Теория вероятностей. Барышева В.К - 107 стр.

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X Y
f
1
(x) = 5
x
ln 5 > 0
f
2
(y) = 5
y
ln 5 > 0.
F
1
(x) · F
2
(y) =
1 5
x
1 5
y
= 1 5
x
5
y
+ 5
xy
= F (x, y) ,
X Y
f(x, y) =
F
00
xy
(x, y) :
f(x, y) = F
00
xy
(x, y) =
(
5
xy
ln
2
5 x 0, y 0
0, x < 0 y < 0
f(2, 1) = 5
3
ln
2
5.
y
Z
0
x
Z
0
5
uv
ln
2
dudv =
y
Z
0
5
v
ln 5 dv ·
x
Z
0
5
u
ln 5 du
= (1 5
y
)(1 5
x
) = 1 5
x
5
y
+ 5
xy
P (1 6 X 6 2, 1 6 Y 6 2) = [F (1, 1) + F (2, 2)] [F (1, 2) F (2, 1)] =
=
1
1
5
1
5
1
5 · 5
+
1
1
25
1
25
+
1
625
1
1
5
1
5
+
1
125
1
1
25
1
5
+
1
125
=
16
625
N