## Revisiting barrier bond

In Analyse einer Barriere-Anleihe I was estimating the present value of a barrier bond emitted by Berliner Landesbank, which referred to the stock price of Daimler: The payout would be 100, if the price was either allways above the barrier of 70%, or, if it dipped below the barrier, but closed above the initial price at the maturity of the bond. In every other case, the buyer would get the value of the stock at maturity. For details on the bond, go here.

Using stochastic modelling on a 13 year history of the Daimler stock price, I estimated a present value at offering of 98.6% assuming a risk free rate of 2% for the 2 year maturity (I checked the sensitivity of the result to the risk free rate, but even at a -0.5% the mean present value is at 99.4%, so a slight loss).

Today I wanted to check what the actual payout would have been.

payoutEnd = function(basis, barriere, kurse)
{
basis = as.numeric(basis)
barriere = as.numeric(barriere)
kurse = as.numeric(kurse)
if(all(kurse >= barriere))
{
res = kurse[1]
} else if( last(kurse)>basis)
{
res = kurse[1]
} else
{
res = last(kurse)
}
return(res)
}

## [1] "DAI.DE"


Okay, so it seems the price never went below the barrier of 42.462, and the final payout was 100%:

payoutEnd(1,0.7, DAI.DE[,3]/DAI.DE[1,3])

## [1] 1


So, using hindsight, this would have been a good investment given the 7% coupon.

## So I finally understood Monty Hall

Lately I have been binge-watching Mythbusters, and one of the more curios myths they took on was the Monty Hall problem. The Monty Hall problem is named after a US TV show, were the candidate had the chance to win whatever price was behind one of three doors, where the other two doors had no price. The twist is that after the candidate choose, the moderator would show what was behind one of the other two doors, obviously one, where no price is, and the candidate now had the chance to switch the door.

Now, intuitively one would say that being shown what is behind a door will not change the chances, and the candidate has a 1 in 3 chance to win the price. Now the myth is, that switching the door will increase the chance to win substantially.

One might say, this is not really a myth, as it can be shown statistically to be true. But I am bad at combinatoric, so after seeing in Mythbusters how far ahead the switching strategy is, I wanted to redo their experiment as a Monte Carlo simulation.

First, we set up the experiment, and sample the winning doors, and the initial selection by the candidate.

# monty hall problem

n = 100000

prices = sample(3,n,1)

selected = sample(3,n,1)

df = data.frame(prices = prices,
selected = selected,
shown = NA,
wins_stay = NA,
wins_switch = NA)


##   prices selected shown wins_stay wins_switch
## 1      2        1    NA        NA          NA
## 2      3        2    NA        NA          NA
## 3      3        2    NA        NA          NA
## 4      3        2    NA        NA          NA
## 5      1        1    NA        NA          NA
## 6      2        3    NA        NA          NA


Next, we define how the moderator has to choose, which door to show in each case. And this is the first hint to why the likelihood to win is higher if the candidate switches: We need to differ between the cases were the candidate chose the winning door or not, because in the case of the candidate choosing a losing door, the door to be opened by the moderator is predetermined – it’s the one which is not winning.

shown = apply(df, 1, function(x){
x = unlist(x)

# x[1] - winning door, x[2] - choosen door
# candidate choose winning door

if(x[1]==x[2]){
return(sample((1:3)[-x[1]],1))
} else {
return((1:3)[-c(x[1], x[2])])
}
})
df$shown = shown head(df)  ## prices selected shown wins_stay wins_switch ## 1 2 1 3 NA NA ## 2 3 2 1 NA NA ## 3 3 2 1 NA NA ## 4 3 2 1 NA NA ## 5 1 1 2 NA NA ## 6 2 3 1 NA NA  Next, we calculate the winning likelihood, if the candidate always stays with the initial selection selected_stay = selected df$wins_stay = prices == selected_stay

##   prices selected shown wins_stay wins_switch
## 1      2        1     3     FALSE          NA
## 2      3        2     1     FALSE          NA
## 3      3        2     1     FALSE          NA
## 4      3        2     1     FALSE          NA
## 5      1        1     2      TRUE          NA
## 6      2        3     1     FALSE          NA

sum(df$wins_stay)/n  ## [1] 0.33196  It’s not very surprising that the percentage is 1 in 3, which is the initial likelihood without any additional information. Finally, we have to compute the door the candidate chooses if he switches. selected_switch = apply(df,1,function(x){ x = unlist(x) (1:3)[!(1:3)%in%c(x[2], x[3])] }) df$wins_switch = prices == selected_switch

##   prices selected shown wins_stay wins_switch
## 1      2        1     3     FALSE        TRUE
## 2      3        2     1     FALSE        TRUE
## 3      3        2     1     FALSE        TRUE
## 4      3        2     1     FALSE        TRUE
## 5      1        1     2      TRUE       FALSE
## 6      2        3     1     FALSE        TRUE

sum(df\$wins_switch)/n

## [1] 0.66804


Following the switching strategy, the candidates chances are 2 in 3, which counter-intuitively is quite logical: The candidate will loose in each case where his initial selection was correct (1 in 3), but will win in each case where his initial selection was wrong (2 in 3).

Oh, and here is a nice clip explaining it much better:

## What happens with my AAA-rated bond portfolio in the next 30 years?

One question a client asked was how rating migrations would effect their portfolio in the long term, and how to adjust the asset allocation to keep a stable average rating. As a small demo and proof of concept, I wrote a small shiny app. It allows you to adjust the portfolio weights based on the rating categories, the duration of the portfolio, and  the growth rate of the portfolio.

## Scratching that itch from ifelse

Okay, as I wrote yesterday, ifelse is rather slow, at least compared to working in C++. As my current project is using ifelse rather a lot, i decided to write a small utility function. In the expectation that I will collect a number of similar functions, I made a package out of it and posted it on github: https://github.com/ojessen/ojUtils

I get a speedup of about 30 times, independent of the target type.

Feedback and corrections greatly appreciated.

Thanks to the people at Travis for providing a free CI server which works directly with github. This of course is a tiny example, but it is good to know that the workflow to set this up can be done in 5 minutes.

And thanks to Romain Fraoncois for showing some Rcpp sugar:

Some data:

require(ojUtils)
## Loading required package: ojUtils
require(microbenchmark)
## Loading required package: microbenchmark
test = sample(c(T,F), size = 1e5, T)
yes = runif(1e5)
no = runif(1e5)

microbenchmark(ifelse(test, yes, no), ifelseC(test, yes, no))
## Loading required package: Rcpp
## Unit: microseconds
##                    expr   min      lq  median      uq    max neval
##   ifelse(test, yes, no) 31925 33404.8 34065.1 58083.5  71891   100
##  ifelseC(test, yes, no)   620   647.5   721.8   817.7 209254   100
test = sample(c(T,F), size = 1e5, T)
yes = rep("a", 1e5)
no = rep("b", 1e5)

microbenchmark(ifelse(test, yes, no), ifelseC(test, yes, no))
## Unit: milliseconds
##                    expr    min     lq median     uq   max neval
##   ifelse(test, yes, no) 57.313 58.763 59.626 72.435 87.92   100
##  ifelseC(test, yes, no)  1.747  1.837  1.926  2.749 29.56   100
test = sample(c(T,F), size = 1e5, T)
yes = rep(1L, 1e5)
no = rep(2L, 1e5)

microbenchmark(ifelse(test, yes, no), ifelseC(test, yes, no))
## Unit: microseconds
##                    expr     min      lq  median      uq   max neval
##   ifelse(test, yes, no) 30747.6 31868.5 32274.8 32829.0 59412   100
##  ifelseC(test, yes, no)   453.7   548.9   581.5   646.2 27575   100
test = sample(c(T,F), size = 1e5, T)
yes = rep(T, 1e5)
no = rep(F, 1e5)

microbenchmark(ifelse(test, yes, no), ifelseC(test, yes, no))
## Unit: microseconds
##                    expr     min      lq  median      uq   max neval
##   ifelse(test, yes, no) 29331.2 31167.3 31719.7 32455.3 60589   100
##  ifelseC(test, yes, no)   460.1   537.1   566.8   640.7 27118   100

## Comparing ifelse with C++ for loop with ifs

I currently am reading a bit about using Rcpp and its potential for speeding up R. I found one unexpected example in the lecture from Hadley Wickham:

require(Rcpp)
signR <- function(x) {
if (x > 0) {
1
} else if (x == 0) {
0
} else {
-1
}
}

cppFunction('int signC(int x) {
if (x > 0) {
return 1;
} else if (x == 0) {
return 0;
} else {
return -1;
}
}')

require(microbenchmark)
microbenchmark(signC(rnorm(1)), signR(rnorm(1)),times = 1e5)

## Unit: microseconds
##             expr   min    lq median   uq  max neval
##  signC(rnorm(1)) 2.832 3.186  3.540 3.54 4130 1e+05
##  signR(rnorm(1)) 2.478 3.186  3.186 3.54 2641 1e+05

As expected, the two versions perform nearly identical. Now for the surprise: I changed the scalar version of signC into a vectorized version:

library(Rcpp)

cppFunction('IntegerVector signCVec(NumericVector x){
int n = x.size();
IntegerVector out(n);
for(int i = 0; i < n; i++){
if(x[i] > 0 ){
out[i] = 1;
} else if(x[i] == 0){
out[i] = 0;
} else {
out[i] = -1;
}
}
return out;
}

')

signRVec <- function(x) {
ifelse(x > 0,1, ifelse(x == 0,0,-1))
}


Now I would have expected the two functions also to be rather similar in execution, but see for yourself:

x = rnorm(1e6)

microbenchmark(signCVec(x), signRVec(x),times = 10)

## Unit: milliseconds
##         expr    min      lq  median      uq     max neval
##  signCVec(x)   8.07   8.103   8.311   8.761   8.952    10
##  signRVec(x) 571.91 581.988 607.664 620.322 743.546    10

Wow: A 60-odd-times reduction using Rcpp.