The recent work on investigations in the criminal justice field has my head turning about potential quantitative applications in this area (check out the John Eck & Kim Rossmo podcasts on Jerry’s site first, then check out the recent papers in Criminology and Public Policy on the topic for a start). One particular problem that was presented to me was detective case loads — detectives are humans, so can only handle so many cases at once. Triage is typically taken at the initial crime reporting stage, with inputs such as seriousness of the offense, the overall probability of the case being solved, and future dangerousness of folks involved being examples of what goes into that calculus to assign a case.

Here I wanted to focus on a different problem though — how long to keep cases open? There are diminishing returns to keeping cases open indefinitely, and so PDs should be able to right size the backend of detective open cases as well as the front end triaging. Here my suggested solution is to estimate a survival model of the probability of a case being solved, and then you can estimate an expected return on investment given the time you put in.

Here is a simplified example. Say the table below shows the (instantaneous) probability of a case being solved per weeks put into the investigation.

```
Week 1 20%
Week 2 10%
Week 3 5%
Week 4 3%
Week 5 1%
```

In survival model parlance, this would be the hazard function in discrete time increments. And then we have diminishing probabilities over time, which should also be true (e.g. a higher probability of being solved right away, and gets lower over time). The expected return of investigating this crime at time t is the cumulative probability of the crime being solved at time t, multiplied by whatever value you assign to the case being solved. The costs of investigating will be fixed (based on the detective salary), so is just a multiple of `t*invest_costs`

.

So just to fill in some numbers, lets say that it costs the police department $1,000 a week to keep an investigation going. Also say a crime has a return of $10,000 if it is solved (the latter number will be harder to figure out in practice, as cost of crime estimates are not a perfect fit). So filling in our table, we have below our detective return on investment estimates (note that the cumulative probability of being solved is not simply the sum of the instantaneous probabilities, else it would eventually go over 100%). So return on investment (ROI), at week 1 is `10,000*0.2 = 2,000`

, at week 2 is `10,000*0.28 = 2,800`

, etc.

```
h(t) solved% cum-costs ROI
Week 1 20% 20% 1,000 2,000
Week 2 10% 28% 2,000 2,800
Week 3 5% 32% 3,000 3,200
Week 4 3% 33% 4,000 3,300
Week 5 1% 34% 5,000 3,400
```

So the cumulative costs outweigh the total detective resources devoted to the crime by Week 4 here. So in practice (in this hypothetical example) you may say to a detective you get 4 weeks to figure it out, if not solved by then it should be closed (but not cleared), and you should move onto other things. In the long run (I think) this strategy will make sure detective resources are balanced against actual cases solved.

This right sizes investigation lengths from a global perspective, but you also might consider whether to close a case on an individual case-by-case basis. In that case you wouldn’t calculate the sunk cost of the investigation so far, it is just the probability of the case being solved going forward relative to future necessary resources. (You do the same table, just start the cum-costs and solved percent columns from scratch whenever you are making that decision.)

In an actual applied setting, you can estimate the survival function however you want (e.g. you may want a cure mixture-model, so not all cases will result in 100% being solved given infinite time). It is also the case that different crimes will not only have different survival curves, but also will have different costs of crime (e.g. a murder has a greater cost to society than a theft) and probably different investigative resources needed (detective costs may also get lower over time, so are not constant). You can bake that all right into this estimate. So you may say the cost of a murder is infinite, and you should forever keep that case open investigating it. A burglary though may be a very short time interval before it should be dropped (but still have some initial investment).

Another neat application of this is that if you can generate reasonable returns to solving crimes, you can right size your overall detective bureau. That is you can make a quantitative argument I need X more detectives, and they will help solve Y more crimes resulting in Z return on investment. It may be we should greatly expand detective bureaus, but have them only keep many cases open a short time period. I’m thinking of the recent officer shortages in Dallas, where very few cases are assigned at all. (Some PDs have patrol officers take initial detective duties on the crime scene as well.)

There are definitely difficulties with applying this approach. One is that getting the cost of solving a crime estimate is going to be tough, and bridges both quantitative cost of crime estimates (although many of them are sunk costs after the crime has been perpetrated, arresting someone does not undo the bullet wound), likelihood of future reoffending, and ethical boundaries as well. If we are thinking about a detective bureau that is over-booked to begin with, we aren’t deciding on assigning individual cases at that point, but will need to consider pre-empting current investigations for new ones (e.g. if you drop case A and pick up case B, we have a better ROI). And that is ignoring the estimating survival part of different cases, which is tricky using observational data as well (selection biases in what cases are currently assigned could certainly make our survival curve estimates too low or too high).

This problem has to have been tackled in different contexts before (either by actuaries or in other business/medical contexts). I don’t know the best terms to google though to figure it out — so let me know in the comments if there is related work I should look into on solving this problem.

## 2 Comments