Теория вероятностей. Барышева В.К - 109 стр.

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f
1
(x) =
+∞
R
−∞
f (xy) dy =
1
2
π/2
R
0
sin (x + y) dy =−
1
2
cos (x + y)
π/2
0
=
= −
1
2
h
cos
x +
π
2
− cos x
i
=
1
2
(sin x + cos x)
f
2
(x) =
+∞
Z
−∞
f (xy) dx =
1
2
π/2
Z
0
sin (x + y) dx =
1
2
(sin x + cos x)
f
1
(x) =
1
2
(sin x + cos x) , x ∈ [0, π/2]
0, [0, π/2]
f
2
(y) =
1
2
(sin y + cos y) , y ∈ [0, π/2]
0, [0, π/2]
.
f
1
(x) ·f
2
(y) 6= f (xy) X Y
f (x|y) =
f (x, y)
f
2
(y)
=
sin (x + y)
sin y + cos y
0 6 x 6 π/2, 0 6 y 6 π/2
0
f (y|x) =
f (x, y)
f
1
(x)
=
sin (x + y)
sin x + cos x
0 6 x 6 π/2, 0 6 y 6 π/2
0
N
(X, Y )
Y
1 2 3
0 0, 2 0, 1 0, 1
1 0, 2 0, 1 0, 1
2 0, 1 0, 1 0

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