practica 6

1560 - An´lisis de Se˜ ales Aleatorias.
a
n

Soluci´n Tarea 3
o

Abril del 2013.

1.1

1

(4 − y 2 − 0.25x2 ) dy dx

1=k
−1

→

k=

−1
1

fXY (x, y) dy =

6 11 1 2
− x
433
4

fXY (x, y) dx =

6 47
− y2
43 12

fX (x) =
−1
1

fY (y) =
−1

3
43

−1≤x≤1

−1≤y ≤1

1

E{X} =

xfX (x) dx = 0
−1
1

E{Y } =

yfY (y) dy = 0
−1
1

2
σX

22

x2 fX (x) dx = 0.327

= E{(X − µX ) } = E{X } =
−1
1

2
σY

2

2

y 2 fY (y) dy = 0.3085

= E{(Y − µY ) } = E{Y } =
−1

El centroide es
C = (µX , µY ) = (0, 0)
Lacorrelaci´n
o
1

1

−1

−1

RXY = E{XY } =

xy fXY (x, y) dy dx = 0

La covariancia
CXY = RXY − µX µY = 0
El coeficiente de correlaci´n
o
ρ=

CXY
=0
σX σY

ˆ
El estimador Y = aX esRXY
ˆ
Y = aX =
X =0
E{X 2 }
La probabilidad del evento {X 2 − Y ≤ 0} es
1

1

−1

x2

2

P ({X − Y ≤ 0}) =

UNAM - FI

fXY (x, y) dy dx = 0.328

Prof. C. Rivera

1560 - An´lisisde Se˜ ales Aleatorias.
a
n

Soluci´n Tarea 3
o

Abril del 2013.

2.1

1

1

1=

1

(4 − m2 − 0.25n2 ) = 28.5k

P (X = n, Y = m) = k
n=−1 m=−1

n=−1 m=−1

2
57
Lasprobabilidades marginales de X son
k=

1

P (X = n) =

P (X = n, Y = m) =
m=−1

2
(4 − m2 − 0.25n2 )
57

Para n = −1
1

P (X = −1) =

P (X = −1, Y = m) =
m=−1

37
2
(4 − m2 −0.25(−1)2 ) =
57
114

Para n = 0
1

P (X = 0, Y = m) =

40
2
(4 − m2 − 0.25(0)2 ) =
57
114

P (X = 1, Y = m) =

P (X = 0) =

37
2
(4 − m2 − 0.25(1)2 ) =
57
114

m=−1

Para n = 1
1P (X = 1) =
m=−1

Las probabilidades marginales de Y son
1

P (Y = m) =

P (X = n, Y = m) =
n=−1

2
(4 − m2 − 0.25n2 )
57

Para m = −1
1

P (Y = −1) =

P (X = n, Y = −1) =
n=−12
17
(4 − (−1)2 − 0.25(n)2 ) =
57
57

Para m = 0
1

P (X = n, Y = 0) =

23
2
(4 − (0)2 − 0.25(n)2 ) =
57
57

P (X = n, Y = 1) =

P (Y = 0) =

17
2
(4 − (1)2 − 0.25(n)2 ) =...