ACTEX Practice Problem

[Jy88]

Practice Problem 32 (Quantile risk measure and CTE)

The quantile risk measure vs the CTE.
My understanding is that quantile is percentile, referring to one point estimate. CTE(alpha) is the average of all the losses that fall in (1-alpha) of the distribution.

My question is:
What do they mean by V(alpha) falls in probability mass then CTE would give bad results?

If alpha < P(L(0)=0; then the quantile falls withthin the probability mass at zero and we adjust by sing CTE(alpha)(X) = 1-e/(1-alpha) CTE (alpha)(X)?
CTE(alpha)(L0) = V(1-alpha)/2?

And for quantile risk measure, property bounded below by mean loss and subadditive is not satisfy? Can this be proven using the question stated:

100 simulations produce future claim liability of zero in 88 of the trials. PV for the remaining 12 trials are:

1, 35, 14, 3, 78, 51, 215, 11, 79, 120, 19, 15

No guarantee charge.

Thanks.

[Jy88]

Practice Problem 27

Use of policyholder behavior model in embedded options.

1) For the minimum guaranteed crediting rate – What is segmented approach? What is portfolio approach? It says that with portfolio approach of setting rates, two phenomena would be expected if rates fall
(a) Lapses will slow as co. offer competitive rate compared to those set on segmented approach.
(b) Expect increased cash inflows from other companies due to the competitive rates.
2) Book value withdrawal – If rates rise, using a segmented approach, lapse rates are likely to increase for (policies that have past the surrender charges) and without an MVA, is difficult to dissuade alert policyholders. What is a MVA?

[campbell]

1. When V(alpha) falls within a probability mass, CTE is adjusted. As a very simplified example:

You want CTE(90) for a loss distribution. Suppose the loss distribution is 0 for 95% of the probability and 100 for 5% of the probability. Then CTE(90) = 0/0.1*(.05*100 + .05*0) = 50. Basically, you take up to the amount of probability you need to get to your CTE level.

2. Quantile risk measure (aka Value at Risk) satisfies neither subadditivity nor bounded by mean loss. To show that for the example you've given, the bound is easily broken by picking the level low enough: VaR(80) will be 0, but the mean loss is obviously greater than 0.

Subadditivity is a little harder. Let me think -- suppose X = the loss if it's positive and even, and 0 otherwise; and suppose Y = the loss if it's positive and odd, and 0 otherwise. So X + Y will give you distribution given.

But X is simulated with these losses:
14, 78, 120, and 97 0s.

Y is simulated with these losses:
1, 35, 3, 51, 215, 11, 79, 19, 15 and 91 0s.

So VaR(90) = 0 for X, and VaR(90) = 0 for Y, but VaR(90) > 0 for X + Y.

Yeah, it's a little strained, which is why VaR is usually okay for use for risk management (at least for the banking use of a 10-day horizon).



Practice Problem 32 (Quantile risk measure and CTE)

The quantile risk measure vs the CTE.
My understanding is that quantile is percentile, referring to one point estimate. CTE(alpha) is the average of all the losses that fall in (1-alpha) of the distribution.

My question is:
What do they mean by V(alpha) falls in probability mass then CTE would give bad results?

If alpha < P(L(0)=0; then the quantile falls withthin the probability mass at zero and we adjust by sing CTE(alpha)(X) = 1-e/(1-alpha) CTE (alpha)(X)?
CTE(alpha)(L0) = V(1-alpha)/2?

And for quantile risk measure, property bounded below by mean loss and subadditive is not satisfy? Can this be proven using the question stated:

100 simulations produce future claim liability of zero in 88 of the trials. PV for the remaining 12 trials are:

1, 35, 14, 3, 78, 51, 215, 11, 79, 120, 19, 15

No guarantee charge.

Thanks.

[yanz]

question for you guys: all these actex problems you're refering to - did they come in your manual? As a sitter for one of the other exams, I'm wondering how much I'm missing by having no practice problems in our manual...especially since the topics that I'm seeing above correspond to our syllabus as well.

[campbell]

I don't have the Actex manual. I just read the old 8V exams (for stuff still relevant), and think up my own questions.

[Jy88]

Campbell,

Think I need to refresh my memory on CTE. How did you calculate the CTE as illustrated in your example. When you said CTE(90), you want to calculate the average of the 10th percentile of the distribution. In this case, you have 5% of 0 and 5% 100. So the average is just 50. And you divide zero by 10% as we are working in the condition of the 10th percentile.

However, on the quantile risk measure if you have a VAR(90) on X that would be the 4.5th ranked results which is not zero?

Thanks again.


1. When V(alpha) falls within a probability mass, CTE is adjusted. As a very simplified example:

You want CTE(90) for a loss distribution. Suppose the loss distribution is 0 for 95% of the probability and 100 for 5% of the probability. Then CTE(90) = 0/0.1*(.05*100 + .05*0) = 50. Basically, you take up to the amount of probability you need to get to your CTE level.

2. Quantile risk measure (aka Value at Risk) satisfies neither subadditivity nor bounded by mean loss. To show that for the example you've given, the bound is easily broken by picking the level low enough: VaR(80) will be 0, but the mean loss is obviously greater than 0.

Subadditivity is a little harder. Let me think -- suppose X = the loss if it's positive and even, and 0 otherwise; and suppose Y = the loss if it's positive and odd, and 0 otherwise. So X + Y will give you distribution given.

But X is simulated with these losses:
14, 78, 120, and 97 0s.

Y is simulated with these losses:
1, 35, 3, 51, 215, 11, 79, 19, 15 and 91 0s.

So VaR(90) = 0 for X, and VaR(90) = 0 for Y, but VaR(90) > 0 for X + Y.

Yeah, it's a little strained, which is why VaR is usually okay for use for risk management (at least for the banking use of a 10-day horizon).

[campbell]

There are one hundred numbers for the simulation of X. Ninety-seven of them are 0, and then there's 14, 78, and 120. How is VaR(90) not 0 for this example? Likewise for Y - it's got nine non-zero losses and ninety-one 0 losses. The VaR(90) is 0 for Y as well.

It doesn't matter which of the various estimates for VaR you use (whether the 10th highest, 11th highest, or 10.5th highest), they're all going to be 0.

The X and Y come from the original loss distribution, just the non-zero losses are allocated depending on even/odd (if it were continuous, I'd have them divided up by even number +/- 1/2 and odd number +/- 1/2). So there will be exactly the same number of Xs and Ys as the original.

[Jy88]

Thanks Campbell. Think I missed out the 97 in ur explanation.

Now, I get it.

On the subadditvity part, you said that VAR(90)> X+Y. Hmm..if you don mind explanining this? As if you have 97 zeros and 91 zeros, when you add them up this would give you 188 zeros. And VAR(90) would also be zero for X+Y isn't it?

Thanks a lot for that once again. :-p

There are one hundred numbers for the simulation of X. Ninety-seven of them are 0, and then there's 14, 78, and 120. How is VaR(90) not 0 for this example? Likewise for Y - it's got nine non-zero losses and ninety-one 0 losses. The VaR(90) is 0 for Y as well.

It doesn't matter which of the various estimates for VaR you use (whether the 10th highest, 11th highest, or 10.5th highest), they're all going to be 0.

The X and Y come from the original loss distribution, just the non-zero losses are allocated depending on even/odd (if it were continuous, I'd have them divided up by even number +/- 1/2 and odd number +/- 1/2). So there will be exactly the same number of Xs and Ys as the original.

[campbell]

No.
X+Y equals the original one hundred losses.

X+Y does not mean concatenate the sequences, it means add the losses.

You have to remember which X goes with which Y -- the way I defined them, they cannot both be non-zero at the same time.

So you have eighty-eight zeros and the original 12 losses.

[Jy88]

Thanks Campbell :-p