to Answers Practice Questions 1 ACTSC 431/831, FALL 2011
1. (a)
FY (y ) = Pr{Y y |N = 0} Pr{N = 0} + Pr{Y y |1 N 5} Pr{1 N 5}
+ Pr{Y y |N > 5} Pr{N > 5}
0,
y < 0,
y/100
y/200
1 0.5 e
0.2 e
, y 0.
=
(b) The probability is Pr{Y > 500} = 1 FY (500) = 0.019786.
(c) Pr{0 < Y < 300} = FY (300) FY (0) = 0.63048.
(d) The mean of the total loss is E (Y ) = 0.3 0 + 0.5 100 + 0.2 200 = 90.
(e) The 20th percentile of Y is 0.2 = 0.
(f) The 80th percentile of Y is the root of equation: FY (y ) = 0.8, y > 0. Solving the
equation, we obtain that the 80th percentile of Y is 0.8 = 153.87.
2. (a) Note that 0 Y L 70.
FY L (y ) =
0,
Pr{X 80 y } = 80+y ,
200
Pr{40 + 0.75(X 120) y } =
1,
y < 0,
0 y < 40,
y +50
, 40 y < 70,
150
y 70.
(b) Note that Y P = Y L |X > 80 and 0 < Y P 70.
FY P (y ) =
0,
y
Pr{X 80 y | X > 80} = 120 ,
Pr{40 + 0.75(X 120) y | X > 80} =
1,
y 0,
0 < y < 40,
y 10
, 40 y < 70,
90
y 70.
(c) The probability is FY L (50) FY L (40) = 1/15.
(d) The probability is FY P (70) FY P (30) = 5/12.
(e) The mean of the per loss of the policy is
E (Y L ) =
or E (Y L ) =
0
120
160
1
1
(40 + 0.75(x 120))
dx +
dx
200
200
80
120
200
1
+
70
dx = 29
200
160
(x 80)
SY L (y )dy = 29.
(f) The median of Y P is the solution to FY P (y ) = 0.5, 40 < y < 70. Hence 0.5 = 55.
(g) The 90th percentile of Y P is 0.9 = 70.
3. Note that the cdf and sf of X are
0,
x < 0,
x
, 0 x < 200,
F ( x) =
200
1,
x 200,
1,
x < 0,
200 x
and S (x) =
, 0 x < 200,
200
0,
x 200.
1
(a) The cdf of Y P is given by
0,
y 0,
y
S (y + 50)
, 0 < y < 150,
=
F Y P (y ) =
150
1
, y > 0,
S (50)
1,
y 150,
y 0,
0,
which is the cdf of the uniform distribution U (0, 150).
(b) The cdf of Y L is given by
FY (y L ) =
0,
y < 0,
0,
y < 0,
y + 50
=
, 0 y < 150,
200
F (y + 50), y 0,
1,
y 150.
(c) The cdf of Y is given by
FY (y ) =
0,
y < 0,
y
F (y ), y < 50,
, 0 y < 50,
=
200
1,
y 50,
1,
y 50.
(d) The expectation of the per-loss variable is given by E (Y L ) =
56.25
200
1
50 (x 50) 200 dx
=
(e) We have E (Y ) = E (X 50) = 050 S (y )dy = 43.75 or E (Y ) = E (X 50) =
50
50
2
2
0 xf (x)dx +50 (1 F (50)) = 43.75, and E (Y ) = E [(X 50) ] = 2 0 yS (y )dy =
2083.33 or E (Y 2 ) = E [(X 50)2 ] = 050 x2 f (x)dx + 502 (1 F (50)) = 2083.33.
Hence, the variance of the limit loss variable is V ar(Y ) = 169.27.
(f) By (a), we know that Y P has the uniform distribution U (0, 150), hence e(50) =
L)
X 50)
E (Y P ) = 75 or e(50) = E [(F (50)+ ] = 1E (Y(50) = 75.
1
F
(g) We have V ar[(X 50)2 |X > 50] = V ar[(Y P )2 ] = E [(Y P )4 ] (E [(Y P )2 ])2 .
By (a), we know that Y P has the uniform distribution U (0, 150), hence E [(X
50)2 |X > 50] = E [(Y P )2 ] = V ar(Y P ) + E [(Y P )]2 = 7500 or E [(X 50)2 |X >
(x50)2 f (x)dx
50] = E [(Y P )2 ] = 0 y 2 fY P (y )dy = 7500 or E [(X 50)2 |X > 50] = 50 1F (50)
=
3
P4
7500. Furthermore, E [(Y ) ] = 4 0 y SY P (y )dy = 101250000. Thus, V ar[(X
50)2 |X > 50] = 4.5 107 .
4. (a) We know that X1 + X2 still has a normal distribution. Thus, V aR (X1 + X2 ) =
2
2
1 + 2 + 1 + 2 + 21 2 1 ().
(b) Note that 1 < < 1. Thus, it is easy to verify that V aR (X1 + X2 )
V aR (X1 ) + V aR (X2 ) for 0.5 < 1, and V aR (X1 + X2 ) > V aR (X1 ) +
V aR (X2 ) for 0 < < 0.5.
(c) Since X1 + X2 still has a normal distribution, we have T V aR (X1 + X2 ) =
1 (
2
2
1 + 2 + 1 + 2 + 21 2 ( )) .
1
(d) Since 1 < < 1, (x) > 0 for all x (, ), it is easy to see that
T V aR (X1 + X2 ) T V aR (X1 ) + T V aR (X2 ) for all 0 < < 1.
2