Documents

HJK

Description
Description:
Categories
Published
of 29
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
    1 Supplementary notes for Exam C Overview 1.1    Introduction In July and August 2013, the SoA added a number of questions to the sample exam questions document for Exam C on the Be-an-Actuary website. These were to cover syllabus items recently added to Exam C. The attached note covers the additional material needed for these syllabus items. There are three sections in this note. The first looks at the idea of an extreme value distribution. The second describes an alternative approach to dealing with large data sets. The final section introduces a number of additional simulation techniques in various situations. As you read this material, you should have in mind as far as possible the material in Chapter 2 for the first section, that in Chapter 7 for the second section, and that in Chapter 12 for the final section. We have given the syllabus items themselves in an appendix to this note.  Supplementary note 2 1.2    Extreme value distributions The following section should be read in conjunction with Chapter 2 of the textbook. There are some areas of insurance work where it is useful to model quantities using distributions with particularly heavy tails. One example of a situation like this would be when constructing a model for the largest value in a set of identical and independently distributed random variables. If we are trying to model the maximum value from a random sample, intuitively this maximum value is likely to be in some sense large. We may therefore want to model it with a distribution which is heavy-tailed. Distributions of this type are known as extreme value distributions . In the situation outlined above, the inverse Weibull distribution is often use as a model. This is related to the Weibull distribution studied earlier as follows. Inverse Weibull distribution (Fr é chet distribution) If a random variable X   has a Weibull distribution, then the random variable 1/ YX    is said to have an inverse Weibull distribution. The inverse Weibull distribution has the following attributes: PDF:   (/) /() x xe fxx               CDF:   (/) x Fxe          Moments:      1/ kk EXk        This distribution is sometimes known as the Fréchet distribution. More details for the Inverse Weibull distribution are given in the Tables for Exam C. The process of finding the distribution for the random variable 1/ YX    can be applied to other distributions. Examples of the inverse exponential distribution and the inverse gamma distribution are given in the Tables for Exam C. Other distributions which are sometimes used in this context are the Gumbel distribution , and the Weibull distribution itself (without inversion). Details for the Gumbel distribution are given below. Gumbel distribution A random variable X   is said to have a Gumbel distribution if: 1()exp yy  fxee          where xy      , and x    The distribution function is: ()exp y Fxe       The details for the Weibull distribution itself are given in Chapter 2 of the textbook. The Pareto distribution also has a thick tail, and can sometimes be used in these situations.   Supplementary note 3 1.3    Large data sets – an alternative approach The following section should be read in conjunction with Chapter 7 of the textbook. We have seen in Chapter 7 how the Kaplan-Meier method can be adapted for use with large data sets. Here we look at another approach for calculating mortality rates when large numbers of lives are involved. To help to describe the main features of the method, we shall use a small sample of six lives to demonstrate the approach. The exact exposure method A company is trying to estimate mortality rates for the holders of a certain type of policy. It has the following information about a group of 6 lives, who all hold a policy of this type. The investigation ran for a three-year period, from Jan 1 2010 to Dec 31 2012.  Life Date of birth Date of purchase Mode of exit Date of exit 1 Mar 1 1965 Jul 1 2009 Alive Dec 31 2012 2 Jul 1 1965 Nov 1 2009 Death Mar 1 2011 3 Aug 1 1965 Apr 1 2010 Surrender Feb 1 2012 4 Apr 1 1965 Jun 1 2011 Alive Dec 31 2012 5 May 1 1965 Aug 1 2010 Surrender Jun 1 2012 6 Oct 1 1965 May 1 2010 Death Apr 1 2012 We see that of the 6 lives, two survived within the population to the end of the investigation, two of the lives surrendered their policies while the investigation was in progress, and two died during the period of the investigation. We wish to use the information in the table above to estimate mortality rates at various ages. We shall assume that each month is exactly one-twelfth of a year, to simplify the calculations. We start by finding the ages at which each life started to be observed, and the age at which life ceased to be observed. Note that although Life 1 purchased his policy on July 1 2009, the investigation had not started at that point. So the date on which Life 1 is first observed is  January 1 2010. Life 2 is also first observed on this date.  Supplementary note 4 This gives us the following table of ages:  Life Age at first observation Age at last observation 1 1012 44 1012 47 2 612 44 812 45 3 812 44 612 46 4 212 46 912 47 5 312 45 112 47 6 712 44 612 46 In order to estimate the mortality rates, we need to find out the length of time for which each life was alive, and for which they were a member of the investigation. We need to subdivide these periods by age last birthday. So, for example, we shall use 44 e  for the period of time during which a life (or group of lives) was aged 44 last birthday, 45 e  for the corresponding period of time for which lives were aged 45 last birthday, and so on. From the table above, we can now find the contribution of each life to each of 44 e , 45 e , 46 e  and 47 e . This gives us the following table of figures (in months):  Life Age at first observation  Age at last observation 44 e   45 e   46 e   47 e   1 1012 44 1012 47 2 12 12 10 2 612 44  812 45 6 8 - - 3 812 44 612 46 4 12 6 - 4 212 46 912 47 - - 10 9 5 312 45 112 47 - 9 12 1 6 712 44 612 46 5 12 6 - This gives us totals in each of the k e  columns of 17, 53, 46 and 20 respectively.
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks
SAVE OUR EARTH

We need your sign to support Project to invent "SMART AND CONTROLLABLE REFLECTIVE BALLOONS" to cover the Sun and Save Our Earth.

More details...

Sign Now!

We are very appreciated for your Prompt Action!

x