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April 2008 - #7 part 1

M

MindFull

Ton up Member
I know that this part of the question was only worth 2 marks but I feel like I'm still missing somthing. Here it is.
In a certain small country all listed companies are required to have their accounts audited on an annual basis by one of the three authorised audit firms (A, B and C). The terms of engagement of each of the audit firms require that a minimum of two annual audits must be conducted by the newly appointed firm. Whenever a company is able to choose to change auditors, the likelihood that it will retain its auditors for a further year is (80%, 70%, 90%) where the current auditor is (A,B,C) respectively. If changing auditors a company is equally likely to choose either of the alternative firms. (i) A company has just changed auditors to firm A. Calculate the expected
number of audits which will be undertaken before the company changes
auditors again.

In the solution, they get 1 + 1 + 0.8 + (0.8)^2 + ...
I was just wondering why the prob. for all the years isn't just 0.8, i.e., why do we multiply by 0.8 ea. time?

Thanks much.
 
After just reading the report,
The examiners found the expected number of audits by summing the probabilities that the ith audit would be completed by A before the auditor is changed.

You can't change the auditor in first two years so first two numbers are 1.
Third year - 0.8 chance of NOT changing (or equivalently 3rd audit completed before change)
Fourth year - 0.8^2 chance of not changing by fourth year (because you need to have not changed in the third year and fourth year) and so on

ie each term is

P(not changed BY (i+1)th year) = P(not changed BY ith year) * P(does not change IN (i+1)th year given not changed BY ith year)

Hope this helps.

(Interesting twist on the expectation of life = sum of tpx for a non-mortality model in my opinion (but maybe its just been a while since I did this stuff). My initial thought was along the lines of sum the number of audits by probability of that number of audits - same result but not as neat.)
 
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