I was thinking about this last week at work when I was coding part of a model involving the parameters of a truncated lognormal distribution. The lognormal distribution definitely feels like it was named the wrong way round.
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What is a Log-normal Distribution?

In other words, a Log-normal distribution is a distribution such that the log of the distribution is a normal distribution. It is not, as you might think, a distribution which is the log of the normal distribution.

So if $Y \sim N( \mu , {\sigma}^2 ) $ then $Log ( Y ) $ is not a lognormal distribution, instead $ e ^ Y $ is a lognormal distribution.

So to create a lognormal distribution, we don't take the log of the normal distribution, we take the exponential!
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Why does this matter?

The Reinstatement Premium payable following a loss to an Excess of Loss contract, is related to the ceded loss to the contract by a simple formula. Therefore, it seems reasonable that we should be able to come up with a simple formula relating the price charged for the Excess of Loss contract to the price charged for a Reinstatement Premium Protection (RPP) cover. I was in a meeting last week with two brokers who were trying to do just this. They had come up with an indicative price for an Excess of Loss programme and were trying to use this to price the equivalent RPP cover.

At the time I didn't have an answer for them, and when I did a quick Google, nothing came up. When thinking about it subsequently though, there are a couple of easy approximate methods we can use. Below I discuss three different methods for how you can price an RPP cover, two of which do not require any stochastic modelling assuming you already know the price of the Excess of Loss layers.

Let's quickly review what we mean by all these terms so that we are starting from the same point.

What is a Reinstatement Premium?

If you already understand how a Reinstatement Premium works, then feel free to skip this section.

Most Excess of Loss contracts will have some form of reinstatement premium. This is a payment from the Insurer to the Reinsurer to reinstate the protection in the event of a loss. In the London market, most contracts willl have either $1$, $2$, or $3$ reinstatements and generally these will be payable at $100 \%$.

What is a Reinstatement Premium Protection Cover?

The reinstatement premiums can be quite a large proportion of the overall premium paid for the Excess of Loss reinsurance. This is especially the case for lower level, working layers. Furthermore, the Reinstatement Premium will be payable when the insurer has just suffered a loss. From the point of view of the insurer, this additional payment comes at the worst possible time. The Insurer is being asked to fork over another large premium to the Reinsurer, just after having suffered a loss.

To address some of these concerns, Reinsurers developed a product called a Reinstatement Premium Protection cover (RPP cover). This cover pays the Insurer's Reinstatement Premium for them, which gives the insurer further indemnification in the event of a loss. Here's an example of how it works in practice:

Let's suppose we are considering a $5m$ xs $5m$ Excess of Loss contract, there is one reinstatement at $100 \%$ (written $1$ @ $100 \%$), and the Rate on Line is $25 \%$. The Rate on Line is just the Premium divided by the Limit. So here, the Premium can be found by multiplying the Limit and the RoL:

$$5m* 25 \% = 1.25m$$

So we see that the Insurer will have to pay the Reinsurer $1.25m$ at the start of the contract. Now let's suppose there is a loss of $7m$. The Insurer will recover $2m$ from the Resinsurer, but they will also have to make a payment  to cover the reinstatement premium of: $\frac {2m}  {5m} * (5m * 25 \% ) = 2m * 25 \% = 0.5m$ to reinstate the cover. So the Insurer will actually have to pay out $5.5m$. The RPP cover, if purchased by the insurer, would pay the additional $0.5m$ on behalf of the insurer

Now that we know how it works, how would we price the RPP cover?

Three methods for pricing an RPP cover

Method 1 - Full stochastic model

If we have priced the original Excess of Loss layer ourselves using a Monte Carlo model, then it should be relatively straight forward to price the RPP cover. We can just look at the expected Reinstatements, and apply a suitable loading for profit and expenses. This loading will probably be broadly in line with the loading that is applied to the expected losses to the Excess of Loss layer, but accounting for the fact that the writer of the RPP cover will not receive any form of Reinstatement for their Reinsurance.

What if we do not have a stochastic model set up to price the Excess of Loss layer? What if all we know is the price being charged for the Excess of Loss layer?

Method 2 - Simple formula

This was the situation I was in when I was asked to price the RPP cover last week. The broker had come up with some very approximate pricing for a programme of Excess of Loss layers. This pricing was driven by the burning cost, and other commercial factors rather than any actuarial modelling. The brokers wanted to come up with an approximate price for the RPP cover, just based on the price of the Excess of Loss layer. The two should be related, as they pay out dependant on the same underlying losses. So what can we say?

If we denote the Expected Losses to the layer by $EL$, then the Expected Reinstatement Premium should be:
$$EL * ROL $$
To see this is the case, I used the following reasoning; if we had losses in one year equal to the $EL$ (I'm talking about actual losses, not expected losses here), then the Reinstatement Premium for that year would be the proportion of the layer which had been exhausted $\frac {EL}  {Limit} $ multiplied by the Deposit Premium $Limit * ROL$ i.e.:

$$ RPP = \frac{EL}  {Limit} * Limit * ROL  = EL * ROL$$

Great! So we have our formula right? The issue now is that we don't know what the $EL$ is. We do however know the $ROL$, does this help?

If we let $DP$ denote the deposit premium, which is the amount we initially pay for the Excess of Loss layer and we assume that we are dealing with a working layer, then we can assume that:

Which is our first very simple formula for the price that should be charged for an RPP. Was there anything we missed out though in our analysis? 
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Method 3 - A more complicated formula:

There is one subtlety we glossed over in order to get our simple formula. The writer of the Excess of Loss layer will also receive the Reinstatement Premiums during the course of the contract. The writer of the RPP cover on the other hand, will not receive any reinstatement premiums (or anything equivalent to a reinstatement premium). Therefore, when comparing the Premium charged for an Excess of Loss layer against the Premium charged for the equivalent RPP layer, we should actually consider the total expected Premium for the Excess of Loss Layer rather than just the Deposit Premium.

What will the additional premium be? We already have a formula for the expected Reinstatement premium:

$$EL * ROL $$

Therefore the total expected premium for the Excess of Loss Layer is the Deposit Premium plus the additional Premium:

$$ DP + EL * ROL $$

This total expected premium is charged in exchange for an expected loss of $EL$.

So at this point we know the Total Expected Premium for the Excess of Loss contract, and we can relate the expected loss to the Excess of Loss layer to the Expected Loss to the RPP contract. 

i.e. For an expected loss to the RPP of $EL * ROL$, we would actually expect an equivalent premium for the RPP to be:

$$ RPP =  (DP + EL * ROL) * ROL $$

This formula is already loaded for Profit and Expenses, as it is based on the total premium charged for the Excess of Loss contract. It does however still contain the $EL$ as one of its terms which we do not know.

We have two choices at this point. We can either come up with an assumption for the profit and expense loading (which in this hard market might be as little as only be $5 \% - 10 \%$ ). And then replace $EL$ with a scaled down $DP$:

$$RPP =  \frac{DP} {1.075} * ( 1 + ROL) * ROL $$

Or we could simply replace the $EL$ with the $DP$, which is partially justified by the fact that the $EL$ is only used to multiply the $ROL$, and will therefore have a relatively small impact on the result. Giving us the following formula:

$$RPP =  DP ( 1 + ROL) * ROL $$

Which of the three methods is the best?

The full stochastic model is always going to be the best in my opinion. If we do not have access to one though, then out of the two formulas, the more complicated formula we derived should be more accurate (by which I mean more actuarially correct). If I was doing this in practice, I would probably calculate both, to generate some sort of range, but tend towards the second formula.

That being said, when I compared the prices that the Brokers had come up with, which is based on what they thought they could actually place in the market, against my formulas, I found that the simple version of the formula was actually closer to the Broker's estimate of how much these contacts could be placed for in the market. Since the simple formula always comes out with a lower price than the more complicated formula, this suggests that there is a tendency for RPPs to be under-priced in the market.

This systematic under-pricing may be driven by commercial considerations rather than faulty reasoning on the part of market participants. According to the Broker I was discussing these contracts with, a common reason for placing an RPP is to give a Reinsurer who does not currently have a line on the underlying Excess of Loss layer, but who would like to start writing it, a chance to have an involvement in the same risk, without diminishing the signed lines for the existing markets. So let's say that Reinsurer A writes $100 \%$ of the Excess of Loss contract, and Reinsurer B would like to take a line on the contract. The only way to give them a line on the Excess of Loss contract is to reduce the line that Reinsurer A has. The insurer may not wish to do this though if Reinsurer A is keen to maintain their line. So the Insurer may allow Reinsurer B to write the RPP cover instead, and leave Reinsurer A with $100 \%$ of the Excess of Loss contract. This commercial factor may be one of the reasons that traditionally writers of an RPP would be inclined to give favourable terms relative to the Excess of Loss layer so as to encourage the insurer to allow them on to the main programme and to encourage them to allow them to wrte the RPP cover at all.

Moral Hazard

​One point that is quite interesting to note about how these deals are structured is that RPP covers can have quite a significant moral hazard effect on the Insurer. The existence of Reinstatement Premiums is at least partially a mechanism to prevent moral hazard on the part of the Insurer. To see why this is the case, let's go back to our example of the $5m$ xs $5m$ layer. An insurer who purchases this layer is now exposed to the first $5m$ of any loss. But they are indemnified for the portion of the loss above $5m$, up to a limit of $5m$. If the insurer is presented with two risks which are seeking insurance - one with a total sum insured of $10m$, and another with a total sum insured of $6m$, the net retained exposure is the same for both risks from the point of view of the insurer. By including a reinstatement premium as part of the Excess of Loss layer, an therefore ensuring that the insurer has to make a payment any time a loss ceded to the layer, the reinsurer is ensuring that the insurer keeps their financial incentive to not have losses in this range. 

By purchasing an RPP cover, the insurer is removing their financial interest in losses which are ceded to the layer. There is an interesting conflict of interest in that the RPP cover will almost always be written by a different reinsurer to the Excess of Loss layer. The Reinsurer that is writing the RPP cover is therefore increasing the moral hazard risk whichever Reinsurer has written the Excess of Loss layer. Which will almost always be business written by one of the Reinsurer's competitors! 

Working Layers and unlimited Reinstatements

Another point to note is that this pricing analysis makes a couple of implicit assumptions. The first is that there is a sensible relationship between the expected loss to the layer and the premium charged for the layer. This will normally only be the case for 'working layers'. These are layers to which a reasonable amount of loss activity is expected. If we are dealing with clash or other higher layers, then the pricing of these layers will be more heavily driven by considerations beyond the expected loss to the layer. These might be capital considerations on the part of the Reinsurer, commercial considerations such as 

Another implicit assumption in this analysis is that the reinstatements offered are unlimited,. If this is not the case, then the statement that the expected reinstatement is $EL * ROL$ no longer holds. If we have limited reinstatements (which is the case in practice most of the time) then we would expect the expected reinstatement to be less than or equal to this.
​

Over the Christmas break, I finally got around to playing some Computer Games, which I unfortunately haven't done in ages. One really fun game (if you're like me and love difficult puzzles) is a game called Myst.

It was originally released in 1993 as a point and click puzzle adventure game, but it has subsequently been remade, using the Unity engine, as a full 3d interactive game. I really liked the fact that there is no time limit, no danger of dying, just a cool story, interesting puzzles and relaxing music.

One puzzle that I thought was cool involved aligning gear wheels. If you haven't played the game and don't want spoilers, then stop reading now.

The Gear Puzzle

​We are given the contraption in the image below, and are told that we need to end up with the numbers 2,2,1 (reading from top to bottom). We have three levers we can pull, the left lever moves the top two gears left once, the right lever moves the bottom two gears right once. The lever on the back wall resets the machine. The starting values of the gears are 3,3,3. So how many times do we need to pull each lever in order to end up at 2,2,1?

After playing around with it for a while I began to think that it might not even be possible to get to 2,2,1 using the two levers. Then I thought if that's true, can I prove it as well?

Proof that the puzzle is impossible

Instead of thinking of the value as $3$, let's call it $0$ instead. We can then think of the starting state as a vector:

To see that we are not able to do this with just the basic operations from the starting point we are given, note that the the sum of the first and third elements of our starting vector is equal to the second vector mod $3$, that is, $0 + 0 \equiv 0$ $(mod$ $3)$. For the vector we trying to create, the first and third is not congruent to the second element mod $3$. Instead we have $ 2 + 1 \equiv 0$ $(mod$ $3) $ rather than $2$.

The two operations we are applying to our starting vector preserve this congruence property, as it applies to both the vectors we are adding on as well. Therefore, we can conclude that this property is an invariant of the system, and because our goal does not satisfy this condition, we will never be able to reach:

by just using simple pulls of the lever. We need to realise this in order to force us to look for an alternative way to solve the puzzle. Otherwise we would probably just keep yanking on the levers in the hope that we will eventually get there. Once we realise this and start looking around, we are able to find another way to manipulate the machine, allowing us to make the numbers we need.

"The safest way to double you money is to fold it in half" Kin Hubbard

"I like to play Blackjack. I'm not addicted to gambling. I'm addicted to sitting in a semi-circle." Mitch Hedberg

Gambling is a risky business, in the long run everyone loses. Well not everyone. Bookmakers always seem to do alright. How is it that bookmakers make so much money out of gambling then? Is it through their superior wit and street smarts, or is there something else going on? It turns out that the method used by bookmakers involves a lot less insight and risk than you might think. The process is mathematical, and is guaranteed to turn a profit, and actually quite interesting.

Making a Dutch Book

Let's start with an example, suppose I am a bookmaker offering odds on the Super Bowl coin toss. We all know that for a fair coin, the probability of getting a heads is 50%, and the probability of getting tails is 50%.

We might suppose therefore that we would set our odds at 1/1, meaning for a bet of £1 if you win, you get £1 back plus your original £1. This means that you would double your money if you win (which happens about half the time) and lose all your money if you lose (which happens about half the time). Over time, if you made lots of bets of £1 at these odds, you would expect to break even.

Let's suppose though, that for reasons unknown to us, far more money is being bet on heads than tails. For example, suppose £1,000 is put on heads, and only £500 is put on tails. We can then examine our position under two scenarios, one where heads wins, and one where tails wins.

We have taken in £1,500 in total, and our outgoings are either £2,000 or £1,000. Making us a profit of £500 half the time, and -£500 half the time. Given each outcome is equally likely, over the long run if we keep offering odds like these we will expect to break even.

We can do better than this though. We can actually set the odds so that we break even every single time, not just in the long run. In order to see this, let's suppose that we change the odds so that they are 1/2 for heads, and 2/1 for tails. So that means, for every £2 bet on heads, we will pay out £1, plus the original £2 bet, and for every £1 bet on tails, we will pay out £2, plus the original £1 bet. Under these new odds, we would then end up with the following position:

We see now that because of how we have amended the odds, we don't actually care who wins. We will always pay out the same amount. The Bookmaker can then adjust the odds slightly, so that they will always pays out less than they have taken in. For example, they might offer 5/11 for heads, and 9/5 for tails. Under these odds, not only will the Bookmaker not care who wins, but they will always make a small margin on all bets placed. The exact figures in this case would be:

So we see that the Bookmaker has a foolproof system of always making money from gambling, and plus, they don't even need to be good at predicting who will win.

Irrational odds?

Isn't it crazy to offer odds of 2/1 on a coin toss though? We all know​ that the actual odds for a fair coin should be 1/1! The answer is that as soon as people see these odds, they should start betting on tails, and by supply and demand the odds for tails will start to move down, and the odds for heads will start to move up.

The bookmaker themselves might also decide to take a position at these odds and effectively bet money themselves by allowing the payout to be skewed towards a 50/50 split, giving their payout a positive Expected Value.

The first time I read about this system of bookmaking was in the book "Financial Calculus: an introduction to derivative pricing" by Baxter and Rennie, which I was reading in preparation for the IFoA ST6 exam. Baxter and Rennie brought it up because Bookmakers are actually undertaking a form of arbitrage, similar to the series of  notional trades used when deriving the price of futures contracts. You can see the similarities by simply noting that both the bookmaker and the derivatives trader are acting so as to not have any exposure to the actual value of the underlying event. By doing this, they don't actually need to take a view at all on what the expected outcome is, but they can instead exploit the relative values of different parts of the market.

What margin to online bookmakers charge?

I had a quick look at some odds offered by online bookmakers on a couple of events. One of the events I looked at was UFC 218. BestFightOdds.com gives a comparison of the odds offered by various betting websites. This is mainly to help people pick which website to bet on in order to get the best odds, but we can use it to compare the margins that the different websites charge. Assuming an equal bet is placed on each fight on the card, I calculated the following margins for the websites listed by BestFightOdds:

They range from a low of 2.62%, up to about 6%. The individual margin on each fight varies quite a bit, and also I suspect each website will run different margins on different sports and different events. The reason for this is that some of the higher margin websites might offer more incentives, and more free bets than the lower margin websites., plus they might be more established and spend more on advertising allowing them to charge more and still retain sufficient numbers of customers.

The fact that websites run margins does mean that if you are going to make money gambling you need to first beat the margin before you can even break even. Let's say you're really good at gambling and you can out-predict the market by 2%, you would still lose money overall gambling on any of these websites, because you also need to factor in the margin (also called the vig) that the websites charge.

Sources:

(1) BestfightOdds.com:
www.bestfightodds.com/events/ufc-218-holloway-vs-aldo-2-1368

Last week it was announced that UK Rail Fares were to increase once again at the maximum allowed rate - 3.4%, corresponding to the RPI increase in July 2017.

News article:
​www.bbc.co.uk/news/business-42234488

When reading this it got me thinking - why is RPI even being used any more? Aren't we supposed to be using CPI now?

In 2013 the ONS stated that:

"Following a consultation on options for improving the Retail Prices Index (RPI), the National Statistician, Jil Matheson, has concluded that the formula used to produce the RPI does not meet international standards and recommended that a new index be published."

So basically the ONS no longer endorses RPI as the best indicator of the level of inflation in the economy. The ONS instead supports the use of CPI. So why does it matter that some organisations are still using RPI?  To see why, let's take a look at a chart showing the historic RPI and CPI increases in the UK:

Source:
www.ons.gov.uk/economy/inflationandpriceindices

RPI has been greater than CPI in every single month since 2010. In fact, in this time period, RPI has been an average of 0.8% higher than CPI. This fact might go some way to explain why the Government is so slow to move rail increases from RPI to CPI. This way the Government and Rail Companies can claim that they are only increasing their costs in line with inflation, which seems fair, yet the index they are using is actually higher than the usual inflation index used in the UK.

The Government also indexes some of it's outgoings by an inflation index, for example the State Pension, so at least this is also being consistently overstated right?

Well actually no! Wherever the government is using an inflation index to increase payments, it seems to have already transitioned to a CPI increase. Let's look at the list of items which use the inflation index which is more beneficial to the Government: (remember that CPI is almost always lower than RPI):

There are some pretty hefty items on the list, including, the State Pension, Benefit Payments, Rail Fares,  Utility bills, Student Loans. The ones that decrease government spending seem to have already transitioned over, and the ones that increase the amount of tax collected, or the cost of regulated items all seem to still be using the higher RPI index.

Now let's look at the list of items which use the inflation index which is to the disadvantage of the Government.

I was attempting to set up a Loss Model in VBA at work yesterday. The model was a Compound-Poisson Frequency-Severity model, where the number of events is simulated from a Poisson distribution, and the Severity of events is simulated from a Severity curve.

There are a couple of issues you naturally come across when writing this kind of model in VBA. Firstly, the inbuilt array methods are pretty useless, in particular dynamically resizing an array is not easy, and therefore when initialising each array it's easier to come up with an upper bound on the size of the array at the beginning of the program and then not have to amend the array size later on. Secondly, Excel has quite a low memory limit compared to the total available memory. This is made worse by the fact that we are still using 32-bit Office on most of our computers (for compatibility reasons) which has even lower limits. This memory limit is the reason we've all seen the annoying 'Out of Memory' error, forcing you to close Excel completely and reopen it in order to run a macro.

The output of the VBA model was going to be a YLT (Yearly Loss Table), which could then easily be pasted into another model. Here is an example of a YLT with some made up numbers to give you an idea:

It is much quicker in VBA to create the entire YLT in VBA and then paste it to Excel at the end, rather than pasting one row at a time to Excel. Especially since we would normally run between 10,000 and 50,000 simulations when carrying out a Monte Carlo Simulation. We therefore need to create and store an array when running the program with enough rows for the total number of losses across all simulations, but we won't know how many losses  we will have until we actually simulate them.
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And this is where we come across our main problem. We need to come up with an upper bound for the size of this array due to the issues with dynamically resizing arrays, but since this is going to be a massive array, we want the upper bound to be as small as possible so as to reduce the chance of a memory overflow error.

Upper Bound

What we need then is an upper bound on the total number of losses across all the simulations years. Let us denote our Frequency Distribution by $N_i$, and the number of Simulations by $n$. We know that $N_i$ ~ $ Poi( \lambda ) \: \forall i$. 

Lets denote the total size of the YLT array by $T$. We know that $T$ is going to be:

$$T = \sum_{1}^{n} N_i$$
​We now use the result that the sum of two independent Poisson distributions is also a Poisson distribution with parameter equal to the sum of the two parameters. That is, if $X$ ~ $Poi( \lambda)$ , and $Y$ ~ $Poi( \mu)$, then $X + Y$ ~ $Poi( \lambda + \mu)$. By induction this result can then be extended to any finite sum of independent Poisson Distributions. Allowing us to rewrite $T$ as:

$$ T \sim Poi( n \lambda ) $$

$$ T \sim N ( n \lambda , n \lambda ) $$
Remember that $T$ is the distribution of the total number of losses in the YLT, and that we are interested in coming up with an upper bound for $T$.

Let's say we are willing to accept a probabilistic upper bound. If our upper bound works 1 in 1,000,000 times, then we are happy to base our program on it. If this were the case, even if we had a team of 20 people, running the program 10 times a day each, the probability of the program failing even once in an entire year is only 4%.

I then calculated the $Z$ values for a range of probabilities, where $Z$ is the unit Normal Distribution, in particular, I included the 1 in 1,000,000 Z value.

We then need to convert our requirement on $T$ to an equivalent requirement on $Z$.

$$ P ( T \leq x ) = p $$ 

If we now adjust $T$ so that it can be replaced with  a standard Normal Distribution, we get:

Now replacing the left hand side with $Z$ gives:

Hence, our upper bound is given by:

Dividing through by $n \lambda $ converts this to an upper bound on the factor above the mean of the distribution. Giving us the following:

We can see that given $n \lambda$ is expected to be very large and the $Z$ values relatively modest, this bound is actually very tight.

For example, if we assume that $n = 50,000$, and $\lambda = 3$, then we have the following bounds:

So we see that even at the 1 in 1,000,000 level, we only need to set the YLT array size to be 1.2% above the mean in order to not have any overflow errors on our array.


References
(1) Proof that the sum of two independent Poisson Distributions is another Poisson Distribution
math.stackexchange.com/questions/221078/poisson-distribution-of-sum-of-two-random-independent-variables-x-y

(2) Normal Approximation to the Poisson Distribution.
stats.stackexchange.com/questions/83283/normal-approximation-to-the-poisson-distribution