SOLUTION FOR HOMEWORK 12, ACTS Welcome to your last homework. Here you will see some actuarial notions that sometimes used on actuarial Exam 1.

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1 SOLUTION FOR HOMEWORK 2, ACTS 36 Welcome to your last homework. Here you will see some actuarial notions that sometimes use on actuarial Exam.. Notation: B is benefit. Given: B = min(y, ), f Y (y) = 2y 3 I(y > ). Fin: E(B). Solution: Write, E(B) = min(y, )f Y (y)y = = 2y 2 y + 2y 3 y yf Y (y)y + f Y (y)y = [ 2y ] y= + [ y 2 ] y= = (2 /5) + () 2 =.9 Answer: D 2. Given: Y = min(x, ), f X (x) = (/5)I( < x < 5). Fin: Var(Y ). Solution: Remember that Var(Y ) = E(Y 2 ) [E(Y )] 2. Let us calculate the moments. Write 5 E(Y ) = xf X (x)x + f X (x)x Further, E(Y 2 ) = We conclue that = (/5)[x 2 /2] + (/5) = 2/5. 5 x 2 f X (x)x + 2 f X (x)x = (/5)[x 3 /3] + (6)(/5) = 2/5. Var(Y ) = (2/5) (2/5) 2 = Notation: X is the amount of loss, Y is the payment, is an unknown euctable. Given: Y = max(, X ), f X (x) = (/)I( < x < ). Fin: such that E(Y ) = (.25)E(X). Solution: Let us euce a general formula for a mean payment on a policy with euctable. Write, E(Y ) = E(max(, X )) = () f X (x)x+ (x )f X (x)x = (x )f X (x)x. ()

2 Remark: Using integration by parts we get another formula, (x )f X (x)x = [ (x )( F X (x)] x= + ( F X (x))x = ( F X (x))x. (2) This formula may be useful if the cf is given. Its rawback is that you nee to memorize it. Thus, I prefer to stick with the iea of () but it is goo to know the alternative (2). In our particular case we use () an write 3 E(Y ) = 3 (x )x = 3 (/2)( 3 ) 2. At the same time E(X) = 5 so we get the equation whose solution yiels = 5. (/2)( 3 ) 2 = (.25)(5). Notation: N - amount of loss, L - the event that loss occurre, - euctable, m - net premium, k - an unknown constant. Given: P(L) =.5, p N L (x) = kx I(x {, 2, 3,, 5}), = 2. Fin: Net premium m which is the expecte premium pai to a policyholer. Solution: Net premium is m = E(max(, N )) = ()P(N ) + We nee to calculate the pmf p N (x). Write l=+ (x )p N (x). p N (x) = P(N = x) = P((N = x) L) + P((N = x) L ) = P(L)P(N = x L) + P(L )P(N = x L ) =.5p N L (x) + =.5kx I(x {, 2, 3,, 5}). Now we can finish the calculation of the net premium, 5 m = (.5)k (x 2)x = (.5)k[(/3) + (2/) + (3/5)] = k(.77). x=3 To fin k we note that x p N L (x) = which is equivalent in our case to k[ + /2 + /3 + / + /5] = implying k =.38. We conclue that m = (.38)(.77) =.3. Answer: A 5. Notation: X is the amount of loss, C is the amount of claim. Given: F X (x) = for x <, F X (x) =.2 +.3x for x < 2, F X (x) = for x 2, C = αx, 2

3 α is an unknown constant from (, ), E(C) =.5. Fin: α. Solution: Attention: The istribution of X is a mixe istribution where P(X = ) = P(X = 2) =.2 an over the interval [, 2) the istribution of X is continuous with the pf f X (x) =.3I( x < 2). Please plot the cf an see this! Note that it is always absolutely pruent to check a istribution on the presence of iscrete components whenever the istribution is given via its cf! Now let us solve the problem at han. Since E(C) = αe(x), we calculate the expectation of X. Write, E(X) = ()(.2) + 2 Then the equation (α)() =.5 implies α =.5. Answer: D x(.3)x + (2)(.2) = (.3)(/2) +. =. 6. Notation: is the amount of euctable, PD is the event of Partial Damage, TD is the event of Total Damage, ND is the event of No Damage, X is the amount of loss. All amounts are in thousans! Given: P(PD) =., P(TD) =.2, P(ND) = (.2 +.) =.9, f X PD (x) =.53e x/2 I( < x < 5), =, p X TD (x) = I(x = 5). Fin: E(max(, X )) Solution: Remember that the units of losses are in thousans. Write using the total probability theorem, E(max(, X )) = ()P(X )+()P(ND)+P(PD) (x )f X PD (x)x+p(td)(5 ) (plug in the given information) (using integration by parts) 5 = (.) (x )(.53)e x/2 x + (.2)(5 ) 5 = (.){[(x )(.6e x/2 ] 5 x= (.6)e x/2 x} +.28 = (.)[ ] +.28 =.3282 This is the answer in thousans, so the aske quantity is Answer: B 7. Solution: Let X be the insurance payment, Y is the repair cost, is the euctable. Then, given the car is amage, X = max(, Y ). The question is about the stanar 3

4 eviation, an we can calculate it only as a square root of the variance. In its turn, we can calculate the variance via the first an secon moments. Let us calculate the moments. Write, Further, E(X 2 ) = E(X) = (y )f Y (y)y = 5 25 (y 25)(/5)y = (/5)[(/2)(y 25) 2 ] 5 y=25 = (25) 2 /3 = (y ) 2 f Y (y)y = (/5)[(/3)(y 25) 3 ] 5 y=25 = (25) 3 /5 = 35, 28. Now we calculate the variance Var(X) = E(X 2 ) [E(X)] 2 = 3, 28 [52.83] 2 = 62, 76. an then the stanar eviation is [62, 76] /2 = 3. Answer: B 8. Notation: N is the number of storms, X is the amount pai. Given: N is Poisson with the mean λ =.5, an X =, max(, N ). Fin: E(X). Solution: This is a rewore euctable problem. Write E(X) = 5 (x )p N (x) = 5 (x ) e λ λ x x= x= = 5 [ e λ λ x (x )! e λ λ x ] x! (in what follows I use the fact that the Poisson pmf is summable to ) x= x! = 5 e λ λ x {λ x= (x )! [ e λ λ ]} = 5 e λ λ y {λ! y= y! [ e λ ]} = 5 {λ + e λ } (I use the given λ =.5) = 5 [ ] = Notation: L is the amount of loss, is euctable. Given: f L (x) = (/3)e x/3 I(x > ), =. Fin: a constant c such that P( < X < c X > ) =.95. Solution: Write,.95 = P( < X < c X > ) = P( < X < c) P(X > )

5 = c (/3)e x/3 x (/3)e x/3 x = e /3 e c/3 e /3 = e (c )/3. Now plug in =, solve the obtaine equation an get c = 999. Answer: E. Solution: Since X is exponential we immeiately get c =.. Let Y enote the benefit. Then F Y (y) = for y, F Y (y) = y f(x)x for < y < 25, F Y (25) = 25 f(x)x, an F Y (y) = for y > 25. For this cf we must fin a meian m which is efine as solution of the equation F Y (m) = /2. Let us check if m (, 25). Write for some z (, 25), F Y (z) = z z f(x)x =. e.x x = [ e.x ] z x= = [ e.z ]. Equate the right sie of the last equality to /2 an get z = 73. Because 73 (, 25), we conclue that the meian m = 73. 5