Problem source

The life of a certain species of elementary particles
can be described as follows. Each particle has a life time 
of $T$ seconds, after which it disintegrates into $X$ particles
of the same species, where $X$ is a random variable with 
binomial distribution $\mathrm{B}(2,p)\,$.
A population of these particles starts with the creation
of a single such particle at $t=0\,$. Let $X_n$ be 
the number of particles in existence in the time interval 
$nT < t < (n+1)T\,$, where $n=1\,$, $2\,$, $\ldots$.  
Show that $\P(X_1=2 \mbox { and } X_2=2) = 6p^4q^2\;$, where $q=1-p\,$.
Find the possible values of $p$ if it is known that 
$\P(X_1=2 \vert X_2=2) =9/25\,$.
Explain briefly why  $\E(X_n) =2p\E(X_{n-1})$ and hence determine $\E(X_n)$
in terms of $p$.
Show that for one of the values of $p$ found above  
$\lim_{n \to \infty}\E(X_n) = 0$ 
and that for the other
$\lim_{n \to \infty}\E(X_n) = + \infty\,$.

Solution source

Notice that we can see the total number generated as $X_n \sim B(2X_{n-1},p)$, since a Binomial is a sum of independent Bernoullis, and there are two Bernoullis per particle.

\begin{align*}
&& \mathbb{P}(X_1=2 \mbox { and } X_2=2) &= \underbrace{p^2}_{\text{two generated in first iteration}} \cdot \underbrace{\binom{4}{2}p^2q^2}_{\text{two generated from the first two}} \\
&&&= 6p^4q^2
\end{align*}

\begin{align*}
&& \mathbb{P})(X_1 = 2 |X_2 = 2) &= \frac{ \mathbb{P}(X_1=2 \mbox { and } X_2=2) }{ \mathbb{P}( X_2=2) } \\
&&&= \frac{6p^4q^2}{6p^4q^2+2pq \cdot p^2} \\
&&&= \frac{3pq}{3pq+1} \\
\Rightarrow && \frac{9}{25} &= \frac{3pq}{3pq+1} \\
\Rightarrow && 27pq + 9 &= 75pq \\
\Rightarrow && 9 &= 48pq \\
\Rightarrow && pq &= \frac{3}{16} \\
\Rightarrow && 0 &= p^2 - p + \frac3{16} \\
\Rightarrow && p &= \frac14, \frac34
\end{align*}

By the same reasoning about the Bernoullis, we must have $\E[X_n] = \E[\E[X_n | X_{n-1}]] = \E[2pX_{n-1}] = 2p \E[X_{n-1}]$ therefore $\E[X_n] = (2p)^n$.

If $p = \frac14$ then $\E[X_n] = \frac1{2^n} \to 0$
If $p = \frac34$ then $\E[X_n] = \left(\frac32 \right)^n \to \infty$