#3. 3. You are given the following:
• The losses in 2004 follow a LogNormal Distribution, with parameters μ = 5 and σ = 2.5.
• Assume that losses increase by 5% from 2004 to 2005, 3% from 2005 to 2006,
and 6% from 2006 to 2007.
What is the increase between 2004 and 2007 in the expected number of claims exceeding a 10,000 deductible?
Solution: The inflation factor is: (1.05)(1.03)(1.06) = 1.1464.
In 2007, the losses follow a LogNormal Distribution, with parameters μ = 5 + ln(1.1464) = 5.137, and σ = 2.5.
In 2004, S(10000) = 1 - Φ[ln(10,000) - 5)/2.5] = 1 - Φ(1.68) = 1 - 0.9535 = 0.0465.
In 2007, S(10000) = 1 - Φ(ln(10,000) - 5.137)/2.5) = 1 - Φ(1.63) = 1 - 0.9484 = 0.0516.
The increase in the expected number of claims exceeding a 10,000 deductible is: 0.0516/0.0465 - 1 = 11%.
I don't understand why you are just comparing the S(x). I thought the expected number of claims exceeding a 10000 deductible is E(X-10000)+ = E(x) - E(X^10000) ? i did it like that and got a different answer.
#6.6. William M. Lowe, consulting actuary, works on each assignment in intervals.
The length in hours of these intervals has an Exponential Distribution with a mean of 2. William bills each work interval at $500 per hour, excluding any fraction of an hour.
So if for example, a work interval lasts 2.7 hours, then the client is only billed $1000. The number of work intervals per assignment is distributed as
a zero-truncated Geometric Distribution with β = 4.
Determine the average amount William bills per assignment.
I don't understand S(1) + S(2) + S(3) + ... = e-1/2 + e-2/2 + e-3/2 + ... = e-1/2 / (1 - e-1/2) = 1 / (e0.5 - 1) = 1.541.
is there a formula for S(1) + .......?
Thanks.
• The losses in 2004 follow a LogNormal Distribution, with parameters μ = 5 and σ = 2.5.
• Assume that losses increase by 5% from 2004 to 2005, 3% from 2005 to 2006,
and 6% from 2006 to 2007.
What is the increase between 2004 and 2007 in the expected number of claims exceeding a 10,000 deductible?
Solution: The inflation factor is: (1.05)(1.03)(1.06) = 1.1464.
In 2007, the losses follow a LogNormal Distribution, with parameters μ = 5 + ln(1.1464) = 5.137, and σ = 2.5.
In 2004, S(10000) = 1 - Φ[ln(10,000) - 5)/2.5] = 1 - Φ(1.68) = 1 - 0.9535 = 0.0465.
In 2007, S(10000) = 1 - Φ(ln(10,000) - 5.137)/2.5) = 1 - Φ(1.63) = 1 - 0.9484 = 0.0516.
The increase in the expected number of claims exceeding a 10,000 deductible is: 0.0516/0.0465 - 1 = 11%.
I don't understand why you are just comparing the S(x). I thought the expected number of claims exceeding a 10000 deductible is E(X-10000)+ = E(x) - E(X^10000) ? i did it like that and got a different answer.
#6.6. William M. Lowe, consulting actuary, works on each assignment in intervals.
The length in hours of these intervals has an Exponential Distribution with a mean of 2. William bills each work interval at $500 per hour, excluding any fraction of an hour.
So if for example, a work interval lasts 2.7 hours, then the client is only billed $1000. The number of work intervals per assignment is distributed as
a zero-truncated Geometric Distribution with β = 4.
Determine the average amount William bills per assignment.
I don't understand S(1) + S(2) + S(3) + ... = e-1/2 + e-2/2 + e-3/2 + ... = e-1/2 / (1 - e-1/2) = 1 / (e0.5 - 1) = 1.541.
is there a formula for S(1) + .......?
Thanks.
Mahler PE1