8
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $c_k$? $\endgroup$ – Neil Strickland Sep 4 '15 at 16:45
  • $\begingroup$ I have checked it for a several distinct distributions. One of the simplest probably is P(X_i=0)=1/2 and P(X_i=1)=1/2, i=1,2,3. In that case a_j=b_j=c_j=1/2, j=0,1. It works with Poisson distribution for example also. $\endgroup$ – Fancier of Mathematica Sep 4 '15 at 18:24
  • $\begingroup$ Are $X_i,\ i=1,2,3$ independent? $\endgroup$ – Samrat Mukhopadhyay Sep 8 '15 at 12:29
  • 1
    $\begingroup$ 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 $n$. $\endgroup$ – Mike Jury Sep 8 '15 at 16:49
  • 1
    $\begingroup$ @FancierofMathematica Please add the relevant hypotheses to your question by using the edit feature. It is not good enough to state some of them in the comments. The clearer the question the more likely you will get a good answer. $\endgroup$ – Chris Ramsey Sep 8 '15 at 22:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.