Distribution of a random variable obtained by chaining distribution

Question

Let us define $M_0=2^n$ for $n\in\mathbb{N}$. Let $\ell\in\mathbb{N}$ be the number of random variables we are working with. For $1\leqslant i\leqslant\ell$, we define $M_i$ to be a random variable following a binomial distribution with parameters $M_{i-1}$ and $2^{-n}$.

I'm interested in computing $\mathbb{P}\left[M_\ell=1\middle|M_\ell\geqslant1\right]$. To be fair, I only needed to lower-bound this probability, which is easy to do by computing $\mathbb{P}\left[M_1=1\middle|M_1\geqslant1\right]\approx\frac{\mathrm{e}^{-1}}{1-\mathrm{e}^{-1}}>\frac12$, but, out of curiosity, I was wondering if it was possible to get a closed-form for $\mathbb{P}\left[M_\ell=1\middle|M_\ell\geqslant1\right]$, and I do not know of any theorem that would help me with this.

Intuitively, this probability should decrease exponentially fast to $1$, and it might be possible to get the closed-form expression of the probability for $\ell\leqslant3$ by hand, but it will quickly become tedious to operate that way. Is there any clever way I'm missing for this?

Answer 1

Abbreviate $N=M_0=2^n$ and $b=1/N$. As described, the conditional distribution of $M_i$, given that $M_{i-1}=n$, is binomial with parameters $n$ and $b$. The conditional generating function of $M_i$ is therefore $$ E[s^{M_i}\mid M_{i-1}]=(1-b+bs)^{M_{i-1}}. $$ Using $g_i$ to denote the generating function of $M_i$, this leads to the recursion $$ g_{i}(s) = g_{i-1}(1-b+bs), $$ and, because $g_0(s) = s^N$, by recursion, $$ g_i(s) = (1-b^i+b^is)^N, \qquad0\le s\le 1. $$ In other words, $M_i$ has the binomial distribution with parameters $N$ and $b^i$.

In particular, $$ P[M_i=1\mid M_i\ge 1] = {Nb^i(1-b^i)^{N-1}\over 1-(1-b^i)^N}. $$