Hope this helps. Let me know if you have trouble understanding my notation
or solutions.
If anyone has simpler solutions to any of these, I'd love to see them.
(Especially #13 and #22).
Dave Kennerud
kennerud@concentric.net
#2) Given lambda (I'll use l for lambda), HM = l. So VHM = Var(l) = u
Similarly, given l, PV = l. So EVPV = E(l) = u.
So k = EVPV / VHM = u / u = 1.
Answer: A
#4) I'll use l for lamda, and I'll use a and b for the gamma parameters (a
= alpha and b = lambda as in Hogg and Klugman)
First, calculate a and b: a / b = mean of gamma = 0.05
a / (b^2) = var. of gamma = 0.01
Solving for a and b yields: a = 0.25 b = 5
The posterior distribution will also be gamma (always the case for
Poisson/gamma problems), with parameters
a' = a + number of claims = 0.25 + n
b' = b + number of exposures = 5 + 10 = 15
Since the posterior variance = prior variance: a' / (b'^2) = 0.01
0.25 + n = 0.01 * 15^2 = 2.25
n = 2
Answer: C
#13) The probability of a single claim being less than k is: F(k) = 1 - e^(-k)
So if there are n claims, the prob. all are less than k is: (1 - e^(-k))^n
The probability of n claims is p(n) as given: p(n) = (e^l - 1)^(-1) *
(l^n) / n!
So the prob. of having exactly n claims, all < k is: (e^l - 1)^(-1) *
(l^n) / n! * (1 - e^(-k))^n
= (e^l - 1)^(-1) * { [ l * (1 - e^(-k) ]^n / n! }
(This is equivalent to the probability of exactly n claims, the largest
of which is less than k)
Now, to get the prob. that the largest claim is less than k, you need to
sum the above over all n = 1 to infinity.
Forget about the (e^l - 1)^(-1) for the moment, since that has no n's in it.
Let y = [ l * (1 - e^(-k) ]. So [ l * (1 - e^(-k) ]^n / n! = y^n / n!,
and what you need to sum is:
y^n / n! n = 1 .. infinity
But summing y^n / n! from 0 to inf. is e^y, so summing from 1 to inf. is
(e^y - 1) = (e^[ l * (1 - e^(-k) ] -1)
So the final answer (remember the e^l - 1 ) is: (e^l - 1)^(-1) * (e^[ l *
(1 - e^(-k) ] -1)
Answer: D
#20) Calculate the MLE for l.
L(x1,...,xn) = l^(-n) * e^(- (x1 + . . . + xn) / l )
ln(L) = -n * ln(l) - (x1 + . . . + xn) / l )
d(ln(L))/dl = -n/l + (x1 + . . . + xn)/l^2
Setting this to 0 and solving for l yields: l = Sample mean (I'll use X
for sample mean).
So e^(-1/l) is apx. e^(-1/X).
Answer: B
#21) Since there is only one comparison point, the minimum distance
statistic is just:
( F(1) - Fn(1) )^2
Since this has only one term, to minimize it all you need to do is to set
F(1) = Fn(1)
But F(1) = 1 - e^(-1/l), and Fn(1) = 1 - p, so: F(1) - Fn(1) = (e^(-1/l)) - p
So e^-(1/l) = p.
Answer: C
#22) We need to estimate the variance of: h(l) = e^(-1/l)
To do this you need the formula for the approximate variance of a function
of a parameter (p. 93 of H&K)
Since there is only one parameter, the apx. variance will be: h'(l)^2 * s^2,
where s is the apx. variance of the parameter l.
To calculate s^2, need to use Rao-Cramer: ln(f(x)) = -ln(l) - x/l
Taking 2 derivative with respect to l yields: 1/l^2 - 2*x/l^3
Taking the expected value (w.r.t. x): 1/l^2 - 2*E(x)/l^3 = 1/l^2 -
2*l/l^3 = -1/l^2
So s^2 will be: -1/(n* -1/l^2) = l^2 / n
Now, h'(l) = 1/l^2 * e^(-1/l)
So, finally, h'(l)^2 * s^2
= [1 /l^2 * e^(-1/l) ] ^ 2 * ( l^2 / n )
= 1/(n * l^2) * e^(-2/l)
Answer: D
#23) If the sample r.v. are xi, let yi = { 1 , if xi > 1
{ 0 , if xi <= 1
Then the yi are Bernouilli r.v. with prob. of success = prob(xi > 1) =
e^(-1/l)
Also, p = (proportion of xi greater than 1) = (y1 + . . . + yn)/n, so p
is the average of n Bernouilli r.v.
Since var(yi) = e^(-1/l) * (1 - e^(-1/l)), the variance of p will be:
e^(-1/l) * (1 - e^(-1/l) ) /n
Answer: E
#26) First, calculate the following assuming m is fixed. (N = claim ct
r.v., S = claim size r.v.)
E(N) = m; Var(N) = m; E(S) = 20m; Var(S) = 400m^2
Then the PV|m = E(N)*Var(S) + Var(N)*E(S)^2 = m*400m^2 + m*(20m)^2 = 800m^3
So EVPV = 800*E(m^3)
Now, E(m^3) = integral from 0 to infinity of m^5*e^(-m)/2. Since
m^5*e(-m)/(5!) is the pdf of a gamma r.v. with parameter lambda = 1 and
alpha = 6, this integral will be equal to: 5! / 2 = 60.
Thus, EVPV = 800*60 = 48,000
Answer is D
#29) The posterior p.d.f. will be proportional to the Likelihood function,
which is:
g(l) * f(50) = 500,000*(1/l^5)*e^(-150/l)
The only solution that has this form ( C * (1/l^5) * e(-150/l) for some
constant C) (Note that you don't actually have to find the constant C)
Answer. E
>X-Lotus-FromDomain: SUNLIFE
>From: Carrie_Cheung/Actuarial/Corporate/SunLife@SunLifeOfCanada.com
>To: studygroup4b@lists.casact.org
>Date: Mon, 19 Oct 1998 09:26:39 -0400
>Subject: SPRING 98 Exam
>
>Hi,
>
>Everyone must be study hard. I wonder whether there is any one who want to
>try the SPRING 98 Exam. I have problem with quiet a number of questions.
>
>Question 2, 4, 13, 20-23( they are multiple question, which I have no where
>to start ), 26 and 29.
>
>If some one prefer to fax their work ( instead of typing) , my fax number
>is (416) 979-6111. Or if you have problems with questions other than the
>above, I might be able to help you.
>
>Thanks and Good Luck to your Exams,
>Carrie
>
>
>
>