Category: Actuarial Modelling

Converting a Return Period to a RoL

15/3/2019

I came across a useful way of looking at Rate on Lines last week, I was talking to a broker about what return periods to use in a model for various levels of airline market loss (USD250m, USD500m, etc.). The model was intended to be just a very high level, transparent market level model which we could use as a framework to discuss with an underwriter. We were talking through the reasonableness of the assumptions when the broker came out with the following:

'Well, you’d pay about 12.5 on line in the retro market at that attachment level, so that’s a 1 in 7 break-even right?'

My response was:

'ummmm, come again?'

His reasoning was as follows:

Suppose the ILW pays $1$ @ $100$% reinstatements, and that it costs $12.5$% on line.
Then if the layer suffers a loss, the insured will have a net position on the contract of $75$%. This is the $100$% limit which they receive due to the loss, minus the original $12.5$% Premium, minus an additional $12.5$% reinstatement Premium. The reinsurer will now need another $6$ years at $12.5$% RoL $(0.0125 * 6 = 0.75)$ to recover the limit and be at break-even.

Here is a breakdown of the cashflow over the seven years for a $10m$ stretch at $12.5$% RoL:

So the loss year plus the six clean years, tells us that if a loss occurs once every 7 years, then the contract is at break-even for this level of RoL.

So this is kind of cool - any time we have a RoL for a retro layer, we can immediately convert it to a Return Period for a loss which would trigger the layer.

Generalisation 1 – various rates on line

We can then generalise this reasoning to apply to a layer with an arbitrary RoL. Using the same reasoning as above, the break-even return period ends up being:

$RP= 1 + \frac{(1-2*RoL)}{RoL}$

Inverting this gives:

$RoL = \frac{1}{(1 + RP)}$

So let's say we have an ILW costing $7.5$% on line, the break-even return period is:

$1 + \frac{(1-0.15)}{0.075} = 11.3$

Or let’s suppose we have a $1$ in $19$ return period, the RoL will be:

$0.05 = \frac{1}{(1 + 19)}$

Generalisation 2 – other non-proportional layers

The formula we derived above was originally intended to apply to ILWs, but it also holds any time we think the loss to the layer, if it occurs, will be a total loss. This might be the case for a cat layer, or a clash layers (layers which have an attachment above the underwriting limit for a single risk), or any layer with a relatively high attachment point compared to the underwriting limit.

Adjustments to the formulas

There are a few of adjustments we might need to make to these formulas before using them in practice.

Firstly, the RoL above has no allowance for profit or expense loading, we can account for this by converting the market RoL to a technical RoL, this is done by simply dividing the RoL by $120-130$% (or any other appropriate profit/expense loading). This has the effect of increasing the number of years before the loss is expected to occur.

Alternately, if layer does not have a paid reinstatement, or has a different factor than $100$%, then we would need to amend the multiple we are multiplying the RoL by in the formula above. For example, with nil paid reinstatements, the formula would be:

$RP = 1 + \frac{(1-RoL)}{RoL}$

Another refinement we might wish to make would be to weaken the total loss assumption. We would then need to reduce the RoL by an appropriate amount to account for the possibility of partial losses. It’s going to be quite hard to say how much this should be adjusted for – the lower the layer the more it would need to be.

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Extending the Copula Method

26/8/2018

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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.

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Correlations, Friedrich Gauss, and Copulas

29/7/2018

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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.

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Correlation Matrix - Positive Semi-Definite Requirement

22/6/2018

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If you have played around with Correlating Random Variables using a Correlation Matrix in [insert your favourite financial modelling software] then you may have noticed the requirement that the Correlation Matrix be positive semi-definite.

But what exactly does this mean? And how would we check this ourselves in VBA or R?

Mathematical Definition

Let's start with the Mathematical definition. To be honest, it didn't really help me much in understanding what's going on, but it's still useful to know.

A symmetric $n$ x $n$ matrix $M$ is said to be positive semidefinite if the scalar $z^T M z $ is positive for every non-zero column vector $z$ of $n$ real numbers.

If I am remembering my first year Linear Algebra course correctly, then Matrices can be thought of as transformations on Vector Spaces. Here the Vector Space would be a collection of Random Variables. I'm sure there's some clever way in which this gives us some kind of non-degenerate behaviour. After a bit of research online I couldn't really find much.

Intuitive Definition

The intuitive explanation is much easier to understand. The requirement comes down to the need for internal consistency between the correlations of the Random Variables.

For example, suppose we have three Random Variables, A, B, C.

Let's suppose that A and B are highly correlated, that is to say, when A is a high value, B is also likely to be a high value. Let's also suppose that A and C are highly correlated, so that if A is a high value, then C is also likely to be a high value. We have now implicitly defined a constraint on the correlation between B and C. If A is high both B and C are also high, so it can't be the case that B and C are negatively correlated, i.e. that when B is high, C is low.

Therefore some correlation matrices will give relations which are impossible to model.

Alternative characterisations

You can find a number of necessary and sufficient conditions for a matrix to be positive definite, I've included some of them below. I used number 2 in the VBA code for a real model I set up to check for positive definiteness.

1. All Eigenvalues are positive.

If you have studied some Linear Algebra, then you may not be surprised to learn that there is a characterization using Eigenvalues. It seems like just about anything to do with Matrices can be restated in terms of Eigenvalues. I'm not really sure how to interpret this condition though in an intuitive way.

2. All leading principal minors are all positive

This is the method I used to code the VBA algorithm below. The principal minors are just another name for the determinant of the upper-left $k$ by $k$ sub-matrix. Since VBA has a built in method for returning the determinant of a matrix this was quite an easy method to code.

3. It has a unique Cholesky decomposition

I don't really understand this one properly, however I remember reading that Cholesky decomposition is used in the Copula Method when sampling Random Variables, therefore I suspect that this characterisation may be important!

Since I couldn't really write much about Cholesky decomposition here is a picture of Cholesky instead, looking quite dapper.

All 2x2 matrices are positive semi-definite

Since we are dealing with Correlation Matrices, rather than arbitrary Matrices, we can actually show a-priori that all 2 x 2 Matrices are positive semi-definite.

Proof

Let M be a $2$ x $2$ correlation matrix.

$$M = \begin{bmatrix} 1&a\\ a&1 \end{bmatrix}$$

And let $z$ be the column vector $M = \begin{bmatrix} z_1\\ z_2 \end{bmatrix}$

Then we can calculate $z^T M z$

$$z^T M z = {\begin{bmatrix} z_1\\ z_2 \end{bmatrix}}^T \begin{bmatrix} 1&a\\ a&1 \end{bmatrix} \begin{bmatrix} z_1\\ z_2 \end{bmatrix} $$

Multiplying this out gives us:

$$ = {\begin{bmatrix} z_1\\ z_2 \end{bmatrix}}^T \begin{bmatrix} z_1 & a z_2 \\ a z_1 & z_2 \end{bmatrix} = z_1 (z_1 + a z_2) + z_2 (a z_1 + z_2)$$

We can then simplify this to get:

$$ = {z_1}^2 + a z_1 z_2 + a z_1 z_2 + {z_2}^2 = (z_1 + a z_2)^2 \geq 0$$

Which gives us the required result. This result is consistent with our intuitive explanation above, we need our Correlation Matrix to be positive semidefinite so that the correlations between any three random variables are internally consistent. Obviously, if we only have two random variables, then this is trivially true, so we can define any correlation between two random variables that we like.

Not all 3x3 matrices are positive semi-definite

The 3x3 case, is simple enough that we can derive explicit conditions. We do this using the second characterisation, that all principal minors must be greater than or equal to 0.

Demonstration

Let M be a $3$ x $3$ correlation matrix:

$$M = \begin{bmatrix} 1&a&b\\ a&1&c \\ b&c&1 \end{bmatrix}$$

We first check the determinant of the $2$ x $2$ sub matrix. We need that:

$ \begin{vmatrix} 1 & a \\ a & 1 \end{vmatrix} \geq 0 $

By definition:

$ \begin{vmatrix} 1 & a \\ a & 1 \end{vmatrix} = 1 - a^2$

We have that $ | a | \leq 1 $, hence $ | a^2 | \leq 1 $, and therefore:

$ | 1- a^2 | \geq 0 $

Therefore the determinant of the $2$ x $2$ principal sub-matrix is always positive.

Now to check the full $3$ x $3$. We require:

$ \begin{vmatrix} 1 & a & b \\ a & 1 & c \\ b & c & 1 \end{vmatrix} \geq 0 $

By definition:

$ \begin{vmatrix} 1 & a & b \\ a & 1 & c \\ b & c & 1 \end{vmatrix} = 1 ( 1 - c^2) - a (a - bc) + b(ac - b) = 1 + 2abc - a^2 - b^2 - c^2 $

Therefore in order for a $3$ x $3$ matrix to be positive demi-definite we require:

$a^2 + b^2 + c^2 - 2abc = 1 $

I created a 3d plot in R of this condition over the range [0,1].

It's a little hard to see, but the way to read this graph is that the YZ Correlation can take any value below the surface. So for example, when the XY Corr is 1, and the XZ Corr is 0, the YZ Corr has to be 0. When the XY Corr is 0 on the other hand, and XZ Corr is also 0, then the YZ Corr can be any value between 0 and 1.

Checking that a Matrix is positive semi-definite using VBA

When I needed to code a check for positive-definiteness in VBA I couldn't find anything online, so I had to write my own code. It makes use of the excel determinant function, and the second characterization mentioned above. Note that we only need to start with the 3x3 sub matrix as we know from above that all 1x1 and all 2x2 determinants are positive.

This is not a very efficient algorithm, but it works and it's quite easy to follow.



Function CheckCorrMatrixPositiveDefinite()

Dim vMatrixRange As Variant
Dim vSubMatrix As Variant
Dim iSubMatrixSize As Integer
Dim iRow As Integer
Dim iCol As Integer
Dim bIsPositiveDefinite As Boolean

bIsPositiveDefinite = True


vMatrixRange = Range(Range("StartCorr"), Range("StartCorr").Offset(inumberofrisksources - 1, inumberofrisksources - 1))

' Only need to check matrices greater than size 2 as determinant always greater than 0 when less than or equal to size 2'

If iNumberOfRiskSources > 2 Then
    For iSubMatrixSize = iNumberOfRiskSources To 3 Step -1
        ReDim vSubMatrix(iSubMatrixSize - 1, iSubMatrixSize - 1)
        For iRow = 1 To iSubMatrixSize
            For iCol = 1 To iSubMatrixSize
                vSubMatrix(iRow - 1, iCol - 1) = vMatrixRange(iRow, iCol)
            Next
        Next
        'If the determinant of the matrix is 0, then the matrix is semi-positive definite'
        If Application.WorksheetFunction.MDeterm(vSubMatrix) < 0 Then
            CheckCorrMatrixisPositiveDefinite = False
            bIsPositiveDefinite = False
        End If
    Next
End If

If bIsPositiveDefinite = True Then
    CheckCorrMatrixPositiveDefinite = True
Else
    CheckCorrMatrixPositiveDefinite = False
End If

End Function

Checking that a Matrix is positive semi-definite in R

Let's suppose that instead of VBA you were using an actually user friendly language like R. What does the code look like then to check that a matrix is positive semi-definite? All we need to do is install a package called 'Matrixcalc', and then we can use the following code:


is.positive.definite( Matrix )

That's right, we needed to code up our own algorithm in VBA, whereas with R we can do the whole thing in one line using a built in function! It goes to show that the choice of language can massively effect how easy a task is.

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Was the Lognormal Distribution misnamed?

20/2/2018

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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.

What is a Log-normal Distribution?

We say that a Random Variable $X$ has a Log-Normal Distribution that is:

$$ X \sim LogN( \mu , { \sigma }^2 ) $$

if: $$ Log (X) \sim N( \mu , { \sigma }^2 ) $$

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!

Why does this matter?

Definitions are just definitions after all, and as long as everyone knows how something is defined and there is no ambiguity one definition is usually as good as another. In this case though, defining it in this way does have some ugly and unnatural consequences. For example, if we take the result that the sum of two independent normal distributions is also a normal distribution, i.e.

If: $$ X \sim N( {\mu}_1 , {{\sigma}_1}^2 ) , Y \sim N( {\mu}_2 , {{\sigma}_2}^2 ) $$ Then: $$ X + Y \sim N( {\mu}_1 + {\mu}_2 , {{\sigma}_1}^2 + {{\sigma}_2}^2 ) $$

Then applying this result to the lognormal distribution, we get:

If $ X \sim LogN( {\mu}_1 , {{\sigma}_1}^2 ) $ and $ Y \sim LogN( {\mu}_2 , {{\sigma}_2}^2 ) $ assuming independence,

Then:

$$ XY \sim LogN( {\mu}_1 + {\mu}_2 , {{\sigma}_1}^2 + {{\sigma}_2}^2 ) $$

Maybe this doesn't look too bad to you. But what if I replace $X$ and $Y$ with ${LogN}_1$ and ${LogN}_2$? Then we get:

$$ {LogN}_1 * {LogN}_2 \sim LogN( {\mu}_1 + {\mu}_2 , {{\sigma}_1}^2 + {{\sigma}_2}^2 ) $$

This should definitely look wrong to you! Remember that for a standard logarithm:

$$ Log (AB) = Log(A) + Log(B) $$

Instead we have an identity that looks much more like an exponential:

$$ e^A * e^B = e^{ (A + B ) } $$

And that's precisely because we are dealing with an exponential! The lognormal distribution is simply the exponential of the normal, which is a much more natural way of phrasing it than to say that the lognormal distribution is a distribution such that the logarithm of the distribution is a normal distribution. So we have two reasons why the Lognormal Distbribution should have been called the Exponential Normal Distribution (Or possibly the X-Normal Distribution for short). The identity above makes perfect sense when using exponentials, and we would have a naming convention that is much more natural.

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Pricing a Reinstatement Premium Protection Cover (RPP)

19/2/2018

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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:

$$DP = EL * (1 + \text{ Profit and Expense Loading } ) $$

Plugging this into our formula above, we can then conclude that the expected Reinstatement Premiums will be:

$$\frac {DP} { \text{ Profit and Expense Loading } } * ROL $$

In order to turn this into a price (which we will denote $RPP$) rather than an expected loss, we then need to load our formula for profit and expenses i.e.

$$RPP = \frac {DP} {\text{ Profit and Expense Loading }} * ROL * ( \text{ Profit and Expense Loading } ) $$

Which with cancellation gives us:

$$RPP = DP * ROL $$

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?

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.

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