i a_i(t) i^2\le\sum_i a_i(0) i^2.$$

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.
