Let's assume we are in the insurance business and the values we are observing are losses.

So there is a general statement that says the Conditional Tail Expectation at percentile $p$ is usually greater than the Value at Risk (percentile value) of $p+\frac{1}{2}(100-p)$

In other words, $CTE(p)> VaR(p+\frac{1}{2}(100-p))$

For example, let $p$ be $90%$. Then this statement becomes $CTE(90)>VaR(95) $

I have performed quite a few simulations based on an exponential distribution and have not found a case for which this is false.

Thus, I am wondering:

  1. What is the basis for this statement?

  2. When will this be false (Cases where there aren't many values beyond $p$)?

  • $\begingroup$ Value at RIsk has different (but related) definitions in banking and insurance. Are you referring to the Tail VaR common in insurance applications? (It might be even less ambiguous to actually give the integrals for Var and CTE) $\endgroup$ – Glen_b Apr 14 '18 at 23:14
  • $\begingroup$ @Glen_b Yes you are right. I am referring to Value at Risk from banking and insurance. Can you enlighten me what is different in the definition of Value at Risk for banking and insurance? $\endgroup$ – user101998 Apr 15 '18 at 2:46
  • $\begingroup$ Value at risk for a bank typically relates to underperforming assets, but for an insurer it's larger than expected liabilities. The bank is looking at the left tail of the asset performance while the insurer is looking at the right tail of the aggregate claims distribution. This difference is why I was asking which one you were doing, since it would affect the definitions required to try to answer the question. $\endgroup$ – Glen_b Apr 15 '18 at 11:49
  • $\begingroup$ Great observation. So if these were losses from an insurer's perspective, this question becomes clear right? Let me edit it first. $\endgroup$ – user101998 Apr 16 '18 at 2:20

If we define the Conditional Tail Expectation at $t$ as $E[X|X>t],$ then $\mathrm{CTE}(90)$ will be $$E[X|X>x_{90}],$$ where $x_{90}$ is the 90th percentile of $X.$ For the exponential distribution, this expectation is very simple: $$E[X|X>x_{90}] = x_{90} + \beta,$$ where $\beta$ is the scale parameter (and also the mean). This is from the memoryless property.

Also for an exponential distribution, we know the quantiles are given by $x_{p} = -\beta \ \mathrm{ln} \left(1-p \right), $ where $p$ is in decimal form.

So we have $$E[X|X>x_{90}]=x_{90} + \beta=-\beta \ \mathrm{ln} \left(1-0.9 \right)+\beta \approx 3.303 \beta$$

Now using the quantile formula, we have for Value-at-Risk that $$\mathrm{VaR}(95)=-\beta \ \mathrm{ln} \left(1-0.95 \right) \approx 2.996 \beta $$

So in the case of the exponential distribution the claim of $$\mathrm{CTE} \left( 90 \right) > \mathrm{VaR} \left(95 \right)$$ will always be true.

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  • $\begingroup$ Thanks, but there is no way in general to assert the truth value of this statement? $\endgroup$ – user101998 Apr 17 '18 at 15:45
  • $\begingroup$ It is not true in general. Suppose $X$ is uniform on $[0,1].$ Then the two quantities are equal. Suppose the density is $f(x)=2x$ on $[0,1].$ Then you will find that the CTE is slightly smaller than the VaR. $\endgroup$ – soakley Apr 17 '18 at 22:00

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