Egzamin dla Aktuariuszy z 11 października 2003 r.
Prawdopodobieństwo i Statystyka
Zadanie 1
10 ⋅ 5 !
4
10
P( X = )
1 =
=
5 !
5
55
9 ⋅ 5 !
4
9
P( X = 2) =
=
5 !
5
55
.......
5 !
4
1
P( X = 10) =
=
5 !
5
55
10
EX
∑ i 1(1−
=
i) = 1 (11⋅55 − ) = 220
385
= 4
i=
55
55
55
1
Zadanie 2
=
=
P( S = k A =
n
) P( AS k
n
) P( S k
n
)
P( )
A
n
n
!
P( A) = ∑ k n k n− k
k
n
k
n k
a
p 1
( − p)
= ∑
−
a
p 1
( − p)
=
n k
n k!( n
k)!
k =0
k =1
−
n
n−1
= ∑
( n − )
1 !
n 1
k
n−
a
p 1
( −
k
p)
= k −1 = l = ∑ − l+1
n− l−1
ap
a
p
1
( − p)
=
( k
)
1 !( n
k)!
l
1
p
k =1
−
−
l =0
−
2
1
1
1
E(
k
n
p
p
n
S A
a
p 1
(
p)
kp 1
(
p)
k
1
l
n
) n
n
= ∑
k
n− k
−
−
k
n k
−
=
∑ −
−
−
= − = =
n
k
ap
p
k
1
k =0
k =1
−
n−
1 −
1
=
p ∑ n −
1
l +1
n− l−1
1 − p
p
( l + )
1 p
1
( − p)
=
(( n − )1 p + )1 = np − p +1
p
l
p
p
l =
1
0
−
Zadanie 3
n
t
P( M ≤ t) =
θ
n
θ
1
100
20
n
n
P θ
( < aM ) = P M >
= 1− = ,
0 05 → a =
=
a
a
95
19
n
θ
1
n
n
P M <
= = ,
0 05 → b = 20 → b = 20
b
b
20
n
n
bM − aM = M 20 −
19
2
2
2
2
2
2
2
ES = var S + ( ES ) = d µ + σ
m
+ µ m
ES = m
µ
2
2
E( SN ) = cov( S, N ) + ESEN = d µ
+ m
µ
cov( S, N ) = E(cov( S, N N ) + cov( E( S N ), E( N N ) = cov( Nµ N ) 2
,
= d
µ
E( 2 2
a S + 2 abS − 2 aSN − 2 bN +
2
N ) =
2
= a ( 2 2
2
2
2
d µ + σ
m
+ µ m )+ 2 ab( m µ )
2
+ b − 2 a( 2
2
d
µ
+ m
µ
)−2
2
2
bm + d + m → min
∂ = 2( 2 2
2
2
2
d µ + σ
m
+ µ m ) a + 2 b m µ
− 2 2
d
µ
− 2
2
m
µ
= 0
∂ a
∂ = 2 a m
µ
+ 2 b − 2 m = 0
∂ b
b = m − a m
µ
(2 2 2
d µ + 2
2
σ
m
+ 2 2 2
µ m ) a + 2 m
µ ( m − a m
µ ) − 2
2
d
µ
− 2
2
m
µ
= 0
2
2
2
2
2 d
µ
+ 2 m
µ
− 2 m
µ
d
µ
a =
=
2
2
2
2
2
2
2
2
2
2
2 d µ + 2 σ
m
+ 2 µ m − 2 µ m
d µ + σ
m
2
2
2
2
2
2
2
2
2
d
µ
m
µ
md µ + m σ − µ d m m σ
b = m −
=
=
2
2
2
2
2
2
2
2
2
d µ + σ
m
d µ + σ
m
d µ + σ
m
2
∂
2
2
2
2
2
= 2 d µ + 2 σ
m
+ 2 µ m
2
a
∂
2
∂ = 2
2
∂ b∂ =2 mµ
a
∂ b
∂
2 2 2
d µ + 2
2
σ
m
+ 2 2 2
µ m
2 m
µ
= 4 2 2
d µ + 4
2
σ
m
+ 4 2 2
µ m − 4 2 2
µ m > 0 bo m>0 bo N
2 m
µ
2
nieujemne
Z tego wynika, że jest to minimum Zadanie 5
10
L =
6
P ( X > 10 ⋅ ∏ 1
)
θ
i=1
θ −10
ln L = 6 ln
−10ln θ
θ
∂
6 θ
10
10
60 −10( θ −10)
ˆ
=
−
=
= 0 → θ = 16
∂ θ θ −10 2
θ
θ
θ( θ −10)
∑( Xi− )2
1
−
2
e
∑ Xi−1
H
H
=
n
2
e
X
N
n
X
N n n
2
∑
0
1
i ≅
( ,
0
∑
)
i ≅
∑
( , )
X
−
i
2
e
1
≅ }
N (0 )
1
,
∑ X − n
i
1
1 1
1
1
P e
2
> t = P ∑ X − n > ln t = P
X
> ln t + n
= → ln t = − n
0
0
i
0
2
2
n
2
2
1
β = P
ln
0
1 ∑ X
< t + n = P 1 ∑
< = 1
< −
= −
n
i
( Xi ) P ( X
n ) F (
n )
2
− 1
1
− / 2
1
e n
2Π n
β
H
n
2 n
2Π
2Π
lim
= lim
⋅ 2Π =
n
n
lim
=
−
e n / 2 / 2Π n
1 − n/ 2
− n / 2
1
2Π
− e
2Π n − e
2Π
2Π n +
2
2 n
n
2Π n n
2Π n
1
= lim
= lim
= lim
= 1
1
2Π n + 2Π
2Π ( n + )
1
1 + n
Zadanie 7
48
48
+
4
5
4
1 −
52
A = P(
P A
A
5
2
1
≥
A
A
2
≥1 )
( ≥ ∩ ≥ )
=
P( ≥
A 1 )
=
48
5
1 − 52
5
a
2 sy
a
3 sy
8
7
6
8
7
6
}
a
4 sy
48
48
3
+
3
+ 48
3
2
52
B = P(
5
≥
A
A
2
pik )
=
5
1
4
52
5
52
!
48
4 ⋅
!
48
−
−
3 ⋅
!
48
3 ⋅
!
48
+
+ 48
!
47
!
5
!
43
!
5
!
44
!
4
⋅
A =
≈ 1222
,
0
!
45
!
3
2
!
46
B =
≈ ,
0 2214
!
52
!
48
−
!
51
!
47
!
5
!
43
!
5
!
47
!
4
Zadanie 8
2
≅ χ )
1
(
P( X < Y )
}
2
2
= P X < Y
Y = 3 Z Z ≅ N (
)
1
,
0
X
= P( X < 3 Y ) = P
< 3 = P( X < ) 3 X ≅ F
)
1
,
1
(
Y
−1
3
2
1
x
x =
3
∫
dx =
t =
1
2
2
dt
arctg 3
(
)
5
,
0
2
∫
2
Γ ( )
5
,
0
( x + )
1
x
t
(
)
5
,
0
t
1
(
)
5
,
0
0
= 2
Γ2
2
2
0
( + ) =
Γ
= Π
Γ
f ( x) = arctgx 1
f (
′ x) =
2
1 + x
2 x
f (
′ x) = − (
+ x )2
2
1
2
6 x − 2
f (
′ x) = (
1 + x )3
2
12 x(1+ x )3
2
− ( 2
6 x − 2)⋅ 3⋅ (1+ x )2
2
2 x
f IV ( x) =
(
1 + x )6
2
rozwinięcie Taylora wokół x = 1
0
f IV
)
1
(
≈ 0 → arctg 3 ≈ ,105 bo: Π 1
1 ( x − )
1 2
4 ( x − )
1 3
+ ( x − )
1 −
+
4
2
2
2
8
6
N
( k
f
) ( x 0 )( x − x 0 ) k f ( x) ≈ ∑
k!
k =0
Z tego odpowiedź około 0,6667
Zadanie 9
1 1
θ
X
∏ − i
L =
n
θ
1
ln L =
−
1 ∑ ln X
ln
i − n
θ
θ
∑ln X
+ ∑ln
∑ln
i
n
θ
n
X
X
i
= −
− = 0 →
= → θ
i
ˆ
0
= −
θ
∂
θ 2
θ
θ 2
n
2
∂
2∑ ln X
n
2
X
θ
n
i
∑ln i +
=
+
=
< 0 → max
ˆ
2
3
2
3
∂
θ
θ
θ
θ
θ
1 −
1
1
1
θ
1
t
P(
−
−
x
ln X < t)
1
= P
< t
e = P( X > − t e ) = ∫
dx = θ
x
= − θ
e
≅ wykl(1
1
θ )
X
−
θ
t
− t
e
e
− ∑
1
ln X
,
i ≅ Γ n
θ
+
+ −
+
2
2
n 1 2
2
2
2
n 1 2 n
n
θ 2
R θ
(
= Eθˆ
)
− E
θ θˆ
2
+ θ =
θ − θ
2
+ θ = θ
=
n
n
n
+
2
1
2
n 1 2
ˆ
Eθ =
n( n + )
1 θ =
θ
2
n
n
Eθ = 1
ˆ
⋅ nθ = θ
n
Zadanie 10
EY = βx
i
i
2
2
var Y = x σ
i
i
∑ βc x β
c x
i
i =
→ ∑ i = 1
i
2
2
1
1
βˆ
var
= ∑( c x σ
mi
n m
in g
dy c
x
c
i
i )
→
i
i =
→ i =
n
nxi
1
1
Y
Z tego: βˆ = ∑ c Y
Y
i i = ∑
= ∑ i
i
nx
n
x
i
i