# “Nested” Binomial Distribution

I am currently working with a statistical distribution, and I'm wondering if any exploration has been done on this.

The distribution is denoted $\xi$. To construct $\xi$ we use auxillary random variables $X_1,X_2, \ldots, X_k$

For indexing purposes we let the random variable

$X_1 = 1$

$X_2 \sim \mathrm{Bin}(4, \frac{1}{2})$

$X_3 \sim \mathrm{Bin}(4X_2, \frac{1}{2})$

and in general

$X_k \sim \mathrm{Bin}(4X_{k-1}, \frac{1}{2})$

So that the number of trials of the binomial distribution is itself a random variable.

One can show that the sum of $P(X_k = 0)$ over all $k$ is about $0.08737$.

The distribution $\xi$ is given by $P(\xi=k) := \dfrac{P(X_{k+1} = 0) - P(X_k = 0)}{0.08737}$ (we divide because probabilities have to add up to $1$).

• How exactly is $\xi$ defined, with regard to $X_1,\dots,X_k,\dots$? – Clement C. Nov 3 '15 at 13:35
• $P(\xi = k):=P(X_{k+1} = 0) - P(X_{k} = 0)$, with suitable division to force the probabilities to add to 1. – user242594 Nov 3 '15 at 14:09
• One should say $X_3\mid X_2 \sim \mathrm{Bin}(4X_2, \frac{1}{2})$, i.e. this is a conditional distribution. $\qquad$ – Michael Hardy Sep 9 '16 at 17:50
• "the sum of $P(X_k = 0)$ over all $k$ is about $0.08737$" You mean, the limit of $P(X_k = 0)$ when $k\to\infty$, right? – Did Sep 9 '16 at 19:22
• For your interest, the number approximately equal to $0.08737$ (actually, $0.08738$, see below), which is the extinction probability of the underlying branching process, is the real root of the polynomial $q^3+5q^2+11q-1$, that is, $$q=\frac23 \left(\sqrt[3]{17+3 \sqrt{33}}-\frac2{\sqrt[3]{17+3 \sqrt{33}}}\right)-\frac53\approx0.0873780253841527$$ – Did Sep 9 '16 at 19:32

I think it would be illustrative to rephrase the question more precisely. Let $X_0 = 1$, and define the sequence of random variables $$X_k \mid X_{k-1} \sim \operatorname{Binomial}(n = 4X_{k-1}, p = \tfrac{1}{2}), \quad k = 1, 2, \ldots.$$ Then define a random variable $\xi$ by $$\Pr[\xi = k] = \frac{\Pr[X_{k+1} = 0] - \Pr[X_k = 0]}{\lim_{n \to \infty} \Pr[X_n = 0]}, \quad k = 0, 1, 2, \ldots$$
It is worth noting that $$\begin{array}{c|c|c|c}k & \Pr[X_k = 0] & \Pr[\xi = k] & \Pr[\xi = k-1]/\Pr[\xi = k] \\ \hline 0 & 0 & 0.715283 & -- \\ 1 & 0.0625 & 0.196295 & 3.64393 \\ 2 & 0.0796518 & 0.0603027 & 3.25515 \\ 3 & 0.084921 & 0.019112 & 3.15523 \\ 4 & 0.0865909 & 0.00611566 & 3.12509 \\ 5 & 0.0871253 & 0.00196292 & 3.1156 \\ 6 & 0.0872968 & 0.000630643 & 3.11256 \\ 7 & 0.0873519 & 0.000202675 & 3.11159 \\ 8 & 0.0873696 & 0.0000651421 & 3.11128 \\ 9 & 0.0873753 & 0.0000209381 & 3.11118 \\ 10 & 0.0873772 & 6.73002 \times 10^{-6} & 3.11115 \\ 11 & 0.0873777 & 2.1632 \times 10^{-6} & 3.11114 \\ 12 & 0.0873779 & 6.9531 \times 10^{-7} & 3.11113 \end{array}$$ so that it looks like $\xi$ asymptotically behaves much like a geometric distribution. But other than this, I doubt that there is a nice closed form for $\xi$.
• Cool, thanks so much! Do you think it is possible to rigorously prove that the ratio is, say, greater than $3$? – user242594 Nov 4 '15 at 1:07
This is an example of a Galton–Watson branching process. These originated in Francis Galton's statistical study of the extinction of family names. In your example of such a process, a man could have as many as four sons. Your random variable $X_k$ is the number of male descendants in the $k$th generation.
Your random variable $\xi$ is the generation in which the name becomes extinct.
If the expected number of sons of each man is more than $1$, then the probability of ultimate extinction is less than $1$. In your case the expected number of sons of each man is $2$. That means the probability that the name survives forever is more than $0$.