Category Archives: Decision Theory

Las Vegas and financial institutions

Exactly one month ago, I entered the Bellagio casino to gamble at the roulette. It was actually a request from my daughter’s godfather (who happens to be a probabilist, actually). On a comment on a previous post, he suggested the following deal,

In the Bellagio you put 10$ for me on the 33 and 10$ for you as well. If 33 shows up, you bring me to a French “3 étoiles” restaurant next time you stop by in France. If 33 doesn’t shows up, I bring you to MacDonald…

I have to admit that I like eating in French “3 étoiles” restaurants, so I did gamble. Well, I could not remember the terms of the agreement very well (neither the number to select, nor the amount to put on the table). So I did ask my daughter which number I was supposed to pick, and she choose 22. Anyway, the number that came up was neither the 33 nor the 22, so we lost. And roulette tables at the Bellagio were down to a $15 minimum (from what I remember, I was supposed to play $5 or $10). And I have seen tables with $100 minimum (probably more, but I am not sure, and I could not take much pictures inside) ! So I did play $15 (I did keep chips as souvenirs), and I have to admit that I was excited during a few seconds. I really enjoyed that thrilling sensation ! And I was playing only $15 !

Later on, in the room of the hotel – while we’ve been watching TV – we saw some poker games where people where putting $200,000 on the table (there was almost a million in the pot) ! I tried to explain my kids that this was a reason why there were so many signs on the walls claiming that kids were not supposed to enter casinos, and so many ads about gambling being an addiction. It is not reasonable to put so much on a table ! $200,000 on the table ? This is probably more than all I could possibly own !

I thought about all that yesterday, when I discovered the following Table, about leverage ratios of bank,see  http://fool.com/investing/…

Company

Leverage Ratio (Assets-to-Equity), 2007

Bear Stearns 34:1
Morgan Stanley 33:1
Merrill Lynch 32:1
Lehman Brothers 31:1
Goldman Sachs 26:1

(with similar values in U.K., according to http://voxeu.org/…)

What does 30 mean ? It means that a company with $1 in capital holds $30 in (various) financial positions (see http://newleftreview.org/II/… for a discussion). If you think about it, with a relative decline of 3.5%, the absolute loss is larger than the capital hold by the company… Now, if we forget about Lehman, and focus on me, gambling in Las Vegas, we can try to illustrate this 30:1 leverage ratio as follows. The way I see it is that, if I were a bank with $200,000 equity (equity being, from my understanding, everything I own), I would be able to borrow 30 times this amount, and put this money on some table in Las Vegas. OK, there might be a big difference, since in Vegas, on average I will loose money, while most models in finance claim that (on average) we should gain money (somehow, since it might depend on your reference level). And no one really own the casino in real life. But still. A 30 leverage ratio means that I would be playing more than $6 million on a table in Las Vegas ! How should I understand that 30 leverage ratio ? Am I really such a small player ? Are banks really too big players ? Or perhaps they do not hold enough capital to play that big….

From Simpson’s paradox to pies

Today, I wanted to publish a post on economics, and decision theory. And probability too… Those who do follow my blog should know that I am a big fan of Simpson’s paradox. I also love to mention it in my
econometric classes. It does raise important questions, that I do relate to multicolinearity, and interepretations of regression models, with multiple (negatively correlated) explanatory variables. This paradox has amazing pedogological virtues. I did mention it several times on this blog (I should probably mention that I discovered this paradox via Marco Scarsini, who did learn me a lot of things, in decision theory and in probability). For those who do not know this paradox, here is an example that Marco gave in one of his talk, a few years ago. Consider the following statistics, when healthy people entered in some hospital

hospital total survivors deaths survival
rate
hospital A 600 590 10 98%
hospital B 900 870 30 97%

while, when sick people entered in the same hospitals

hospital total survivors deaths survival
rate
hospital A 400 210 190 53%
hospital B 100 30 70 30%

Somehow, whatever your health situation, you should choose hospital A. Now, if we agregate

hospital total survivors deaths survival
rate
hospital A 1000 800 200 80%
hospital B 1000 900 100 90%

i.e. without any doubts, people should choose hospital B.

Actually, Simpson’s paradox is called Simpson’s paradox because Colin Blyth named it that way in 1972, in his paper entitled on Simpson’s paradox and the sure-thing principle (an economic article in a statistical journal), that can be downloaded from http://www.stat.cmu.edu/~fienberg/…. He found this paradox in a paper published in 1951 by Edward Simpson, even if other papers actually did mention it earlier. The most popular application is probably admission at Berckley’s graduate studiesprograms, and sex bias, see Bickel, Hammel & O’Connell (1975), that can be downloaded from http://www.unc.edu/~nielsen/…. I also mentioned a geometric interpretation of this paradox a few years ago on my blog, which is so simple to understand that the paradox is no longer a paradox actually, since on the example above, we had

and

while

With symbolic notations, one can have at the same time

and

with also

as shown on the graph below

There should be connection between Simpson’s paradox and the ecological fallacy (which is an issue I recently discovered and that I found extremely interesting, related again to difficulties of interpreting
regressions). But that’s another story. My point today is that Colin Blyth did mention another nice paradox, that is related, this time, to stochastic orderings. The idea is the following. Consider the three spinners drawn below (imagine some arrows in those circles)

  • spinner A: no matter where the arrow stops, the gain is 3,
  • spinner B: 56% chances to gain 2, 22% chances to gain 4, and 22% chances to gain 6,
  • spinner C: 51% chances to gain 1, 49% chances to gain 5.

Instead of spinners, it is also possible to consider three different lotteries,

You play against a friend, you pick a spinner, while the friend picks another. Everyone flick his arrow, the highest number wins (no matter the difference). Let us compute the odds. First case, A against B, from
A’s perspective

B-2 B-4 B-6
A-3 56%
+1
win
22%
-1
lose
22%
-3
lose

In that case, A has 56% chance of beating B. Second case, A against C, from A’s perspective,

C-1 C-5
A-3 51%
+1
win
49%
-2
lose
In that case, A has 51% chance of beating C. Third (an final) case, B against C, from B’s perspective. Assuming independence between the spinners, joint probabilities can easily be computed,
C-1 C-5
B-2 28.56%
+1
win
27.44%
-3
lose
B-4 11.22%
+3
win
10.78%
-1
lose
B-6 11.22%
+5
win
10.78%
+1
win
In that case, B has 61.78% chance of beating C. So, if we try to summarize,
  • A is the best choice, since it beats both with – always – more than 50% chance,
  • C is the worst choice, since it is beaten by both with – always – more than 50% chance,
Now, assume that you play not against one friend, but two friends. An everyone picks a different spinner. Let
us compute the odds, one more time. First case, A against B and C, from A’s perspective
B-2
C-1
B-2
C-5
B-4
C-1
B-4
C-5
B-6
C-1
B-6
C-5
A-3 28.56%
+1
win
27.44%
-2
lose
11.22%
-1
lose
10.78%
-1
lose
11.22%
-3
lose
10.78%
-3
lose
In that case, A has 28.56% chance of beating B and C. Second case, B against A and C, from B’s perspective,
A-3
C-1
A-3
C-5
B-2 28.56%
-1
lose
27.44%
-2
lose
B-4 11.22%
+1
win
10.78%
-1
lose
B-6 11.22%
+3
win
10.78%
+1
win
In that case, B has 33.22% chance of beating A and B.Third (an final) case, C against A, from C’s perspective,
A-3
B-2
A-3
B-4
A-3
B-6
C-1 28.56%
-2
lose
11.22%
-3
lose
11.22%
-5
lose
C-5 27.44%
+2
win
10.78%
+1
win
10.78%
-1
lose

In that case, C has 38.22% chance of beating A and B. So, if we try to summarize, this time

  • C is the best choice, since has (strictly) more than 1/3 chances to win, which the highest probability
  • A is the worst choice, since has (strictly) less than 1/3 chances to win, which the lowest probability

Odd isn’t it ? Now, is there an interpretation of that paradox ? Yes, Martin Gardner, in his paper on induction and probability, mentioned the case of drug testing. The value we had with the spinner is the health level, rated from 1 to 6. Thus, taking drug A, you always get an average health level of 3. With drug C, on the other hand, you get either very sick (level 1) or very well (level 5). Consider now a doctor who wants to maximize the patient’s chance of being well. If only pills A and C are available, then the doctor should choose A. This is what we’ve seen in the first part. Assume that now a company delivers a third pill, called drug B. Then the doctor should find C more interesting…. Odd, isn’t it ?

Colin Blyth gave a more amusing application. Assume that you like to go to the restaurant, and you like get a dessert there. Dessert A – the apple pie – is the average one, with a standard level, that you rank 3 (on a scale from 1 to 6). Dessert C – the cheese cake – can either be awfull (ranked 1) or delicious (ranked 5). You’d better go for the apple pie if you want to maximize the probability of not being disappointed (i.e. maximizing your “best chance” according to Colin Blyth, but I guess it can be interpreted as regret minimization too). Now assume that dessert B – the blueberry pie – is available (with ranks given by the spinner). Then you should go for the cheese cake. I let you imagine the discussion that you can have, then, with your favorite waitress

– Hi Mr Freakonometrics, do you want a piece of apple pie ? (yes, actually she also comes frequently on my blog, and knows me from my pseudo…)

– Probably. But actually, I was wondering if you did have your blueberry pie today ?

– Yes, in fact we do….

– Great, in that case, I’ll go for the cheese cake.

She’ll probably think that I am freak… so I hope she’ll come and read my post, to understand that, actually, it does make a lot of sense to go for what was supposed to be my worst case.

Natural Catastrophe Insurance: How Should the Government Intervene?

An updated version of the joint paper with Benoit Le Maux is online on http://papers.ssrn.com/.

The present paper develops a new theoretical framework for analyzing the decision to provide or buy insurance against the risk of natural catastrophes. In contrast with conventional models of insurance, the insurer has a non-zero probability of insolvency that depends on the distribution of the risks, the premium rate, and the amount of capital in the company. Among several results, we show that risk-averse policyholders will accept to pay higher rates for a government-provided insurance with unlimited guarantee. However, depending on the correlation between and within the regional risks, a government program can be more attractive to high-correlation than to low correlation areas, which may lead to inefficiencies if the insurance ratings are not appropriately chosen.

Course on risk measures (in French)

The course on risk measure, in Luminy, starts at 16.00 on Monday (here). The slides can be found here,

Note that additional references can be downloaded on the internet, e.g. the short course on risk measures by Freddy Delbaen (here) or the article from the Encyclopedia of quantitative finance, by Hans Föllmer and Alexander Schied (there). See also here for the paper by Jean Marc Tallon, Johanna Etner and Meglena Jeleva, on decision theory under uncertainty.

Primes, joker et mathématiques financières

Lors du dîner de gala des JEEA jeudi soir, Mohamed m’a posé une question intéressante, et je lui ai promis un billet (ou plutôt deux car son problème est compliqué, et je ne connais la solution qu’à une version simple). Le problème est le suivant: un gros assureur à trois lettres souhaite encourager les agents commerciaux par une prime. On leur donne un joker, et durant une période d’un mois, ils concluent des affaires nouvelles. Ils ont la possibilité de toucher une fois (et un seule) une prime (en utilisant leur joker) qui sera proportionnelle au montant de l’affaire signée. Quelle est la stratégie optimale pour utiliser leur joker ?. Une question plus courte pour résumer cette optimalité: le deuxième jour, un gros contrat (aux yeux du vendeur) est signé: faut-il utiliser son joker ou vaut-il mieux attendre un peu ? Bon, le vrai problème est qu’ils ont 5 jokers, et qu’ils peuvent les utiliser en une seule fois, ou en plusieurs…. Avant de réfléchir à cette histoire de 5 jokers, regardons un peu avec un….

  • Formalisation du problème…

Faisons quelques hypothèses forcément simplificatrices… On suppose que chaque jour, un contrat est signé, et que les montant des contrats sont indépendants et identiquement distribués (on ne fait pas de plus gros deal en début de mois). Soit le montant de la prime associée à la kième affaire (si on utilisait le joker). Il faut alors arbitrer, chaque jour k, entre

  • toucher 
  • ne pas toucher la prime, et espérer que l’on touchera davantage plus tard.

Notons la valeur du joker à la date k. Alors

Aussi

soit

On sait aussi que (le dernier jour, si on a le joker, on l’utilise). Autrement dit, on devrait  y arriver par induction backward… Et la résolution dépend de la loi des montants des affaires.

  • si F est uniforme sur [0,100]

Dans ce cas, l’équation se simplifie. Si ,

et

soit

On peut visualiser cette fonction sur le graphique suivant, en fonction du temps

Autrement dit, on se fixe une stratégie a priori, et on s’y tient ! Sur la simulation suivant, on utilise son joker dès le 4ème jour,

Bon, je suis nul en calculs, mais en faisant du monte carlo, on en déduit la loi de la date optimale d’exercice,

ainsi que le gain espéré (ce qui permettra à Mohamed de se couvrir).

Notons que l’espérance de la date d’exercice est environ le 12ème jour, et le montant moyen est de 95 (contre 50 en exerçant le dernier jour).

  • si on change de loi, une loi exponentielle ?

Je pense qu’on peut faire des calculs fermés…. mais je suis un peu paresseux…. on obtient la courbe suivante

La distribution de la date optimale donne

et pour le montant empoché

Dans ce cas, on exerce en moyenne au bout de 17 jours, pour un gain moyen de 174.

  • L’exercice d’options américaines

Damned, mais tout ça correspond au problème de valorisation des options américaines (ou plutôt Bermudéennes car le temps est discret). Les options dites bermudéennes peuvent être exercée à un ensemble prédéterminé de dates

L’idée de la valorisation est simple: à chaque date, le détenteur de l’option a en effet de choix,

  • exercer son option et en retirer un payoff 
  • conserver son option, de telle sorte que son option vaut  en 

Si on note  le facteur d’actualisation entre les dates  et , on en déduit que la valeur en  de l’option peut s’écrire

est la filtration naturelle, et  est une probabilité risque neutre, sous laquelle la valeur actualisée de l’actif est une martingale, i.e.

Je renvoie à mes notes de cours de méthodes numériques en finance (ici) mais en utilisant les arbres binomiaux, on peut valoriser un put américain, par exemple,

Paradoxe, vote et démocratie

Le paradoxe le plus classique sur les élections est le paradoxe de Condorcet, correspondant à une intransitivité des préférences. Pour rappels, dans le cours d’économie de l’incertain, j’avais postulé que les relations de préférences étaient une relation d’ordre (et donc étaient transitives) sans que cela ne soulève la moindre objection. Prenons un cas pratique: un sondage LH2 pour RMC-BFM TV-20 minutes, diffusé le 12 mars 2007 sur les élections présidentiels, donnait les intentions de vote suivant,

  • au premier tour, Sarkozy (28) > Royal (26) > Bayrou (22)
  • au second tour, Sarkozy (52) > Royal (48) mais  Bayrou (55) > Sarkozy (45)

Autrement dit, on peut A>B, et B>C, mais sans pour autant avoir A>C.

Pour Condorcet, il n’existait pas de système simple assurant cette cohérence, à savoir une fonction de choix social indiscutable, permettant d’agréger des préférences individuelles en préférences sociales. Arrow démontre, sous réserve d’acceptation de quelques hypothèses, qu’il n’existerait pas de système du tout assurant la cohérence, hormis celui où la fonction de choix social coïnciderait avec les choix d’un seul individu, parfois surnommé dictateur, indépendemment du reste de la population.

Le théorème d’Arrow peut être énoncé sous la forme suivante.

Pour au moins 3 options de choix et deux individus, il n’existe pas de fonction de choix social satisfaisant les propriétés suivantes :

  1. Universalité : la fonction de choix social doit être définie pour tout profil de préférences logiquement possible. Cela signifie qu’on ne peut éliminer le choix de certains individus du groupe des individus qui déterminent le choix collectif.
  2. Non-dictature : il n’existe aucun individu pour lequel la liste de ses choix personnels coïncide avec la fonction de choix social, indépendamment des préférences des autres ;
  3. Unanimité : lorsque tous les individus ont les mêmes préférences, la fonction de choix social doit associer ces mêmes préférences à la société.
  4. Indifférence des Options Non-Pertinentes : le classement relatif de deux options ne doit dépendre que de leur position relative pour les individus et non du classement d’options tierces ; si l’on ne considère qu’un sous-ensemble d’options, la fonction ne doit pas aboutir à un autre classement de ce sous-ensemble.

Sur les paradoxes du vote, je renvoie à une très bonne émission qui avait été diffusée dans Archimède sur Arte en janvier 2001. L’intégralité du texte se trouve ici,