We have random variables $X,X_1,X_2,I_1$ and $I_2$ which are independent, $X,X_1$ and $X_2$ have the same distribution and $I_1,I_2 \sim \text{Bernoulli}\left(\frac{1}{2}\right)$. The following holds $$X = 2+I_1X_1+I_2X_2.$$ We need to compute the probability-generating function os $X$.
So $I_1$ and $I_2$ are indicators and we have:
- $X \mid \{I_1 = 1, I_2 = 1\} \sim 2+X_1+X_2$ therefore $\mathbb{E}[s^X\mid I_1 = 1, I_2 = 1] = s^2M_X(s)^2$,
- $X \mid \{I_1 = 0, I_2 = 1\} \sim 2+X_2$ therefore $\mathbb{E}[s^X\mid I_1 = 0, I_2 = 1] = s^2M_X(s)$,
- $X \mid \{I_1 = 1, I_2 = 0\} \sim 2+X_1$ therefore $\mathbb{E}[s^X\mid I_1 = 1, I_2 = 0] = s^2M_X(s)$,
- $X \mid \{I_1 = 0, I_2 = 0\} \sim 2$ therefore $\mathbb{E}[s^X\mid I_1 = 0, I_2 = 0] = s^2$.
From there we get $$M_X(s)=\sum_{i,j\in\{0,1\}}\mathbb{E}[s^X\mid I_1 = i, I_2 = j]\cdot\mathbb{P}(I_1 = i, I_2 = j).$$ Because $I_1$ and $I_2$ are independent and $\sim \text{Bernoulli}\left(\frac{1}{2}\right)$ we get $\mathbb{P}(I_1 = i, I_2 = j)=\frac{1}{4}$. We get the equation $$M_X(s) = \frac{s^2}{4}M_X(s)^2 + \frac{s^2}{2}M_X(s)+\frac{s^2}{4}$$ and from there $$\frac{s^2}{4}M_X(s)^2 + \left(\frac{s^2}{2}-1\right)M_X(s)+\frac{s^2}{4} = 0.$$ Solving the quadratic we get $$M_X(s)=\frac{2-s^2\pm\sqrt{1-s^2}}{s^2}.$$
So now my question; which sign before the square root should I choose to get the PGF of $X$? (If I have made any mistakes, please correct me xD.)
There isn't a distribution for $X, X_1, X_2$ to satisfy the above properties. The series expansions for both roots are \begin{align*} G_1(s) = \frac{2-s^2 + \sqrt{1-s^2}}{s^2} &= 3s^{-2} - \frac{3}{2}s^0 - \frac{1}{8}s^2 - \cdots \\ G_2(s) =\frac{2-s^2 - \sqrt{1-s^2}}{s^2} &= s^{-2} \color{red}{- \frac{1}{2}s^0} + \frac{1}{8}s^2 + \cdots \end{align*} For $G_1(s)$, all coeffients are not bounded between $[0, 1]$, and for $G_2(s)$, the coefficient of $s^0$ is negative.
In general, you verify if the resulting solutions satisfy the properties of a PGF; if multiple solutions satisfy, then you have non-unique distributions which satisfy your original property.