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代做TEST 2代寫Java程序

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MATH 4281, Winter 2020: TEST 2
Instructor: Alexey Kuznetsov
March 26, 2020
There are 4 questions and the maximum mark is 34. Marks are indicated next to the problem number. You
have to send your solutions to the test by email (as a single pdf file) to the instructor before 8pm on March 26
(Thursday). Do not forget to write your name and student number on your solutions.
YOU MUST SHOW ALL WORK TO GET FULL CREDIT. YOU CAN CONSULT ANY MA-
TERIALS (TEXTBOOK, INTERNET) BUT YOU CAN’T GET HELP FROM YOUR CLASS-
MATES OR ANYONE ELSE.
1. [6 marks] For a particular policyholder, the manual premium is $600 per year. The past claims experience is
given in the following table.
Year 1 2 3
total claim amount $500 $570 $450
Assess whether full or partial credibility is appropriate and determine the net premium for next year’s claim
amount assuming the normal approximation. Use r = 0.03 and p = 0.95.
2. [12 marks] You are given:
(i) The number of claims made by an invividual insured in a year has Poisson distribution with mean θ;
(ii) The prior distribution for θ is
pi(θ) = 4θ2e−2θ, θ > 0.
(iii) Three claims are observed in year 1, no claims in year 2 and two claims in year 3.
This is what you need to do:
(1a) Find Buhlmann’ estimate for the number of claims in year 4.
(1b) Without finding Bayesian estimate, show that we have exact credibility (that Bayesian estimate must be
equal to Buhlmann’s estimate)
(1c) Explain why we won’t have exact credibility if the prior distribution is
pi(θ) = (1/C)θ2e−2θ, θ > 1.
and C =
∫∞
1
θ2e−2θdθ.
3. [8 marks total] We have random variables X1 and X2, which are i.i.d. conditional on a r.v. Θ. We know the
unconditional distribution of X1 is
P(X1 = i) = 1/3, i = 1, 2, 3.
1
We also know that E[X2|X1 = 1] = 1.3, E[X2|X1 = 2] = 1.8 and E[X2|X1 = 3] = 2.5. Determine Buhlmann’s
credibility estimate of X2, given that X1 = 1.
Hint: Read again Section 18.4 in 4-th edition of the textbook and focus on the remark right before equation
(18.19). That remark implies that Buhlmann’s estimate of X2 (which is a linear function α0 + α1X1) can also
be found by minimizing
Q1 = E
[(
E[X2|X1]− α0 − α1X1
)2]
over all α0, α1.
4. [8 marks] We have four classes of insureds, each of whom may have zero or one claim in any given year, with
the following probabilities: A class is selected at random (with probability 1/4) and four insureds are selected
probability of no claims probability of one claim
Class 1 0.9 0.1
Class 2 0.7 0.3
Class 3 0.5 0.5
Class 4 0.1 0.9
at random from the class. For this group of four insureds we observe 3 claims in year one. Then we select
three more insureds from the same class, and in the second year (for this combined group of seven insureds) we
observe 4 claims. Then we select additional 8 insureds from the same class. Using Buhlmann-Straub credibility,
cs作业代写_matlab代做_machine learning代写 estimate the total number of claims in year 3 for this combined group of 15 insureds.

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