Let $X_1,X_2,X_3$ be three discrete (integer and non-negative valued) random variables with local probabilities $a_k:=\mathbb{P}(X_1=k)$, $b_k:=\mathbb{P}(X_2=k)$, $c_k:=\mathbb{P}(X_3=k)$ and $s_k:=\mathbb{P}(X_1+X_2+X_3=k)$, $k=0,1,2,\dots$

Let us define two recurrent sequences. The first $(\alpha_n)$ by $$\alpha_0:=1,\\ \alpha_1:=0,\\ \alpha_2:=-\frac{1}{b_0c_0},\\ \alpha_n:=\frac{1}{s_0}\left(\alpha_{n-3}-a_{n-2}-\sum_{k=1}^{n-1}s_k\,\alpha_{n-k}\right), n=3,4,\ldots $$ The second $(\beta_n)$ by $$ \beta_0:=0,\\ \beta_1:=1,\\ \beta_2:=-\frac{c_1}{c_0}-\frac{1}{b_0},\\ \beta_n:=\frac{1}{s_0}\left(\beta_{n-3}-\sum_{k=1}^{n-1}s_k\,\beta_{n-k}-a_{n-2}\,c_0+c_0\sum_{k=0}^{n-1}a_k\,b_{n-1-k}\right), n=3,4,\ldots $$ Let us also define a determinant $$ D_n:=\begin{vmatrix} \alpha_{n}& \beta_{n}\\ \alpha_{n+1}& \beta_{n+1} \end{vmatrix}. $$

Can you show that $D_{n+1}>D_{n}>0$ for every $n=0,1,2,\ldots$?

Such a problem arises in insurance mathematics (calculating a ruin probability if a certain claim appears) and has quite a long context behind it.

Since we have a term $1/s_0$, we have to add that $s_0>0$. This implies that $a_0>0$, $b_0>0$ and $c_0>0$. On the other hand, if $\mathbb{P}(X_i=0)=1$ for $ i=1,2,3$, then (in the context of insurance mathematics) this means that there are no claims at all and that's not a problem to solve.

Numerical calculations show that $D_{n+1}>D_{n}>0$ is verified with certain distributions. Moreover, if we define $$ \widetilde{D}_n:=\begin{vmatrix} \alpha_{n}& \beta_{n}\\ \alpha_{n+2}& \beta_{n+2} \end{vmatrix}, $$ then it seems (by numerical calculations) that $\widetilde{D}_{n+1}<\widetilde{D}_n<0$ for every $n=0,1,2,\ldots$

Thanks in advance.