"continuous" and "discontinuous" phase transitions in branching processes.

Question

Consider a Galton-Watson branching process, with offspring distribution $\mathbf{p}=(p_0, p_1, \dots, p_n, \dots)$. Let $O$ be the root of the branching process.

Write $\eta=P(\text{process survives for ever})$ and $\mathcal{H}=\{\mathbf{p}: \eta>0\}$.

Also write $\beta=P(O \text{ is the root of an infinite binary tree contained in the branching process})$ and $\mathcal{B}=\{\mathbf{p}: \beta>0\}$.

Then it's fairly well known that in some sense the phase transition from $\mathcal H^C$ to $\mathcal H$ is typically "continuous" while the phase transition from $\mathcal B^C$ to $\mathcal B$ is typically "discontinuous".

e.g. suppose the offspring distribution is Poisson with mean $\lambda$ and consider $\eta$ as a function of $\lambda$. Then $\eta(\lambda)=0$ for $\lambda\leq 1$ and $\eta(\lambda)>0$ for $\lambda>1$, and $\eta(.)$ is continuous everywhere, including at 1.

On the other hand, consider $\beta$ as a function of $\lambda$. Then there is a critical point $\lambda_c\approx 3.3509$ with the following property: $\beta(\lambda)=0$ for $\lambda<\lambda_c$, and $\beta(\lambda)>0$ for $\lambda\geq \lambda_c$. In particular, $\beta(.)$ is discontinuous at $\lambda_c$. (In fact, at $\lambda_c$, $\beta$ jumps from 0 to approximately 0.535).

My question: how has this been written as a general statement? (rather than just for particular parametrised families as above). My guess is that one would want to write it something like the following:

Let $M_0$ be the set of offspring distributions with the topology induced by the metric $d_0(\mathbf{p}, \mathbf{q})=\sum|p_n-q_n|$.

Similarly $M_1$ and $M_2$ induced by $d_1(\mathbf{p}, \mathbf{q})=\sum n|p_n-q_n|$ and $d_2(\mathbf{p}, \mathbf{q})=\sum n^2|p_n-q_n|$.

Write $\mathbf{p}^*$ for the degenerate distribution with $p^*_n=0$ for $n=0$ and $n\geq 2$ and $p^*_1=1$.

Then:

(1) $\mathcal{H}\setminus\{\mathbf{p}^*\}$ is open as a subset of $M_0$.

(2) $\mathcal{B}$ is closed as a subset of $M_2$. (Probably also $M_1$?)

Of course (1) is basically trivial, since $\mathcal{H}$ is just the set of distributions with mean greater than 1, along with the single point $\mathbf{p}^*$.

(Note that it's NOT the case that $\mathcal{B}$ is closed as a subset of $M_0$. For example, consider a sequence of distributions indexed by $k$ with $p_0=1-4/k$ and $p_k=4/k$, all other $p_n=0$. Then $\beta(k)>0$ for all $k$, but $\beta(k)\to 0$ as $k\to\infty$, and the sequence of distributions converges in $M_0$ to a limit in which just $p_0=1$, for which of course $\beta=0$.)

I don't think it's hard to write down a proof of (2) above, but I don't want to reinvent the wheel. (I don't need to use the result directly, but I would definitely like to refer to it to illustrate a point). So: anyone know where such things have been nicely developed before?

Answer 1

OK, in my opinion this is a very interesting question and, since it got no answer for three weeks, I feel free to post some comments on it (but not a full answer, I am afraid).

As everybody knows, the property that $\eta > 0$ is equivalent to the fact that the mean of the offspring distribution $p$ is $>1$. This is conveniently expressed if one relies on the generating function $f$ of $p$, defined by $f(s)=\sum p_ns^n$. Then (excluding the case where $p$ is a Dirac mass at $1$), $\eta > 0$ iff $f'(1) > 1$ iff the equation $f(s)=s$ has a solution $s_1 < 1$ and when this is so, $\eta=1-s_1$.

Likewise, the property that $\beta > 0$ is equivalent to the fact that the equation $g(s)=s$ has a solution $s_2 < 1$, where $g(s)=f(s)+(1-s)f'(s)$ (hint: condition on the first generation) and when this is so, $\beta=1-s_2$.

Now, it happens that if $s_1 < 1$ exists, the graph of $f$ on the interval $[0,1]$ (in other words, the curve of equation $t=f(s)$ in the $(t,s)$-square $[0,1]\times[0,1]$) must intersect the line $t=s$ with an order of contact $1$, because $f$ is strictly convex and $f(1)=1$. Hence every small modification of $f$ preserves this intersection and is such that $\eta > 0$ as well.

On the other hand, $g(1)=1$ but $g$ is not in general convex, hence it may well happen that the graph of $g$ intersects the line $t=s$ with an order of contact $2$ (in other words, it is tangent to the line $t=s$ at $s_2$). Then, infinitely small modifications of $f$ may move the graph of $g$ away from the line $t=s$, in which case the only solution of $g(s)=s$ is the obvious one $s=1$, hence $\beta$ jumps to $0$.

In other words, "There exists $s < 1$ such that $f(s)=s$" is an open property (excluding the function $f$ such that $f(s)=s$ for every $s$) but "There exists $s < 1$ such that $f(s)+(1-s)f'(s)=s$" is not.

Finally, note that the geometric distributions provide phase transitions that are located at easily computable points. Namely, assume that $p_n=(1-a)a^n$ for every $n\ge 0$, with $0 < a < 1$. Then, $\eta>0$ iff $a > a_1$ with $a_1=1/2$, in particular $\eta(a_1)=0$, but $\beta > 0$ iff $a\ge a_2$ with $a_2=4/5$ and $\beta(a_2)=1/4$ (I think).

To reiterate: these are just minor remarks on a stimulating question, not an answer to the OP's query.