Probabilities, and opening doors (or boxes)

Recently, while we were in the car to Québec city with some PhD students, someone mentioned  the Monty Hall paradox, and we discussed possible extensions… The Monty Hall paradox is usually presented from tv show,

Craig F. Whitaker wrote the problem as follows, in Parade Magazine, September 1990, « Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice? ». Actually, Bertrand proposed the same problem, but with boxes instead of doors…. but the problem was the same.

Assume that the candidate chooses door 1 (without loss of generality since the problem is clearly symmetric).  The probability that the car is behind door 2 is http://freakonometrics.blog.free.fr/public/maths/montyh52.png. The animator can either open door 2 or door 3:

  • if the car is behind door 3, he has to open door 2,
  • if the car is behind door 2, he has to open door 3,
  • if the car is behind door 1, he can open door 2 or 3, and we assume that the opening is equiprobable,

 

Assume that the animator says “the second box is empty”, what should the candidate do ?
To formalize the problem let http://freakonometrics.blog.free.fr/public/maths/montyh1.png denote the event that the car is behind door http://freakonometrics.blog.free.fr/public/maths/montyh2.png, and http://freakonometrics.blog.free.fr/public/maths/montyh3.png the event that the animator opens door http://freakonometrics.blog.free.fr/public/maths/montyh4.png. So, if he opens door 2, the probability that the car is behind door 3 is
http://freakonometrics.blog.free.fr/public/maths/montyh5.png
where
http://freakonometrics.blog.free.fr/public/maths/montyh7.png
from the previous discussion, since he cannot open door 1 (the candidate chose it) and the cannot open door 3 (since the car is behind). Further
http://freakonometrics.blog.free.fr/public/maths/montyh8.png
from equiprobability. And for http://freakonometrics.blog.free.fr/public/maths/montyh9.png we get, similarly
http://freakonometrics.blog.free.fr/public/maths/montyh10.png
Thus,
http://freakonometrics.blog.free.fr/public/maths/montyh11.png
while
http://freakonometrics.blog.free.fr/public/maths/montyh12.png
So the optimal strategy is to open the third door (even if I chose the first one)… It is usually seen as a paradox, but if you consider a much larger number of doors (say 4),

and that the animator opens 2 doors, then should we still change, and open the door that is still closed ? The higher the number of doors, the higher the probability to have something behind the other door…

For instance, with 4 doors, or boxes, if the candidate still chose the first door, and that the animator opens doors 2 and 3, then, the probability that the car is behind the fourth one is
http://freakonometrics.blog.free.fr/public/maths/montyh15.png
i.e.
http://freakonometrics.blog.free.fr/public/maths/montyh50.png
http://freakonometrics.blog.free.fr/public/maths/montyh52.png that appears at the denominator since we focus on the pair http://freakonometrics.blog.free.fr/public/maths/montyh54.png) while
http://freakonometrics.blog.free.fr/public/maths/montyh55.png
Once again, we have that
http://freakonometrics.blog.free.fr/public/maths/monthyh56.png
i.e. the opening of a doors bring us no information about our choice, but it will after conditional probability for the remaining door.
More generally, with n doors, if the animator opens http://freakonometrics.blog.free.fr/public/maths/monthy20.png doors,

then
http://freakonometrics.blog.free.fr/public/maths/montyh60.png
while
http://freakonometrics.blog.free.fr/public/maths/montyh62.png
And here the result is even more intuitive: we have to open the door that was left closed. Actually, it is possible to see that it can be extend to the case where the are http://freakonometrics.blog.free.fr/public/maths/montyh75.png doors (or boxes), the candidate chooses http://freakonometrics.blog.free.fr/public/maths/montyh73.png doors, and the animator opens http://freakonometrics.blog.free.fr/public/maths/montyh71.png (out of the http://freakonometrics.blog.free.fr/public/maths/montyh95.png remaining doors). Then, behind each door chosen by the candidate, the probability does not change
http://freakonometrics.blog.free.fr/public/maths/montyh70.png
where i goes from 1 to http://freakonometrics.blog.free.fr/public/maths/montyh73.png, while
http://freakonometrics.blog.free.fr/public/maths/montyh80.png
where i goes from http://freakonometrics.blog.free.fr/public/maths/montyh76.png to http://freakonometrics.blog.free.fr/public/maths/montyh75.png.

Is it that stupid to make extremely long term forecast when studying mortality ?

I received recently a comment by FCA (here) who raised an important question, about forecast in dynamic mortality models. (S)he mentioned that from his(her) point of view, the econometric models I considered were “good to predict for the next, say, 3 or 4 years. Not for the next 50 years…”. Which was the message I tried to stress last year in a conference about retirement in France (here). But from a quantitativepoint of view, how inconsistent were forecasts made 35 years ago, or 60 years ago ?

Consider here the Lee Carter model, obtained on the periods 1816-1950 (in black below), 1816-1975 (in red) and 1816-2000 (in blue), unfortunately, it is difficult to compare http://freakonometrics.blog.free.fr/public/maths/viekt.png‘s since we have identifiability problems here. Nevertheless, we if consider affine transformation so that  http://freakonometrics.blog.free.fr/public/maths/viekt.png‘s are equal in 1900 and 1950 (say), we obtain

On that graph, we considered an ETS (AAN) forecast. If we do not consider the entire series for forecasting, but only observations following WWI (1945), we obtain

For sketches of the R code,

T=1980
base0=data.frame(D,E,A,Y,a=as.factor(A),
y=as.factor(Y))
base=base0[base0$Y<=T,]
LC2=gnm(D~a+Mult(a,y),offset=log(E),family=
poisson,data=base)
A=LC2$coefficients[1]+LC2$coefficients[2:110]
B=LC2$coefficients[111:220]
K0=LC2$coefficients[221:length(LC2$coefficients)]
Y=as.numeric(K0)
K1=c(K0,forecast(ets(Y,model="AAN"),h=240)$mean)
K2=c(K0,forecast(auto.arima(Y,allowdrift=TRUE),h=240)$mean)
MU=matrix(NA,length(A),length(K1))
MU1=MU2=MU
for(i in 1:length(A)){
for(j in 1:length(K1)){
MU1[i,j]=exp(A[i]+B[i]*K1[j])
MU2[i,j]=exp(A[i]+B[i]*K2[j])
}}
x=40
s=seq(0,109-x-1)
t=2000
Pxt1=cumprod(exp(-diag(MU1[x+1+s,t+s-base1$Year[1]-1])))
Pxt2=cumprod(exp(-diag(MU2[x+1+s,t+s-base1$Year[1]-1])))
r=.035
m=70
h=seq(0,39)
V1=1/(1+r)^(m-x+h)*Pxt1[m-x+h]
V2=1/(1+r)^(m-x+h)*Pxt2[m-x+h]
M=cbind(V1,V2)
apply(M,2,sum)

Actually, it is not that bad…. even if it is only a qualitative intuition. Again, I am not a demographer, and my interest is more on actuarial science… so if we look at the estimation of annuities (still the same insurance contract, as here) for some insured of age 40 in 2000, we get the following graph (where forecasts http://freakonometrics.blog.free.fr/public/maths/viekt.png‘s were obtained on the complete series, i.e. from 1816 until the year we consider),

(here it means that in 1900, I had to forecast mortality for someone of age 40 in 2000… so we had to forecast mortality with a 150 year horizon). Obviously, even if we are able to forecast improvement of mortality rates, it is not enough since it looks like, each year, improvement are alway higher than what what expected. Note that if we run it twice (since there might be problem with initial values in the econometric procedure) we obtain something similar,

So, the output is consistent. And if we change the way we predict future values, e.g. on focusing only on the past 50 years, i.e.

K1=c(K0,forecast(ets(Y[(length(Y)-50):length(Y)],
model="AAN"),h=240)$mean)
K2=c(K0,forecast(auto.arima(Y[(length(Y)-50):length(Y)],
allowdrift=TRUE),h=240)$mean)

we obtain the following graph for the annuity associated to an insurance contract sold in 2000,

so that relative changes compared with 1980 are (in %)

Hence, over a bit more than 25 years, we underestimated annuities of 25%. We if start to take into account possible investments, it is not so bad, I think….  don’t you think ?