AS1051CITY UNIVERSITYLondonBSc Honours Degree in Actuarial SciencePart 1Mathematics for Actuarial Science: Paper 2(solutions)2014Time allowed: 2 hoursFull marks may be obtained for correct answers toALL of the SIX questions in Section A andTWO of the THREE questions in Section B.If more than TWO questions from Section B are answered,the best TWO marks will be credited.All necessary working must be
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AS1051
CITY UNIVERSITY
London
BSc Honours Degree in Actuarial Science
Part 1
Mathematics for Actuarial Science: Paper 2
(solutions)
2014
Time allowed: 2 hours
Full marks may be obtained for correct answers to
ALL of the SIX questions in Section A and
TWO of the THREE questions in Section B.
If more than TWO questions from Section B are answered,
the best TWO marks will be credited.
All necessary working must be shown.
1 Turn over : : :
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Section A
1. (a) [4 marks] (seen)
A relation on a set A is said to be reflexive if aRa 8a 2 A. A relation
is said to be transitive if 8a; b; c 2 A, aRb and bRc ) aRc
(b) [4 marks] (unseen) 8P 2 M P - P = 0A and 0 2 R
Suppose P - Q = lA for some l 2 R then P - Q = -(P - Q) =
-lA = A(-l) and -l 2 R and so QRP.
Suppose that P - Q = lA and Q - R = mA for some l; m 2 R.
Then P - R = P - Q + Q - R = lA + mA. So P - R = A(l + m)
with l + m 2 R. So if PRQ and QRR then PRR.
2. (a) [4 marks] Solve first An+1 = -4An yields Anh = C(-4)n. Now try a
solution of the form un = an + b. This leads to
an + a + b + 4an + 4b = 2n
so a = 2=5 and b = -2=25.
Thus
A
n = C(-4)n + 2
5
n -
2
25
:
(b) [4 marks]
Using the auxiliary equation
λ2 - λ - 1 = 0
we determine that the solution for the homogeneous equation is
A
n = C 1 + p5
2 !n + D 1 -2p5!n :
Now apply A0 = 1 and A1 = 1 to get:
C = 1 + p5
2p5
D = p5-1
2p5 and so
A
n =
1 + p5
2p5
1 + p5
2 !n + p25p-5 1 1 -2p5!n :
2 Turn over : : :
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