I am a big fan of trees. It is a very nice way to see how financial pricing works, for derivatives. An with a matrix-based language (R for instance), it is extremely simple to compute almost everything. Even options multiple assets. Let us see how it works. But first, I have to assume that everyone knows about trees, and risk neutral probabilities, and is familiar with standard financial derivatives. Just in case, I can upload some old slides of the first course on asset pricing we gave a few years ago at École Polytechnique.
Let us get back on the pricing of (European) call options, with trees.The idea is simple. We have to fix the number of periods. Let us start with only one (as described in the slides above). The stock has price 
and can go either up, and then have price 
or go down, and have price 
. And the fundamental theorem of asset pricing says that we do not really care about probabilities of going up, or down. Assuming that we can buy or sell that stock, and that a risk free asset is available on the market, it is possible to price any contingent financial product, like a financial option. Since we know the final value of the option when the stock goes either up, or down, it is possible to replicate the payoff of that option using the stock and the risk free asset. And we can prove that the price of the option is simply
where the probability is the so-called risk neutral probability
So, we've done it here with only one single period, but it is possible to extend it to multiperiods. The idea is to keep that multiplicative representation of possible values of the stock, and to get a recombinant tree. At step 2, the stock can take only three different values: went up twice, went down twice, or went up and down (or the reverse, but we don't care: this is the point of recombining). If we write things down, then we can prove that
for some probability parameter (the so-call risk neutral probability, if it is unique). But we do not really care about those closed formula, the goal is to write an algorithm which computes the tree, and return the price of a call option (say). But before starting, we have to make a connection between that model with up and down prices, and the parameters of the Black-Scholes diffusion, for the stock price. The idea is to identify the first and the second moment, i.e.
(where, under the risk neutral probability, the trend is the risk free rate) and
The code might look like that
n=5; T=1; r=0.05; sigma=.4;S=50;K=50 price=function(n){ u.n=exp(sigma*sqrt(T/n)); d.n=1/u.n p.n=(exp(r*T/n)-d.n)/(u.n-d.n) SJ=matrix(0,n+1,n+1) SJ[1,1]=S for(i in(2:(n+1))) {for(j in(1:i)){SJ[i,j]=S*u.n^(i-j)*d.n^(j-1)}} OPT=matrix(0,n+1,n+1) OPT[n+1,]=(SJ[n+1,]-K)*(SJ[n+1,]>K) for(i in(n:1)) {for(j in(1:i)){OPT[i,j]=exp(-r*T/n)*(OPT[i+1,j]*p.n+ (1-p.n)*OPT[i+1,j+1])}} return(OPT[1,1]) }
We can plot the evolution of the price, as a function of the number of time periods (or subdivision of the time interval, from now till maturity of the European option),
N=10:400 V=Vectorize(price)(N) plot(N,V,type="l")
Note that we can compare with the Black-Scholes price of this call option, given by
where
and
d1=1/(sigma*sqrt(T))*(log(S/K)+(r+sigma^2/2)*T) d2=d1-sigma*sqrt(T) BS=S*pnorm(d1)-K*exp(-r*T)*pnorm(d2) abline(h=BS,lty=2,col="red")
The code is clearly not optimal, but at least, we see what's going on. For instance, we do not need a matrix when we calculate using backward recursions the price of the option. We can just keep a single vector. But this matrix is nice, because we can use it to price American options. For instance, with the code below, we compare the price of an American put option, and the price of European put option.
price.american=function(n,opt="put"){ u.n=exp(sigma*sqrt(T/n)); d.n=1/u.n p.n=(exp(r*T/n)-d.n)/(u.n-d.n) SJ=matrix(0,n+1,n+1) SJ[1,1]=S for(i in(2:(n+1))) {for(j in(1:i)) {SJ[i,j]=S*u.n^(i-j)*d.n^(j-1)}} OPTe=matrix(0,n+1,n+1) OPTa=matrix(0,n+1,n+1) if(opt=="call"){ OPTa[n+1,]=(SJ[n+1,]-K)*(SJ[n+1,]>K) OPTe[n+1,]=(SJ[n+1,]-K)*(SJ[n+1,]>K) } if(opt=="put"){ OPTa[n+1,]=(K-SJ[n+1,])*(SJ[n+1,]<K) OPTe[n+1,]=(K-SJ[n+1,])*(SJ[n+1,]<K) } for(i in(n:1)) { for(j in(1:i)) {if(opt=="call"){ OPTa[i,j]=max((SJ[i,j]-K)*(SJ[i,j]>K), exp(-r*T/n)*(OPTa[i+1,j]*p.n+ (1-p.n)*OPTa[i+1,j+1]))} if(opt=="put"){ OPTa[i,j]=max((K-SJ[i,j])*(K>SJ[i,j]), exp(-r*T/n)*(OPTa[i+1,j]*p.n+ (1-p.n)*OPTa[i+1,j+1]))} OPTe[i,j]=exp(-r*T/n)*(OPTe[i+1,j]*p.n+ (1-p.n)*OPTe[i+1,j+1])}} priceop=c(OPTe[1,1],OPTa[1,1]) names(priceop)=c("E","A") return(priceop)}
It is possible to compare those price, obtained on trees, with prices given by closed (approximated) formulas.
> d1=1/(sigma*sqrt(T))*(log(S/K)+(r+sigma^2/2)*T) > d2=d1-sigma*sqrt(T) > (BS=-S*pnorm(-d1)+K*exp(-r*T)*pnorm(-d2) ) [1] 6.572947 > N=10:200 > M=Vectorize(price.american)(N) > plot(N,M[1,],type='l',col='blue',ylim=range(M)) > lines(N,M[2,],type='l',col='red') > abline(h=BS,lty=2,col='blue') > library(fOptions) > (am=BAWAmericanApproxOption(TypeFlag = + "p", S = S,X = K, Time = T, r = r, + b = r, sigma =sigma)@price) [1] 6.840335 > abline(h=am,lty=2,col='red')
Another great thing with trees, is that it becomes possible to plot to region where it is optimal to exercise our right to sell the stock.
Let us move now to a model with two assets, as suggested by Rubinstein (1994). First, observe that a discretization of two independent Brownian motions will be based on two independent random walk, taking values
i.e. both went up (NW), both went down (SE), and one went up while the other went down (either NE or SW). With independent and symmetric random walks, the probabilities will be respectively 1/4. An if we move one step foreward, we have the following tree.
 |
|
And again, the idea is then to identify the first two moments. This gives us the following system of equations for the four respective (risk neutral) probabilities
For those willing to do the maths, please do. The answer should be
and for the last one
The code here looks like that
price.spead=function(n){ T=1; r=0.05; K=0 S1=105 S2=100 sigma1=0.4 sigma2=0.3 rho=0.5 u1.n=exp(sigma1*sqrt(T/n)); d1.n=1/u1.n u2.n=exp(sigma2*sqrt(T/n)); d2.n=1/u2.n v1=r-sigma1^2/2; v2=r-sigma2^2/2 puu.n=(1+rho+sqrt(T/n)*(v1/sigma1+v2/sigma2))/4 pud.n=(1-rho+sqrt(T/n)*(v1/sigma1-v2/sigma2))/4 pdu.n=(1-rho+sqrt(T/n)*(-v1/sigma1+v2/sigma2))/4 pdd.n=(1+rho+sqrt(T/n)*(-v1/sigma1-v2/sigma2))/4 k=0:n un=matrix(1,n+1,1) SJ= (S1 * d1.n^k * u1.n^(n-k-1)) %*% t(un) - un %*%t(S2 * d2.n^k * u2.n^(n-k-1)) OPT=(SJ)*(SJ>K) for(k in(n:1)) { OPT0=matrix(0,k,k) for(i in(1:k)) { for(j in(1:k)) {OPT0[i,j]=(OPT[i,j]*puu.n+OPT[i+1,j]*pdu.n+ OPT[i,j+1]*pud.n+OPT[i+1,j+1]*pdd.n)*exp(-r*T/n)}} OPT=OPT0} return(OPT[1,1])}
If we look at the details, consider two periods, like on the figure above, the are nine values for the spread,
> n=2 > SJ [,1] [,2] [,3] [1,] 32.02217 84.86869 119.443578 [2,] -47.84652 5.00000 39.574891 [3,] -93.20959 -40.36308 -5.788184
and the payoff of the option is here
> OPT [,1] [,2] [,3] [1,] 32.02217 84.86869 119.44358 [2,] 0.00000 5.00000 39.57489 [3,] 0.00000 0.00000 0.00000
So if we go backward of one step, we have the following square of values
> k=n > OPT0<-matrix(0,k,k) > for(i in(1:k)) + { + for(j in(1:k)) + { + OPT0[i,j]=(OPT[i,j]*puu.n+OPT[i+1,j]*pdu.n+ + OPT[i,j+1]*pud.n+OPT[i+1,j+1]*pdd.n)*exp(-r*T/n) + } + } > OPT0 [,1] [,2] [1,] 22.2741190 58.421275 [2,] 0.5305465 5.977683
The idea is then to move backward once more,
> OPT=OPT0 > OPT0<-matrix(0,k,k) > for(i in(1:k)) + { + for(j in(1:k)) + { + OPT0[i,j]=(OPT[i,j]*puu.n+OPT[i+1,j]*pdu.n+ + OPT[i,j+1]*pud.n+OPT[i+1,j+1]*pdd.n)*exp(-r*T/n) + } + } > OPT0 [,1] [1,] 16.44106
Here calculations are much (much) longer,
> price.spead(250) [1] 15.66496and again, it is possible to use standard approximations to compare that price with a more standard one,
> (sp=SpreadApproxOption(TypeFlag = + "c", S1 = 105, S2 = 100, X = 0, + Time = 1, r = .05, sigma1 = .4, + sigma2 = .3, rho = .5)@price) [1] 15.65077Well, playing with trees is nice, but it might not be optimal for complex products. Next time, we'll discuss other techniques...
et 
. Car sa première réponse à 
, ou que 
.
N'est ce pas ce qu'on
pense quand on apprend qu'un prix a augmenté de 20% avant de baisser de
20% le lendemain ? Car c'est la même erreur que l'on commet. Et mine de
rien, raisonner avec des rendements, c'est pénible.... (tous ceux qui
ont discuté taux d’intérêt avec un banquier le savent).
, et le lendemain un rendement de 
(ou ce que je voyais avec mon fils quand on notait que 
). En fait, on perd de l'argent... Damned, c'est pénible ça, non ? D'autant plus que si on répète cette alternance (entre une baisse de 10% puis une hausse de 10%) sur un an, au final, on a perdu 85% de valeur... Autrement dit, avec des rendements d'espérance nulle, et symétriques, la volatilité à elle seule fait perdre de l'argent...
. En effet
vaut 0.05, on parlera (abusivement) de hausse de 5%. Si on regarde une évolution sur deux jours,
. 
for some obscure reasons, simple things are usually supposed to be simple. Recently, on the internet, I saw a lot of posts on the "
Via 
Yesterday, I wrote a post (
But as @
Now both stocks are expressed in the same currency. To compare returns volatility, a first idea can be to use GARCH models,
Allez, un rapide billet sur l'or (moins statistique que celui écrit il y a 6 mois sur le prix de l'or, 
J'avoue que ça a
un peu cassé le mythe de plusieurs films américains, où on voit des
montagnes de lingots, car finalement un cube de 18 mètres de côté,
c'est relativement peu...
Pour la petite histoire, il semble qu'il y ait 46 millions de livret ouverts (selon wikipedia, 
(car du 1er janvier au 1er août, on compte 14 quinzaines, et du 1er août au 1er janvier suivant, on en compte 8 - je renvois au commentaire d'Olivier qui m'a fait noté que j'avais du mal avec les soustractions avec retenues, 
ce qui donne, numériquement,
symétrique,
où l'on peut gagner 20 euros si la Corée gagner (et rien
si elle perd) et un autre où l'on peut gagner 20 euros si la
Corée perd. Bref, soit je suis acheter sur la Corée, soit
je suis vendeur. Le prix de ces contrats indiquent la
probabilité qu'a la Corée de gagner. On parle aussi de
back or lay, deux évènements étant possibles.
telle que
est l'ensemble des gains possibles, et 
un équivalent certain.
(mutuellement exclusif comme on dit, formant une partition de 
),
un seul pouvant survenir (en l'occurrence l'équipe qui reçoit gagne, ou
l'équipe qui reçoit perd, ou il y a match nul). On considère un
bookmaker qui accepte
de payer 
si l'évènement i survient à un joueur
qui aura parié 1 euro. On pourra légitimement pense que 
. Notre souhait est de relier ces 
, si un
lien quelconque pouvait exister. Ce problème avait
été soulevé par Ramsey ou de Finetti.
le montant total placé sur
l'évènement 
. On pourra dire que l'on est à
l'équilibre si 
est constant. Autrement dit, le ratio des cotes doit être l'inverse des ratios de montants placés,
sera la part gardée
par le bookmaker. On peut penser qu'un bookmaker fixe 
, et alors
en mathématique
financière). On peut d'ailleurs noter que si on note 
le produit
. On définie alors la probabilité implicite des parieurs sous la forme
Depuis
qu'on peut facilement commander des livres d'occasion outre-atlantique,
j'avoue en abuser un peu... La dernière commande que j'ai faite est le "origins of value" de édité par William Goetzmann et Geert Rouwenhorst. Cet ouvrage, qui est une compilation d'articles d'histoire de la finance, contient un chapitre passionnant sur 
des suites récurrentes pour ceux qui se souviennent de
leurs cours de lycée, avec des suites de la forme
mathématiques arabes,
alors que la norme était plutôt l'arithmétique romaine. Pour ceux qui
ont essayé un jour dans leur vie de lire des nombres latins (même dans
Astérix), la logique est assez déroutante ! Par exemple 888 s'écrit
DCCCLXXXVIII... Essayez avec cette écriture de poser une addition ! Les
nombres arabes - eux - utilisaient le système décimale, ce qui a fait gagné
énormément de temps de calcul (on peut aussi remonter à l'arithmétique
indienne, mais on sort largement du cadre de mon billet). Fibonacci
évoque d'ailleurs constamment l'utilisation de la
Goetzmann reprend tout cela dans un article publié en 2003, "
Fibonacci and the Financial Revolution" en ligne 
Le président de la Société d'Economie Politique de conclure "
I wanted to upload here a small problem I started to work on....
unfortunatley, I could not find (yet) a proper answer. Any comments and
suggestions are welcomed. The problem is simple: 
and 
. From Fréchet-Hoeffding bounds, we know that
. Further, correlation matrices are necessarily positive-semidefinite matrices.
will be a function of 
.
and 
) are functions of 
,
and 
.
, respectively describing
global dependence (that can be related to 
from 
(but actually any
positive distribution should work) and 
)
with a Bernoulli 
distribution
from copula 
if 
and 
if 
and 
, and 
In
order to get a better understanding, I look at slices, for instance
when 
. On the graph on the right, to dark area are values
that can not be obtained since correlation have to be
positive-semidefinite. Dots are points that have been obtained in a
particular scenario. I have also included the convex hull if the points
obtained with all the scenarios (here 50,000 scenarios), since I assume
that the set of positive correlation is necessarily convex. Graphs
below have been obtained when 
and 
.
