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Have you ever tried to simulate a negative binomial random variable in a Spreadsheet?
If the answer to that is ‘nope – I’d just use Igloo/Metarisk/Remetrica’ then consider yourself lucky! Unfortunately not every actuary has access to a decent software package, and for those muddling through in Excel, this is not a particularly easy task. If on the other hand your answer is ‘nope – I’d use Python/R, welcome to the 21st century’. I’d say great, I like using those programs as well, but sometimes for reasons out of your control, things just have to be done in Excel. This is the situation I found myself in recently, and here is my attempt to solve it: Attempt 0 The first step I took in attempting to solve the problem was of course to Google it, then cross my fingers and hope that someone else has already solved it and this is just going to be a simple copy and paste. Unfortunately when I did search for VBA code to generate a negative binomial random variable, nothing comes up. In fact, nothing comes up when searching for code to simulate a Poisson random variable in VBA. Hopefully if you've found your way here, looking for this exact thing, then you're in luck, just scroll to the bottom and copy and paste my code. When I Googled it, there were a few solutions that almost solved the problem; there is a really useful Excel add-in called ‘Real statistics’ which I’ve used a few times: http://www.real-statistics.com/ It's a free excel add-in, and it does have functionality to simulate negative bimonials. If however you need someone to be able to re-run the Spreadsheet, they also will need to have it installed. In that case, you might as well use Python, and then hard code the frequency numbers. Also, I have had issues with it slowing Excel down considerably, so I decided not to use this in this case. I realised I’d have to come up with something myself, which ideally would meet the following criteria
How hard can that be? Attempt 1 I’d seen a trick before (from Pietro Parodi’s excellent book ‘Pricing in General Insurance’) that a negative binomial can be thought of as a Poisson distribution with a Gamma distribution as the conjugate prior. See the link below for more details: https://en.wikipedia.org/wiki/Conjugate_prior#Table_of_conjugate_distributions Since Excel has a built in Gamma inverse, we have simplified the problem to needing to write our own Poisson inverse. We can then easily generate negative binomials using a two step process:
Great, so we’ve reduced our problem to just being able to simulate a Poisson in VBA. Unfortunately there’s still no built in Poisson inverse in Excel (or at least the version I have), so we now need a VBA based method to generate this. There is another trick we can use for this - which is also taken from Pietro Parodi - the waiting time for a Poisson dist is an Exponential Dist. And the CDF of an Exponential dist is simple enough that we can just invert it and come up with a formula for generating an Exponential sample. We then set up a loop and add together exponential values, to arrive at Poisson sample. The code for this is give below:
Function Poisson_Inv(Lambda As Double)
s = 0
N = 0
Do While s < 1
u = Rnd()
s = s - (Application.WorksheetFunction.Ln(u) / Lambda)
k = k + 1
Loop
BH_Poisson_Inv = (k - 1)
End Function
The VBA code for our negative binomial is therefore:
Function NegBinOld2(b, r)
Dim Lambda As Double
Dim N As Long
u = Rnd()
Lambda = Application.WorksheetFunction.Gamma_Inv(u, r, b)
N = Poisson_Inv(Lambda)
NegBinOld2 = N
End Function
Does this do everything we want?
There are a couple of downside of though:
This leads us on to Attempt 2 Attempt 2 If we pass the VBA a random uniform sample, then whenever we hit refresh in the Spreadsheet the random sample will refresh, which will force the Negative Binomial to resample. Without this, sometimes the VBA will function will not reload. i.e. we can use the sample to force a refresh whenever we like. Adapting the code gives the following:
Function NegBinOld(b, r, Rnd1 As Double)
Dim Lambda As Double
Dim N As Long
u = Rnd1
Lambda = Application.WorksheetFunction.Gamma_Inv(u, r, b)
N = Poisson_Inv(Lambda)
NegBinOld = N
End Function
So this solves the refresh problem. What about the random seed problem? Even though we now always get the same lambda for a given rand – and personally I quite like to hardcode these in the Spreadsheet once I’m happy with the model, just to speed things up. We still use the VBA rand function to generate the Poisson, this means everytime we refresh, even when passing it the same rand, we will get a different answer and this answer will be non-replicable. This suggests we should somehow use the first random uniform sample to generate all the others in a deterministic (but still pseudo-random) way. Attempt 3 The way I implemented this was to the set the seed in VBA to be equal to the uniform random we are passing the function, and then using the VBA random number generator (which works deterministically for a given seed) after that. This gives the following code:
Function NegBin(b, r, Rnd1 As Double)
Rnd (-1)
Randomize (Rnd1)
Dim Lambda As Double
Dim N As Long
u = Rnd()
Lambda = Application.WorksheetFunction.Gamma_Inv(u, r, b)
N = Poisson_Inv(Lambda)
NegBin = N
End Function
So we seem to have everything we want – a free, quick, solution that can be bundled in a Spreadsheet, which allows other people to rerun without installing any software, and we’ve also eliminated the forced refresh issue. What more could we want? The only slight issue with the last version of the negative binomial is that our parameters are still specified in terms of ‘b’ and ‘r’. Now what exactly are ‘b’ and ‘r’ and how do we relate them to our sample data? I’m not quite sure.... The next trick is shamelessly taken from a conversation I had with Guy Carp’s chief Actuary about their implementation of severity distributions in MetaRisk. Attempt 4 Why can't we reparameterise the distribution using parameters that we find useful, instead of feeling bound by using the standard statistics textbook definition (or even more specifically the list given in the appendix to ‘Loss Models – from data to decisions’, which seems to be somewhat of an industry standard), why can't we redefine all the parameters from all common actuarial distributions using a systematic approach for parameters? Let's imagine a framework where no matter which specific severity distribution you are looking at, the first parameter contains information about the mean (even better if it is literally scaled to the mean in some way), the second contains information about the shape or volatility, the third contains information about the tail weight, and so on. This makes fitting distributions easier, it makes comparing the goodness of fit of different distributions easier, and it make sense checking our fit much easier. I took this idea, and tied this in neatly to a method of moments parameterisation, whereby the first value is simply the mean of the distribution, and the second is the variance over the mean. This gives us our final version:
Function NegBin(Mean, VarOMean, Rnd1 As Double)
Rnd (-1)
Randomize (Rnd1)
Dim Lambda As Double
Dim N As Long
b = VarOMean - 1
r = Mean / b
u = Rnd()
Lambda = Application.WorksheetFunction.Gamma_Inv(u, r, b)
N = Poisson_Inv(Lambda)
NegBin = N
End Function
Function Poisson_Inv(Lambda As Double)
s = 0
N = 0
Do While s < 1
u = Rnd()
s = s - (Application.WorksheetFunction.Ln(u) / Lambda)
k = k + 1
Loop
BH_Poisson_Inv = (k - 1)
End Function
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I was asked an interesting question a couple of weeks ago when talking through some modelling with a client.
We were modelling an airline account, and for various reasons we had decided to base our large loss modelling on a very basic top-down allocation method. We would take a view of the market losses at a few different return periods, and then using a scenario approach, would allocate losses to our client proportionately. Using this method, the frequency of losses is then scaled down by the % of major policies written, and the severity of losses is scaled down by the average line size. To give some concrete numbers (which I’ve made up as I probably shouldn’t go into exactly what the client’s numbers were), let's say the company was planning on taking a line on around 10% of the Major Airline Risks, and their average line was around 1%. We came up with a table of return periods for market level losses. The table looked something like following (the actual one was also different to the table below, but not miles off): 
Then applying the 10% hit factor if there is a loss, and the 1% line written, we get the following table of return periods for our client:
Hopefully all quite straightforward so far. As an aside, it is quite interesting to sometimes pare back all the assumptions to come up with something transparent and simple like the above. For airline risks, the largest single policy limit is around USD 2.5bn, so we are saying our worst case scenario is a single full limit loss, and that each year this has around a 1 in 50 chance of occurring. We can then directly translate that into an expected loss, in this case it equates to 50m (i.e. 2.5bn *0.02) of pure loss cost. If we don't think the market is paying this level of premium for this type of risk, then we better have a good reason for why we are writing the policy!
So all of this is interesting (I hope), but what was the original question the client asked me? We can see from the chart that for the market level the highest return period we have listed is 1 in 50. Clearly this does translate to a much longer return period at the client level, but in the meeting where I was asked the original question, we were just talking about the market level. The client was interested in what the 1 in 200 at the market level was and what was driving this in the modelling. The way I had structured the model was to use four separate risk sources, each with a Poisson frequency (lambda set to be equal to the relevant return period), and a fixed severity. So what this question translates to is, for small Lambdas $(<<1)$, what is the probability that $n=2$, $n=3$, etc.? And at what return period is the $n=2$ driving the $1$ in $200$? Let’s start with the definition of the Poisson distribution: Let $N \sim Poi(\lambda)$, then: $$P(N=n) = e^{-\lambda} \frac{ \lambda ^ n}{ n !} $$ We are interested in small $\lambda$ – note that for large $\lambda$ we can use a different approach and apply sterling’s approximation instead. Which if you are interested, I’ve written about here: www.lewiswalsh.net/blog/poisson-distribution-what-is-the-probability-the-distribution-is-equal-to-the-mean
For small lambda, the insight is to use a Taylor expansion of the $e^{-\lambda}$ term. The Taylor expansion of $e^{-\lambda}$ is:
$$ e^{-\lambda} = \sum_{i=0}^{\infty} \frac{\lambda^i}{ i!} = 1 - \lambda + \frac{\lambda^2}{2} + o(\lambda^2) $$
We can then examine the pdf of the Poisson distribution using this approximation: $$P(N=1) =\lambda e^{-\lambda} = \lambda ( 1 – \lambda + \frac{\lambda^2}{2} + o(\lambda^2) ) = \lambda - \lambda^2 +o(\lambda^2)$$
as in our example above, we have:
$$ P(N=1) ≈ \frac{1}{50} – {\frac{1}{50}}^2$$
This means that, for small lambda, the probability that $N$ is equal to $1$ is always slightly less than lambda. Now taking the case $N=2$: $$P(N=2) = \frac{\lambda^2}{2} e^{-\lambda} = \frac{\lambda^2}{2} (1 – \lambda +\frac{\lambda^2}{2} + o(\lambda^2)) = \frac{\lambda^2}{2} -\frac{\lambda^3}{2} +\frac{\lambda^4}{2} + o(\lambda^2) = \frac{\lambda^2}{2} + o(\lambda^2)$$
So once again, for $\lambda =\frac{ 1}{50}$ we have:
$$P(N=2) ≈ 1/50 ^ 2 /2 = P(N=1) * \lambda / 2$$
In this case, for our ‘1 in 50’ sized loss, we would expect to have two such losses in a year once every 5000 years! So this is definitely not driving our 1 in 200 result.
We can add some extra columns to our market level return periods as follows: 
So we see for the assumptions we made, around the 1 in 200 level our losses are still primarily being driven by the P(N=1) of the 2.5bn loss, but then in addition we will have some losses coming through corresponding to P(N=2) and P(N=3) of the 250m and 500m level, and also combinations of the other return periods.
So is this the answer I gave to the client in the meeting? …. Kinda, I waffled on a bit about this kind of thing, but then it was only after getting back to the office that I thought about trying to breakdown analytically which loss levels we can expect to kick in at various return periods. Of course all of the above is nice but there is an easier way to see the answer, since we’d already stochastically generated a YLT based on these assumptions, we could have just looked at our YLT, sorted by loss size and then gone to the 99.5 percentile and see what sort of losses make up that level. The above analysis would have been more complicated if we have also varied the loss size stochastically. You would normally do this for all but the most basic analysis. The reason we didn’t in this case was so as to keep the model as simple and transparent as possible. If we had varied the loss size stochastically then the 1 in 200 would have been made up of frequency picks of various return periods, combined with severity picks of various return periods. We would have had to arbitrarily fix one in order to say anything interesting about the other one, which would not have been as interesting. Every year we have an office sweepstake for the Cheltenham horse racing festival. Like most sweepstakes, this one attempts to remove the skill, allowing everyone to take part without necessarily needing to know much about horse racing. In case you’re not familiar with a Sweepstake, here’s a simple example of how one based on the World Cup might work:
Note that in order for this to work properly, you need to ensure that the team draw is not carried out until everyone who wants to play has put their money in – otherwise you introduce some optionality and people can then decide whether to enter based on the teams that are still left. i.e. if you know that Germany and Spain have already been picked then there is less value in entering the competition. The rules for our Cheltenham sweepstake were as follows: The Rules The festival comprises 7 races a day for 4 days, for a total of 28 races. The sweepstake costs £20 to enter, and the winnings are calculated as follows: 3rd place in competition = 10% of pot 2nd place in competition = 20% of pot 1st place in competition = 70% of pot Each participant picks one horse per race each morning. Points are then calculated using the following scoring system:
A running total is kept throughout the competition, and the winner is determined after the final race. The odds the scoring are based on are set using the odds printed in the Metro Newspaper on the morning of the races. (as an example, for a horse which has odds of 11/2 in the Metro - if the horse then places 1st, if we selected this horse, we would win (1+11/2)*5 = 32.5 points) Initial thoughts Any set of betting odds can be converted to an implied probability of winning, these would be the odds which over the long run would cause you to breakeven if the race were repeated multiple times with each horse winning a proportion equal to its probability of winning. Because the scoring in our sweepstake is based on the betting odds, using implied probabilities derived from the odds to help select our horses ends up cancelling itself out (which was the intention when designing the rules). The implied probability can be calculated as one over the odds As an aside, the bookie is indifferent to whether this is the correct probability of winning, they structures the odds purely on the ratio of how their customers are betting. They then structure the odds so that they make money on each race, regardless of which horse wins, for an explanation of this, see my post on creating a Dutchbook: www.lewiswalsh.net/blog/archives/12-2017 Here is an example showing how we would calculate the implied probabilities using some made up odds:  We can then use the implied probabilities we just calculated to see what would happen if each horse finished in the first three positions. Once we have done this, we can then calculate the Expected Value of betting on each horse:  We see that the payout varies for each horse, but the EV is the same. This is by design, the intention is that it should not matter which horse you bet on, the sweepstake rules equalise everyone. So what can we - an actuary who knows nothing much about horse racing - do? I don’t have any special knowledge that would allow me to select a horse which would beat the odds listed in the Metro, we appear to be at an impasse. I could attempt to build a model of which horse will win, and then select my horse based on that, but unless it proves to be more accurate than the Metro odds, I might as well just pick at random. Furthermore, if I could build such a model, then I could just start betting actual money. This probably shows you what a difficult problem this is! There's no such thing as free money. It would be a cool project to try, and it’s something I’ve been meaning to attempt for a while, but that’s best saved for another day. Attempt 1 - Metro vs pre-race odds My first thought was that we can exploit the difference in odds between those published in the Metro in the morning, and the latest odds published closer to the start of the race. It seems reasonable to assume that the odds just before the race should be more accurate than the metro odds. There will have been time for additional information to be included in the more up to date odds, e.g. the weather is worse than expected therefore horse x is expected to be slower than usual. Since the payout will be based on the Metro, we will then be able to maximise our EV by exploiting this differential. Our table will end up looking something like this:  We see that we have very small variations in the EVs for some of the horses. It looks like according to this analysis Horse 3 would be the best selection as it has the highest EVs for 1st, 2nd, and 3rd. Based on this strategy, we would then go through each race and select the horse with the highest EV. Is this what I did? No, for a couple of reasons. The biggest issue was that the Metro did not publish odds in the morning for all races, meaning we couldn’t use the Metro, and therefore the rules of the sweepstake were amended to use the official pre-race odds to calculate the payout instead. This meant there was only one set of odds used, and our edge disappeared! Even if we had used this method, there was a more fundamental issue - the margins we ended up with were tiny anyway. The Metro vs pre-race odds did not swing wildly, meaning that even selecting the horse with the highest EV were only marginally better than picking at random. So, was there an alternative? Attempt 2 - 2nd and 3rd place odds My next attempt at an exploitative strategy was based on the insight that the payout multiplier for 2nd and 3rd place was based on the odds of the horse coming 1st, rather than the odds of the horse coming 2nd or 3rd. The expected value of a horse was not quite as I calculated above, it was actually: $$EV = P(1)*p_1 + P(2)*p_2 + P(3)*p_3$$ Above, we were using the implied probability of the horse coming first as a proxy for the probability it would come second and third. This is not the same, and some betting websites do allow you to bet on whether a horse will come 2nd or 3rd. For websites that do not allow you to bet directly on this, then we may still be able to calculate it from the odds for whether a horse finishes in the top 2 or 3 places. We just need to subtract out the implied probability of coming 1st from the probability of coming in the top 2, and then subtracting this out from coming top 3 etc. I therefore added some more columns to my table above, corresponding to the probability of the horses coming 2nd and 3rd, and then used this to calculate the EV instead.  We see that the final column, Total EV, now has quite different values for each horse. In this case Horse 15, Seddon has an EV of 11.72. The favourite horse on the other hand - number 7 - only has an EV of 6.2. The intuitive explanation of this is that the probability of Seddon coming first is very low – this is the reflected in the long odds of 67, this then gives us a large multiplier, but the odds of the horse coming second or third are actually relatively less far out – the fact that it is not the favourite actually increases the odds of it coming in a position which is not 1st. But then if does come in 2nd or 3rd, we would still apply the same large multiplier for the odds of it coming 1st. This then gives us our 2nd edge – we can gain a positive EV by focusing. As a thought experiment, imagine we have a race with three horses – horse A is a clear favourite, horse B is an average horse, and horse C is clearly the weakest. By betting on horse C – the odds of it winning should be very low, so the multiple should be very high, but then this multiple will be applied even if it comes 2nd or 3rd, which is exactly where it is expected to finish. This therefore suggests our next potential strategy – select horses which maximise our EV using the implied probabilities of the horses coming 2nd, 3rd etc. So is this what I did? Well kind of.... The issue with this approach is that typically the horses that provide the best EV also have very long odds. In the race analysed above, our horse has an EV of 11.7, but it only has a 7% chance overall of coming in the top 3. In race two for example, the horse with the best EV actually only had a 2.7% chance of coming in the top 3. Since there are only 28 races in total, if each horse we selected only had a 2.7% chance of coming in, then the probability of us getting 0 points overall in the entire competition would then be: $(1-2.7 \%)^{28} = 48 \%$ So there is roughly a 50% chance we will get 0 points! Alternatively, if we selected the favourite every time, we could expect it to come top 3 almost every time, and thereby guarantee ourselves points most races, but it also has the lowest EV. So we have what appears to be a risk vs reward trade off. Pick outsiders and give ourselves the highest EV overall, or pick the favourites thereby reducing our overall EV but also reducing our volatility. This leads us neatly to attempt 3 - trying to think about how to maximise our probability of winning the competition rather than simply maximising EV for each race. Attempt 3 - EP curves From the work above, we now have our model of the position each horse will finish in each race – using the 1st, 2nd, and 3rd implied probabilities - and we have the payout for each horse – using the odds of the horse coming 1st. We can then bring our selection of horse and these probabilities together in a summary tab and simulate our daily score stochastically using a Monte Carlo method. To do this we just need to turn the implied probabilities into a CDF, and lookup the value of each position and repeat 10,000 times. The output for this then ends up looking like the following, where the value is the number of points we will for a given race.  So we see that across this sample of 20 simulations, most days we to end up with 0 points overall, but a few days have very high scores. So far so good! The next step is to set up an EP table of the following, which looks like the following:  The EP table gives us the probability of exceeding various scores in the competition based on our horse selections. In this case, we see that there is a 1 in 20 chance of getting 453 points or greater in the day. This is useful even on its own – when I was deciding which horses to bet on, I simply played around with the selections until I got an EP table I was comfortable with. My reasoning was quite basic – I decided I wanted to maximise the 1 in 20 value. I wanted to give myself something like a 1/4 chance of winning the whole competition and a 3/4 chance of getting very few points. Since there were four days of races, dividing the 1/4 by another 4 suggests we should be looking at maximising the 1 in 20 level (I admit this reasoning was a bit rough, but it seemed to serve its purpose) The insight here is that the payout structure of the sweepstake is such that coming in the top 3 is all that matters, and in particular coming 1st is disproportionately rewarded. To see this, we can think of the daily selection of horses as attempting to maximise the EV of the overall prize rather than the EV of our overall score - maximising the EV of each race is only a means to this end. So we are actually interested in maximising the following: $0.7 * P(1st) + 0.2 + P(2nd) + 0.1 * P(3rd)$ Which will largely be dominated by P(1st), given the $0.7$ factor. This is largely the strategy I went for in the end. Attempt 4 - Game theory? I’ve brushed over one difficulty above; in order to maximise our prize EV we need to consider not just which strategy we should take, but how this strategy will fare against the strategies that other people will take. If everyone is maximising their 1 in 20 return period then there’s little point us doing exactly the same. Luckily for me, most people were doing little more than picking horses randomly. We could then formalise this assumption, and come up with a numerical solution to the problem above. To do this, we would simulate our returns for each day across 10,000 simulations as above, but this time we would compare ourselves against a ‘base strategy’ of random selection of horses and we would simulate this base strategy for the approximately 40 people who entered. Each simulation would then give us a ranking we would finish in the competition, Here is an example of what that would look like:  And we could then convert this into an EP table which would look like the following:  So we see that if we select these horses, we end up having somewhere between a 1 in 10 and a 1 in 5 chance of winning the competition. Now that we have all of this set up, we can then optimise our horse selection to target a particular return period. I didn’t actually end up setting up the above for the sweepstake, but I suspect it would have been an improvement on my approach Attempt 5 - Multiple day? There is a further refinement we can make to the above. We have so far only really been focusing on maximising our chance of winning by using a fixed strategy throughout the competition. But there is no reason we have to do this. After the first day, we should really be including the current scores of each competitor as part of our calculation of our ranking. i.e. if person 1 had a great day and now has 200 points but we had a bad day and still have 0 points, by accounting for this, the model should automatically increase our volatility i.e. start picking horses with longer odds so as to increase the chance of us catching up. If on the other hand, we had a really good first day and are now in the lead, the model should then automatically reduce our volatility and start selecting the favourites more often to help safely maintain our lead. How did it work in practice? I ended up placing 2nd, and taking 20% of the prize pot which was great! I was behind for most of the competition but then pulled back on the last day when a 66/1 came in 1st place, and I picked up 330 points off a single race. This may look miraculous, but is exactly how the model is supposed to work. Does that have any applicability to gambling generally? Unfortunately not, basically all of the work above is based on exploiting the specific scoring system of the sweepstake. There's no real way of apply this to gambling generally. If you have ever generated Random Variables stochastically using a Gaussian Copula, you may have noticed that the correlation of the generated sample ends up being lower than the value of the Covariance matrix of the underlying multivariate Gaussian Distribution. For an explanation of why this happens you can check out a previous post of mine: www.lewiswalsh.net/blog/correlations-friedrich-gauss-and-copula. It would be nice if we could amend our method to compensate for this drop. As a quick fix, we can simply run the model a few times and fudge the Covariance input until we get the desired Correlation value. If the model runs quickly, this is quite easy to do, but as soon as the model starts to get bigger and slower, it quickly becomes impractical to run it three of four times just to get the output Correlation we desire. We can do better than this. The insight we rely on is that for a Gaussian Copula, the Pearson Correlation in the generated sample just depends on the Covariance Value. We can therefore create a precomputed table of Input and Output values, and use this to select the correct input value for the desired output. I wrote some R code to do just that, we compute a table of Pearson's Correlations obtained for various Input Covariance values when using the Gaussian Copula.
a <- library(MASS)
library(psych)
set.seed(100)
m <- 2
n <- 10^6
OutputCor <- 0
InputCor <- 0
for (i in 1:100) {
sigma <- matrix(c(1, i/100,
i/100, 1),
nrow=2)
z <- mvrnorm(n,mu=rep(0, m),Sigma=sigma,empirical=T)
u <- pnorm(z)
OutputCor[i] <- cor(u,method='pearson')[1,2]
InputCor[i] <- i/10
}
OutputCor
InputCor
Here is a sample from the table of results. You can see that the drop is relatively modest, but it does apply consistent across the whole table. 
Here is a graph showing the drop in values: 
Updated Algorithm
We can then use the pre-computed table, interpolating where necessary, to give us a Covariance value for our Multivariate Gaussian Distribution which will generate the desired Pearson Product Moment Correlation Value. So for example, if we would like to generate a sample with a Pearson Product Moment value of $0.5$, according to our table, we would need to use $0.517602$ as an input Covariance. We can test these values using the following code:
a <- library(MASS)
library(psych)
set.seed(100)
m <- 2
n <- 5000000
sigma <- matrix(c(1, 0.517602,
0.517602, 1),
nrow=2)
z <- mvrnorm(n,mu=rep(0, m),Sigma=sigma,empirical=T)
u <- pnorm(z)
cor(u,method='pearson')
Analytic Formulas I tried to find an analytic formula for the Product Moment values obtained in this manner, but I couldn't find anything online, and I also wasn't able to derive one myself. If we could find one, then instead of using the precompued table, we would be able to simply calculate the correct value. While searching, I did come across a number of interesting analytic formulas linking the values of Kendall's Tau, Spearman's Rank, and the input Covariance.. All the formulas below are from Fang, Fang, Kotz (2002) Link to paper: www.sciencedirect.com/science/article/pii/S0047259X01920172 The paper gives the following two results, where $\rho$ is the Pearson's Product Moment
$$\tau = \frac{2}{\pi} arcsin ( \rho ) $$ $$ {\rho}_s = \frac{6}{\pi} arcsin ( \frac{\rho}{2} ) $$
We can then use these formulas to extend our method above further to calculate an input Covariance to give any desired Kendall Tau, or Spearman's Rank. I initially thought that they would link the Pearson Product Moment value with Kendall or Spearman's measure, in which case we would still have to use the precomputed table. After testing it I realised that it is actually linking the Covariance to Kendall and Spearman's measures. Thinking about it, Kendall's Tau, and Spearman's Rank are both invariant to the reverse Guassian transformation when moving from $z$ to $u$ in the algorithm. Therefore the problem of deriving an analytic formula for them is much simpler as one only has to link their values for a multivariate Guassian Distribution. Pearson's however does change, therefore it is a completely different problem and may not even have a closed form solution. As an example of how to use the above formula, suppose we'd like our generated data to have a Kendall's Tau of $0.4$. First we need to invert the Kendall's Tau formula: $$ \rho = sin ( \frac{ \tau \pi }{2} ) $$ We then plug in $\rho = 0.4 $ giving:
$$ \rho = sin ( \frac{ o.4 \pi }{2} ) = 0.587785 $$
Giving usan input Covariance value of $0.587785$
We can then test this value with the following R code:
a <- library(MASS)
library(psych)
set.seed(100)
m <- 2
n <- 50000
sigma <- matrix(c(1, 0.587785,
0.587785, 1),
nrow=2)
z <- mvrnorm(n,mu=rep(0, m),Sigma=sigma,empirical=T)
u <- pnorm(z)
cor(z,method='kendall')
Which we see gives us the value of $\tau$ we want. In this case the difference between the input Covariance $0.587785$, and the value of Kendall's Tau $0.4$ is actually quite significant. It's the second week of your new job Capital Modelling job. After days spent sorting IT issues, getting lost coming back from the toilets, and perfecting your new commute to work (probability of getting a seat + probability of delay * average journey temperature.) your boss has finally given you your first real project to work on. You've been asked to carry out an annual update of the Underwriting Risk Capital Charge for a minor part of the company's Motor book. Not the grandest of analysis you'll admit, this particular class only makes up about 0.2% of the company's Gross Written Premium, and the Actuaries who reserve the company's bigger classes would probably consider the number of decimal places used in the annual report more material than your entire analysis. But you know in your heart of hearts that this is just another stepping stone on your inevitable meteoric rise to Chief Actuary in the Merger and Acquisition department, where one day you will pass judgement on billion dollar deals in-between expensive lunches with CFOs, and drinks with journalists on glamorous rooftop bars. The company uses in-house reserving software, but since you're not that familiar with it, and because you want to make a good impression, you decide to carry out extensive checking of the results in Excel. You fire up the Capital Modelling Software (which may or may not have a name that means a house made out of ice), put in your headphones and grind it out. Hours later you emerge triumphant, and you've really nailed it, your choice of correlation (0.4), and correlation method (Gaussian Copula) is perfect. As planned you run extracts of all the outputs, and go about checking them in Excel. But what's this? You set the correlation to be 0.4 in the software, but when you check the correlation yourself in Excel, it's only coming out at 0.384?! What's going on? Simulating using Copulas The above is basically what happened to me (minus most of the actual details. but I did set up some modelling with correlated random variables and then checked it myself in Excel and was surprised to find that the actual correlation in the generated output was always lower than the input.) I looked online but couldn't find anything explaining this phenomenon, so I did some investigating myself. So just to restate the problem, when using Monte Carlo simulation, and generating correlated random variables using the Copula method. When we actually check the correlation of the generated sample, it always has a lower correlation than the correlation we specified when setting up the modelling. My first thought for why this was happening was that were we not running enough simulations and that the correlations would eventually converge if we just jacked up the number of simulations. This is the kind of behaviour you see when using Monte Carlo simulation and not getting the mean or standard deviation expected from the sample. If you just churn through more simulations, your output will eventually converge. When creating Copulas using the Gaussian Method, this is not the case though, and we can test this. I generated the graph below in R to show the actual correlation we get when generating correlated random variables using the Copula method for a range of different numbers of simulations. There does seem to be some sort of loose limiting behaviour, as the number of simulations increases, but the limit appears to be around 0.384 rather than 0.4. 
The actual explanation First, we need to briefly review the algorithm for generating random variables with a given correlation using the normal copula. Step 1 - Simulate from a multivariate normal distribution with the given covariance matrix. Step 2 - Apply an inverse gaussian transformation to generate random variables with marginal uniform distribution, but which still maintain a dependency structure Step 3 - Apply the marginal distributions we want to the random variables generated in step 2 We can work through these three steps ourselves, and check at each step what the correlation is. The first step is to generate a sample from the multivariate normal. I'll use a correlation of 0.4 though out this example. Here is the R code to generate the sample:
a <- library(MASS)
library(psych)
set.seed(100)
m <- 2
n <- 1000
sigma <- matrix(c(1, 0.4,
0.4, 1),
nrow=2)
z <- mvrnorm(n,mu=rep(0, m),Sigma=sigma,empirical=T)
And here is a Scatterplot of the generated sample from the multivariate normal distribution: 
We now want to check the product moment correlation of our sample, which we can do using the following code: cor(z,method='pearson') Which gives us the following result: > cor(z,method='pearson') [,1] [,2] [1,] 1.0 0.4 [2,] 0.4 1.0 So we see that the correlation is 0.4 as expected. The Psych package has a useful function which produces a summary showing a Scatterplot, the two marginal distribution, and the correlation: 
Let us also check Kendall's Tau and Spearman's rank at this point. This will be instructive later on. We can do this using the following code: cor(z,method='spearman') cor(z,method='Kendall') Which gives us the following results: > cor(z,method='spearman') [,1] [,2] [1,] 1.0000000 0.3787886 [2,] 0.3787886 1.0000000 > cor(z,method='kendall') [,1] [,2] [1,] 1.0000000 0.2588952 [2,] 0.2588952 1.0000000 Note that this is less than 0.4 as well, but we will discuss this further later on.
We now need to apply step 2 of the algorithm, which is applying the inverse Gaussian transformation to our multivariate normal distribution. We can do this using the following code:
u <- pnorm(z) We now want to check the correlation again, which we can do using the following code: cor(z,method='spearman') Which gives the following result: > cor(z,method='spearman') [,1] [,2] [1,] 1.0000000 0.3787886 [2,] 0.3787886 1.0000000 Here is the Psych summary again: 
u is now marginally uniform (hence the name). We can see this by looking at the Scatterplot and marginal pdfs above. We also see that the correlation has dropped to 0.379, down from 0.4 at step 1. The Pearson correlation measures the linear correlation between two random variables. We generated normal random variables, which had the required correlation, but then we applied a non-linear (inverse Gaussian) transformation. This non-linear step is the source of the dropped correlation in our algorithm. We can also retest Kendall's Tau, and Spearman's at this point using the following code: cor(z,method='spearman') cor(z,method='Kendall') This gives us the following result: > cor(u,method='spearman') [,1] [,2] [1,] 1.0000000 0.3781471 [2,] 0.3781471 1.0000000 > cor(u,method='kendall') [,1] [,2] [1,] 1.0000000 0.2587187 [2,] 0.2587187 1.0000000 Interestingly, these values have not changed from above! i.e. we have preserved these measures of correlation between step 1 and step 2. It's only the Pearson correlation measure (which is a measure of linear correlation) which has not been preserved. Let's now apply the step 3, and once again retest our three correlations. The code to carry out step 3 is below: x1 <- qgamma(u[,1],shape=2,scale=1) x2 <- qbeta(u[,2],2,2) df <- cbind(x1,x2) pairs.panels(df) The summary for step 3 looks like the following. 
This is the end goal of our method. We see that our two marginal distributions have the required distribution, and we have a correlation between them of 0.37. Let's recheck our three measures of correlation. cor(df,method='pearson') cor(df,meth='spearman') cor(df,method='kendall') > cor(df,method='pearson') x1 x2 x1 1.0000000 0.3666192 x2 0.3666192 1.0000000 > cor(df,meth='spearman') x1 x2 x1 1.0000000 0.3781471 x2 0.3781471 1.0000000 > cor(df,method='kendall') x1 x2 x1 1.0000000 0.2587187 x2 0.2587187 1.0000000 So the Pearson has reduced again at this step, but the Spearman and Kendall's Tau are once again the same.
Does this matter?
This does matter, let's suppose you are carrying out capital modelling and using this method to correlate your risk sources. Then you would be underestimating the correlation between random variables, and therefore potentially underestimating the risk you are modelling. Is this just because we are using a Gaussian Copula? No, this is the case for all Copulas. Is there anything you can do about it? Yes, one solution is to just increase the input correlation by a small amount, until we get the output we want. A more elegant solution would be to build this scaling into the method. The amount of correlation lost at the second step is dependent just on the input value selected, so we could pre-compute a table of input and output correlations, and then based on the desired output, we would be able to look up the exact input value to use. Recently I've been reading The Mathematics of Poker (2009, Bill Chen and Jerrod Ankenman) and I came across an interesting idea that I thought I'd write about. For me, understanding how to analyse this situation really gets to the heart of how I think about poker. I'd love to spend more time playing and studying poker but it's such a time-sink and I don't really have the oodles of free time it would require, but every now and again I'll still open up a poker book and read about something, this is one of those interesting topics I was reading about hopefully you find it as interesting as I do. Calling a shove pre-flop in heads up The scenario being analysed in the book is a relatively common situation, particularly in online poker where people are more inclined to shove than in real life games. The question is: How should we analyse whether to call when we have a moderately strong hand against an opponent who has gone all-in pre flop. Let's set up an example so that we have something concrete to talk about. Say we have pocket Kings pre flop, and our opponent goes all-in, how should we decide whether to call? Obviously without knowing our opponent's hand there is no 100% correct answer. There is however one very useful way of analysing the situation. Equity against a range We need to ask ourselves - what cards would are opponent go all-in with, and how does my current hand fare against that range? i.e. we need to calculate our equity against our opponent's range. We are adding another layer of stochastic uncertainty on the event, instead of trying to guess what hand our opponent has (which is almost impossible) we are trying to guess what kind of hands he might go all-in with (which is very much possible). We then take this extra level of uncertainty and calculate the correct mathematical way to proceed. On the one extreme, let's suppose that based on our read of how our opponent is playing, we might think that they would only go all-in with very strong hands, in this case just pocket Aces. We would then be a 4:1 underdog if we call with Ks, and we should definitely fold. (In order to calculate this we could use any standard poker calculator like the following) www.cardplayer.com/poker-tools/odds-calculator/texas-holdem One the other hand, suppose we know for a fact that our opponent has not looked at their cards at all but has still decided to go all in. In this case we should definitely call. The only cards we will be behind are pocket Aces, which make up a small fraction of the possible hands that our opponent could shove with, and we will be ahead or equal against all other possible hands. Therefore we would have a positive EV when calling. What if our read on our opponent's range is somewhere in between though? What we need to do is calculate our equity against each individual hand in our opponent's range, and then calculate the probability of our opponent having a given hand from that range. That is to say, in order to calculate the conditional expectation, we need to calculate the unconditional expectations against each hand and then multiply by the conditional probability of our opponent having that hand, given our belief about our opponent's range. Numerical Example Let's go back to our numerical example, and suppose that we have pocket Kings, and we put our opponent on either Pocket Aces, Kings, Queens, or Jacks. All of these hands are equally likely, so there is a 25% chance of our opponent having each hand. We can look up our equity against each hand (after you've been playing for a while, you naturally start to memorise hand equities anyway) Our probability of winning is then: $$P(A) * P(beating A) + P(K)*P(beating K) + P(Q)*P(beating Q) + P(J) * P(beating J)$$ Putting in our values: $$ 0.25*0.2 + 0.25*0.5 + 0.25*0.8 + 0.25*0.8 = 0.575.$$ We therefore see we have a positive expectation against this range, and should call. No one actually thinks like this in real games? It is a general misconception that professional poker is a game where players are trying to guess exactly what hand their opponent has, are constantly trying to bluff each other, or trying to pick up on subtle tells or signs that their opponent is or isn't bluffing. The more mundane truth is that poker is ultimately a game of imperfect information, where the better player is the one who can correctly interpret the betting information their opponent is giving them, and can then quickly and accurately make the kind of judgements described above during a game. Obviously poker players are not carrying out these calculations in their head to multiple decimal places in real time, what they will do though is review their hands after a game, calculate exactly what they should have done, and try to build up an intuition as to what the correct answer is, so that in the middle of a game they can quickly make decisions. Software to Analyse this situation Is there an easy software based method way of calculating our equity against a range? After I did a quick google there are a few programs that offer this type of analysis. For example: combonator.com/ www.power-equilab.com/ More interestingly though, I also found the following open source software, that can be adapted to carry out this type of analysis: github.com/zekyll/OMPEval At some point, I might try to use this code to set up a page on this website to let people analyse this situation. |
AuthorI work as a pricing actuary at a reinsurer in London. Categories
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