1. A motor insurer operates a no-claims discount system that has five levels. The percentage of the basic premium paid by the insured in each level is as follows: level % premium charged 5 100 4 90 3 80 2 70 1 60 Insured motorists move between levels depending on the number of claims in the previous year. For each policy holder, the number of claims per year follows a poisson distribution with mean 0.25. For those in levels 2, 3, 4 and 5 at the start of the previous year; i) If no claims are made during the previous year, the insured moves down one level (example, from level 4 to level 3). ii) If one claim is made during the previous year, the insured moves up one level (except those in level 5 at the start of the previous year who will remain in level 5). iii) If two claims are made during the previous year, the insured moves up two levels (except those in level 5 at the start of the previous year who will remain in level 5 and those in level 4, who will move to level 5). iv) If three or more claims are made during the previous year, the insured moves to level 5. For those in level 1 at the start of the previous year a no claims discount protection policy applies whereby they remain in level 1 if they make one claim. If they make two claims, they move to level 5. If they make no claims, they remain in level 1. a) Determine the transition matrix for the no-claims discount system (Assuming that all motorists activate their policies). b) A policy holder is in level 3 for the first year of the policy. Assuming that the policy is maintained, calculate the probability that at the start of the third year the policy holder will be; (i) In level 1 (ii) In level 3 c) (i) state conditions under which the probability of being in a particular state after n years converges as n ? ? to some limit which is independent of the initial state. (ii) verify that the conditions are satisfied in this instance. (iii) Determine the ultimate probability that the insured will be in level 1. d) The insurer suspects that the model used for its calculation may be too simplistic. Given annual data listing numbers of claims per policy, broken down by discount level, state which test would be most appropriate to test the assumption the distribution of the number of claims per policy per year is poisson with mean 0.25.
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The set R2×2 of all 2 × 2 matrices form a linear space in an obvious way. For Q90. to Q92., determine whether the given subset of R2×2 is a linear subspace of R2×2. Explain thoroughly. To show it is, you have to demonstrate that all three properties in the definition of linear subspace hold. To show it is not, you only have to show that one property fails. 90. The set of all X ∈ R2×2 that are symmetric, namely XT = X. 91. The set of all X ∈ R2×2 that satisfies X[1 2; 3 4] = [1 0; 0 1]. 92. The set of all X ∈ R2×2 that satisfies AX − XA = [0 0; 0 0], where A is a constant 2 × 2 matrix. (The exact values of the entries of A don't matter.) 93. Let R3×3 be the linear space of all 3 × 3 matrices. Let W be the set of all symmetric 3 × 3 matrices. Then W is a linear subspace of R3×3. Find a basis for W and identify dim(W). For a positive integer n, define Pn as the set of all polynomials in symbol t, of degree at most n. For example, for P2, we have 2 + 5t + t2 ∈ P2, and 2 − t ∈ P2 as well. The addition and scalar multiplication of polynomials make Pn a linear space. The neutral element is the zero polynomial 0. Remark: The zero polynomial 0 is usually not assigned a degree. But sometimes it is considered of degree −∐. With that convention, then it can be thought of as having a degree ≤ n, and so is included in Pn. 94. Every p(t) ∈ P2 can be uniquely written in the form a01 + a1t + a2t2. This statement is announcing a basis A of P2, formed by 1, t, t2. (A) Therefore, what is dim(P2)? (B) What is [2 + 3t − 5t2]A? (C) What is [3 − t]A? 95. [Some Algebra] Every p(t) ∈ P2 can be uniquely written in the form b01 + b1(t − 2) + b2(t − 2)2. In this problem, let's see how this works via a concrete case. Let p(t) = 4 − 5t + 2t2. Rewrite p(t) in the form b01 + b1(t − 2) + b2(t − 2)2. (Suggestion: A convenient way to do this is to introduce the symbol u = t − 2, thus t = u + 2. Replace t by u + 2 in 4 − 5t + 2t2, which you then simplify to the form b01 + b1u + b2u2. Then replace u by t − 2 to get the answer.) 96. Every p(t) ∈ P2 can be uniquely written in the form b01 + b1(t − 2) + b2(t − 2)2. This statement is announcing a basis B of P2. Identify this basis B. 97. Find the 3 × 3 matrices SAB and SBA, where A and B are the two bases of P2 from Q94. and Q96.. 98. Let A and B be the bases of P2 from Q94. and Q96.. Let p(t) = 4 − 5t + 2t2 ∈ P2. (A) What is [p(t)]A? (B) Use SBA found in Q97. and [p(t)]A above to find [p(t)]B. (C) Show that the answer above is consistent with your answer in Q95.. 99. A special case of Rn is R1. Think of R as R1. So R is a 1-dimensional linear space. Define T : P2 → R as T(p(t)) = p(2). For example, T(18 − 17t + 5t2) = 18 − 17 ⋅ 2 + 5 ⋅ 22 = 4. This T is a linear transformation. In our example, if 18 − 17t + 5t2 has been written as 4 + 3(t − 2) + 5(t − 2)2, it would be much easier to see that T maps it to 4. Use the statement that every p(t) ∈ P2 can be uniquely written in the form b01 + b1(t − 2) + b2(t − 2)2 to quickly identify a basis for ker(T). (If you understand what the question is asking, no computation will be needed.) 100. Let A and B be the bases of P2 from Q94. and Q96.. Let D : P2 → P2 be the ‐differential operator‐, i.e. D(p(t)) = p'(t). (E.g., D(t2) = 2t.) (A) Find [D]A. (B) Find [D]B from [D]A via SBADASA B, using the SBA and SAB obtained in Q97.. (C) Find [D]B more directly by finding [D(1)]B, [D(t − 2)]B, [D((t − 2)2)]B, and recall that [D]B = [[D(1)]B [D(t − 2)]B [D((t − 2)2)]B]. (Note: In Calculus I, when taking the derivative of 3(t − 2)2 + 5(t − 2) + 8, simply use the Chain Rule, with t − 2 as the inner function. This would immediately give the answer as 6(t − 2) + 5.)
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