# Is this 2x2 determinant sequence positive and increasing?

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 $$n=0,1,2,\ldots$$

• What exactly do you mean by "calculations show ..."? Do you just mean that you have checked it for a selection of numerical values of the parameters $a_k$, $b_k$ and $c_k$? – Neil Strickland Sep 4 '15 at 16:45
• Are $X_i,\ i=1,2,3$ independent? – Samrat Mukhopadhyay Sep 8 '15 at 12:29
• It seems you will need some additional hypothesis on the $X_j$; if I've calculated correctly, taking all the $X_j$ equal to $0$ with probability $1$ gives $D_n=1$ for all $n$. – Mike Jury Sep 8 '15 at 16:49