Логика. Множества. Вероятность. Лексаченко В.А. - 106 стр.

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R e x e n i e dl sistemy (
ξ, η)
s zakonom raspredeleni:
ξ η −2 −1 0 p
ξ
(
x
j
)
0 0
, 05
0
, 05
0
, 15
0
, 25
1 0,
25
0,
10
0,
10
0,
45
2 0,
20
0,
05
0,
05
0,
30
p
η
(
y
k
) 0
, 50
0
, 20
0
, 30
.
Zakony raspredeleni
ξ
i
η predstavlts tablicami
x
j
0 1 2
p(
x
j
) 0,
25
0,
45
0,
30
,
y
k
−2 −1 0
p(
y
k
) 0,
50
0,
20
0,
30
.
Mξ
=
P
j
x
j
p
ξ
(x
j
) = 1, 05 , Mη
=
P
k
y
k
p
η
(
y
k
) = −1,
2
,
Mξ
2
=
P
j
x
2
j
p
ξ
(
x
j
) = 1,
65
, Mη
2
=
P
k
y
2
k
p
η
(
y
k
) = 2, 2 ,
Dξ = Mξ
2
−
(Mξ)
2
= 0,
5475
, Dη
=
Mη
2
−
(Mη)
2
= 0
,
76 .
M(ξη) =
P
j
P
k
x
j
y
k
p
ξη
(
x
j
, y
k
) =
−0, 5
− 0
,
1
− 0
, 8 −
0
,
1 = −
1
,
5 ,
K
ξη
=
M
(
ξη) −MξM η
= −1
,
5 −
1,
05
· (−
1,
2) = −0, 24 .
J
13
. Opredelit~ c, f
ξ
(
x), Mξ, Dξ, f
η
(
y), Mη, Dη, K
ξη
, esli
f
ξη
(
x, y
) = g(
x, y
) pri (
x, y) ∈ 4ABC
, f
ξη
= 0 pri (x, y
) /
∈ 4
ABC
.
1)
g
(x, y
) =
c(2
x+2y)
,
A
= (0
, 0), B = (0
, 1), C
= (2
, 0)
.
2)
g(x, y) = 2
cxy , A
= (0
, 0)
, B = (0
,
2), C = (1
, 0)
.
3)
g
(x, y
) =
c(3 + 3y
),
A = (0, 0), B = (0
, 1), C = (2,
0).
4) g
(x, y) = 3cx, A
= (0, 0), B = (0
, 2), C
= (1, 0).
5) g
(x, y
) =
c
(2x
+2y)
,
A
= (0,
2)
, B
= (1,
2)
, C = (1, 0)
.
6) g
(x, y
) = 2
cxy
, A = (0,
1)
, B = (2,
1)
, C = (2,
0).
7)
g(x, y
) =
c(3 + 3
y), A
= (0
, 2), B = (1,
2)
, C
= (1, 0)
.
8) g(x, y) = 3cx
, A
= (0, 1)
, B = (2,
1)
, C = (2, 0).
9)
g
(x, y
) = c
(3
x
+3
y)
,
A = (0
, 0), B = (2
,
1)
, C
= (2, 0).
10) g
(
x, y
) = 2cxy
,
A = (0, 0)
, B
= (1
, 2), C
= (1
,
0)
.
R e x e n i e dl g(x, y
) = c
(2x+3y),
A = (0, 0), B
= (0, 1)
, C = (1, 0).
1) Opredelim postonnu c
iz uslovi normirovki
1 =
∞
R
−∞
∞
R
−∞
f
ξη
(
x, y)dxdy
= c
R R
4ABC
(2
x + 3
y)
dxdy
= c
1
R
0
1
−
x
R
0
(2
x
+ 3
y
)dy
dx
=
= c
1
R
0
2
xy + 1
,
5
y
2
1
−
x
0
dx
= c
1
R
0
(1
,
5
−
x−
0, 5
x
2
)
dx
=
5c
6
, otkuda
c
= 1
, 2
.
2) f
ξ
(
x
) =
∞
R
−∞
f
ξη
(
x, y
)
dy =
R
∞
−∞
f
ξη
(x, y)dy, x ∈
[0; 1]
R
∞
−∞
f
ξη
(
x, y)
dy, x /
∈ [0; 1]
=
=
1
, 2
1
−
x
R
0
(2x
+ 3
y
)
dy, x ∈
[0; 1]
0, x /∈ [0; 1]
=
(
1
, 2
2
xy
+ 1, 5
y
2
1−
x
0
, x ∈
[0; 1]
0 x /
∈ [0; 1]
=
106

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