Category: Actuarial Modelling

Poisson Distribution for small Lambda

23/4/2019

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.

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Converting a Return Period to a RoL

15/3/2019

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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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Poisson Distribution for small Lambda

Converting a Return Period to a RoL

Extending the Copula Method

Correlations, Friedrich Gauss, and Copulas

Correlation Matrix - Positive Semi-Definite Requirement

Was the Lognormal Distribution misnamed?

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