# Talk on bivariate count times series in finance and risk management

I will be giving a talk on May 4th, at the Mathematical Finance Days, at HEC Montréal, on multivariate dynamic models for counts. The conference is organized by IFM2 (Institut de Finance Mathématique de Montréal). I will be chairing some session and I will give a talk based on the joint paper with Mathieu Boudreault.

View more presentations from charthur.

In various situations in the insurance industry, in finance, in epidemiology, etc., one needs to represent the joint evolution of the number of occurrences of an event. In this paper, we present a multivariate integer‐valued autoregressive (MINAR) model, derive its properties and apply the model to earthquake occurrences across various pairs of tectonic plates. The model is an extension of Pedelis & Karlis (2011) where cross autocorrelation (spatial contagion in a seismic context) is considered. We fit various bivariate count models and find that for many contiguous tectonic plates, spatial contagion is significant in both directions. Furthermore, ignoring cross autocorrelation can underestimate the potential for high numbers of occurrences over the short‐term. An application to risk management and cat‐bond pricing will be discussed.

# Even odds

This evening, I found a nice probabilistic puzzle on http://www.futilitycloset.com/“A bag contains 16 billiard balls, some white and some black. You draw two balls at the same time. It is equally likely that the two will be the same color as different colors. What is the proportion of colors within the bag?”
To be honest, I did not understood the answer on the blog, but if we write it down, we want to solve

Let us count: if  is the total number of balls, and if  is the number of white balls then

I.e. we want to solve a polynomial equation (of order 2) in , or to be more precise, in

If  is equal to 16, then  is either 6 or 10. It can be visualized below

```> balls=function(n=16){
+ NB=rep(NA,n)
+ for(k in 2:(n-2)){
+ NB[k]=(k*(k-1)+(n-k)*(n-k-1))
+ }
+ k=which(NB==n*(n-1)/2)
+ if(length(k)>0){
+ plot(1:n,NB,type="b")
+ abline(h=n*(n-1)/2,col="red")
+ points((1:n)[k],NB[k],pch=19,col="red")}
+ return((1:n)[k])}
> balls()
[1]  6 10```

But more generally, we can seek other ‘s and other pairs of solutions of such a problem. I am not good in arithmetic, so let us run some codes. And what we get is quite nice: if  admits a pair of solutions, then  is the squared of another integer, say . Further, the difference between  and  is precisely . And  will be one of the answers when the total number of balls will be . Thus, recursively, it is extremely simple to get all possible answers. Below, we have  and the difference between  and ,

```> for(s in 4:1000){
+ b=balls(s)
+ if(length(b)>0) print(c(s,b,diff(b)))
+ }
[1] 9 3 6 3
[1] 16  6 10  4
[1] 25 10 15  5
[1] 36 15 21  6
[1] 49 21 28  7
[1] 64 28 36  8
[1] 81 36 45  9
[1] 100  45  55  10
[1] 121  55  66  11
[1] 144  66  78  12
[1] 169  78  91  13
[1] 196  91 105  14
[1] 225 105 120  15
[1] 256 120 136  16
[1] 289 136 153  17
[1] 324 153 171  18
[1] 361 171 190  19
[1] 400 190 210  20
[1] 441 210 231  21
[1] 484 231 253  22
[1] 529 253 276  23
[1] 576 276 300  24
[1] 625 300 325  25
[1] 676 325 351  26
[1] 729 351 378  27
[1] 784 378 406  28
[1] 841 406 435  29
[1] 900 435 465  30
[1] 961 465 496  31```

Thus, given , consider an urn with  balls. We draw two balls at the same time. It is equally likely that the two will be the same color as different colors. Then the number of colors within the bag are respectively

Finally, observe that the ‘s are well known, from Pascal’s triangle,

also known as triangular numbers,

Maths can be magic, sometimes…