2009_test1sol - E Y 2 -( E [ Y ]) 2 = 3 20 5. . (a) Let X...

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ACTL1001 Class Test 1 S2 2009 Solutions 1. . (a) l 56 = l 55 - d 55 = 84772 (b) 1 - l 59 l 55 = 1 - 80588 85916 = 0 . 062014 (c) d 58 + d 59 l 55 = 0 . 037129 2. . (a) Substituting gives 40 p 30 = s (70) s (30) = 0 . 515714 (b) s (50) - s (65) s (25) = 0 . 162328 (Remark: An obvious ”reasonableness check” for this answer is that these are between 0 and 1) 3. Since the number of hailstorms N is Poisson, the probability that in one year there are 2 or more hailstorms is 1 - P ( N = 0) - P ( N = 1) = 1 - e - 0 . 5 - 0 . 5 e - 0 . 5 = 0 . 090204 Let X be the number of years with 2 or more hailstorms out of the 40. X will be binomial distributed with n = 40 ,p = 0 . 090204. Hence P ( X 2) = 1 - P ( X = 0) - P ( X = 1) = 1 - p 0 (1 - p ) 40 - 40 p 1 (1 - p ) 39 = 0 . 886823 1
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4. . (a) We need to show two things. Firstly f ( y ) needs to be non-negative which is true as y 2 is positive. Secondly we need the integral over the full range to equal 1: Z 2 0 3 8 y 2 dy = 1 8 ± y 3 ² 2 0 = 1 as required. (b) Expected Value Z 2 0 3 8 y 3 dy = 3 32 ± y 4 ² 2 0 = 3 2 second moment Z 2 0 3 8 y 4 dy = 3 40 ± y 5 ² 2 0 = 2 . 4 and hence variance is
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Unformatted text preview: E Y 2 -( E [ Y ]) 2 = 3 20 5. . (a) Let X be the required payment. We have using net present value, at 1% eective per month. 400000 = Xa 120 X = 5738 . 838 (b) Since interest rates have not changed the loan outstanding is most easily calculated using the backwards recursion result Xa 70 = 287909 (c) As there have been a (unforseen) change in interest rates, and that the payment amount per month does not change, the backwards re-cursion result will not be applicable as the repayments of X for the remaining term of the loan will not be sucient to pay o the loan by the end of the 10 years. Hence we should use forward recursion, giving 287909 (1 . 015)-X = 286488 . 8 2...
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2009_test1sol - E Y 2 -( E [ Y ]) 2 = 3 20 5. . (a) Let X...

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