Cte/tvar Soa

[Edwin]

Hi all,

The 2012 November SOA ERM paper Q2 (a) (ii) (HERE) calculates CTE (Conditional Tail Expectation) by the following formula;-

CTE = E[X|X>Xp]



Xp is the VAR.

I was trying to compute this CTE also TVAR by evaluating the integral in the numerator only. I don't see why division by the denominator is necessary.

Please help, note that TVAR = CTE = E[-X|X<-Xp]



This def of TVAR was used in the CT8 paper i wrote here;- (APRIL 2012)

http://www.actuaries.org.uk/researc...subject-ct8-exam-papers-and-examiners-reports

(Apologies for asking many questions,still more coming)

 

[David Wilmot]

The ActEd Course notes comment that...

... if you refer to other texts , you may spot that there are inconsistencies between published definitions of risk measures that are associated with VaR, of which TVaR is just one. Other measures that can have different definitions include expected shortfall (ES) and conditional tail expectation (CTE). In these Course Notes we follow the definitions used by Sweeting.

i.e. such measures are not 'well defined' and you will see different authors using different formulae for the same named risk measure.

 

[Edwin]

The ActEd Course notes comment that...

... if you refer to other texts , you may spot that there are inconsistencies between published definitions of risk measures that are associated with VaR, of which TVaR is just one. Other measures that can have different definitions include expected shortfall (ES) and conditional tail expectation (CTE). In these Course Notes we follow the definitions used by Sweeting.

i.e. such measures are not 'well defined' and you will see different authors using different formulae for the same named risk measure.

Hi David, I agree with you. However, if you can look at the 2014 May Mock A paper question 1. You will see that the TVAR was calculated as the expected values above the VAR divided by 1 - 95%.

In the same way the integral in the denominator in the SOA exam works out to 1 - 95% as well?

 

[td290]

It's true that different authors use different definitions but that's not what's happening here. If the random variable \(X\) represents a loss distribution, with negative values signifying a profit, use CTE = E[X|X>Xp]. If the random variable \(X\) represents a profit distributions, with negative values signifying a loss, use CTE = E[-X|X<-Xp].

The question as to why the denominator is necessary is a bit tricky. Consider a function \(g(X)\), which is equal to \(X\) if \(X>X_p\) and equal to zero otherwise. Now the numerator calculates \({\mathbb E}(g(X))\), which will be too low because we are given that \(X>X_p\) and so that CTE misses out all the situations in which \(g(X)=0\). The denominator provides the necessary adjustment.

Hope that helps.

 

[didster]

The CT 8 paper had a 1/alpha included which is effectively the same denominator.

A simple example why you need the denominator is E[X |X>5] where X is the value of a rolled die.
Obviously the answer is 6 = 6 *1/6 / (1/6)

The denominator is the probability that the value is above 5, VAR, etc. Conditional expectations need conditional probabilities etc.

 

[Edwin]

The CT 8 paper had a 1/alpha included which is effectively the same denominator.

A simple example why you need the denominator is E[X |X>5] where X is the value of a rolled die.
Obviously the answer is 6 = 6 *1/6 / (1/6)

The denominator is the probability that the value is above 5, VAR, etc. Conditional expectations need conditional probabilities etc.

Hi didster, the CT8 1/alpha was wrong, see the examiners report.

 

[td290]

The 1/alpha in the question was wrong. The 1/alpha in the second line of the solution is correct for the reason didster is saying.

 

[Edwin]

The 1/alpha in the question was wrong. The 1/alpha in the second line of the solution is correct for the reason didster is saying.

didster and td290, you are 100% right. Here is the rigour (I don't believe in anything that cannot be shown);

PROOF;-

Let us realise that by considering the distribution of X|X>Xp, we dealing with a truncated pdf. To derive it’s expectation i.e E[X|X>Xp] we need it’s pdf i.e f(X|X>Xp). To get the f(X|X>Xp)we may want to start with F[X|X>Xp].







Now back to the theory.

 

[r_v.s]

Maybe this reasoning would help:

The CTE is the expected loss given that your losses have exceeded the VaR.

Or if you are looking at returns,
it is the expected value of return given that the return has fallen below some benchmark, usually the VaR.

This is not mathematical, but this helped me understand the CTE risk measure's interpretation.
Thought this might help!

 

[Edwin]

Maybe this reasoning would help:

The CTE is the expected loss given that your losses have exceeded the VaR.

Or if you are looking at returns,
it is the expected value of return given that the return has fallen below some benchmark, usually the VaR.

This is not mathematical, but this helped me understand the CTE risk measure's interpretation.
Thought this might help!

Cheers r_v.s, you are absolutely right same thing as td290's and didter's heuristics, but any intuition works backwards to 'understand' the formula, I wanted to 'come up' with the formula.

Your explanations are great to explain the derivation, deep thanks!

 

[Simon James]

Good to see the rigour! However, in the exam, an understanding of the formulae and an ability to apply them is required. The Examiners will not be expecting you to prove such results.

 

[Edwin]

Good to see the rigour! However, in the exam, an understanding of the formulae and an ability to apply them is required. The Examiners will not be expecting you to prove such results.

Thanks Simon, the rigour though is for my own sake. My mind works by deriving then explaining not explaining without deriving.

It is a condition for belief , I find it hard to rationalize formulae and stop there.