Continuity correction, formula sheet and interpolation

[asingh]

I was just wondering if any of you guys know what the (UK) institutes policy regarding continuity correction and interpolation is. Should we use them always, or only when asked to? Also are we given a formula sheet for the exam, and if we are is this available beforehand? And if so, where? Sorry if these are obvious questions but this (CT8 September) is the first exam I'm sitting externally (not through uni) so I've got no idea what to expect.

Thanks

 

[jensen]

hi asingh

You surprised me by not knowing about the tables provided.

Yes, you do get tables and formulas in the exam; it's the yellow table which you can buy from the institute. It does not contain every formula, but it does have most of them.

 

[Anna Bishop]

Re the continuity correction and interpolation my advice would be to apply them. The application of them by the examiners seems to vary by exam and by subject but:

- When I marked a few years ago, there was a half mark for each interpolation.
- The last CT6 exam required a continuity correction on the last question on the paper.
- You certainly need to be able to interpolate for Black-Scholes formulae questions.

Jensen's advice re the Yellow Tables is sound. You definitely need to get hold of a copy from the Institute of Actuaries. Although in the exam, you get given a copy.

Good luck!

Anna

 

[asingh]

Thanks for the advice and info. I was just looking at the solutions to 2007 September exam questions 7(b) and (c), and the way they've done it seems a bit strange to me (for investor 2). For one thing, they haven't used the continuity correction but on top of that, they haven't even allowed for the fact that the loss r.v. L = 1000 - X is discrete since X ~ Binomial (1000, 0.96)

For part (b), the way I would have done it is as follows:

VaR = min {z = 0,1,...,1000 : P(L<=z) >= 0.95} = 1000 – max{x = 0,1,...,1000 : P(X>=x) >= 0.95}

which can be approximated by 1000 – max {x = 0,1,...,1000 : P(Y>x-0.5) >= 0.95} where Y ~ Normal (960,38.4) This gave me 52 as my answer. Similarly I got 48 for part (c)

Does my answer/reasoning seem ok?

 

[Anna Bishop]

Hello Asingh

This was a devil of a question. I'm sure if you'd (correctly) used a continuity correction for Investor 2, you would not have been penalised. However, I don't quite agree your figures

Here's how I would do it:

Investor 2

On each bond, Investor 2 loses 0 with probability 0.96 and loses 1 with probability 0.04.

Let Li be the loss on the ith bond.

Li has a Bin(1, 0.04) distribution.

Let X be the loss on the total investment.

X = L1 + L2 + ... + L1000

X has an approx Normal(40, 38.4) distribution.

For the 95% VaR, we need to find the smallest loss x such that
P[X > x] <= 0.05

Applying continuity corrections:

P[X > 49] = P[Z > (49.5 - 40)/sqrt(38.4)] = 1 - phi(1.5331) = 0.0626
P[X > 50] = P[Z > (50.5 - 40)/sqrt(38.4)] = 1 - phi(1.6944) = 0.0451

so x = 50.

For the 90% VaR, we need to find the smallest loss x such that
P[X > x] <= 0.1

P[X > 47] = P[Z > (47.5 - 40)/sqrt(38.4)] = 1 - phi(1.2103) = 0.1131
P[X > 48] = P[Z > (48.5 - 40)/sqrt(38.4)] = 1 - phi(1.3717) = 0.0851

so x = 48.

Just to complete the picture:

Investor 1

Let X be the loss on the total investment for Investor 1.

X = 0 with probability 0.96
X = 1000 with probability 0.04

For the 95% VaR, we need to find the smallest loss x such that
P[X > x] <= 0.05

P[X > 0] = P[X = 1000] = 0.04

x = 0

For the 90% VaR, we need to find the smallest loss x such that
P[X > x] <= 0.10

P[X > 0] = P[X = 1000] = 0.04

x = 0

Does this help?

Anna

 

[asingh]

Yeah I wrote down the wrong number. I got 50 for part(b) too. Thanks for your help. Btw do u agree that it seems a bit stupid to have the answers being non integers when the loss r.v. can only take on integer values?