Estadistica y procesos estocasticos

1. Universidad Nacional de San Agustı́n
Facultad de Ingenierı́a de Producción y Servicios
Escuela Profesional de Ingenierı́a Electrónica
Tareas Resueltas
Docente:Dr. Alexander Hilario Tacuri
Frank Jonathan Cruz Huachaca
Arequipa-2022

2. Índice
1. Tarea 1 1
2. Tarea 2 1
3. Tarea 3 1
4. Tarea 4 1
5. Tarea 5 2
6. Tarea 6 2
7. Tarea 7 2
8. Tarea 8 3
9. Tarea 9 3

3. Estadı́stica y procesos estocásticos
Resolucion de Tareas
1. Tarea 1
Resolver:
P(y = 0|e) =
P(e|y = 0)P(y = 0)
P(e)
P(y = 0|e) =
(1 − p + q)(1 − q + p)
1 − p+q
2
P(y = 0|e) =
1 − (p − q)2
1 − p+q
2
2. Tarea 2
Para que si sea una Variable Aleatoria (V.A.):
A3 = {ϕ, Ω, {W3} , {W2, W4} , {W1, W5} , {W6}}
3. Tarea 3
Resolver:
P(x < t + 5) = 1 −
Z 5
∞
Z 0
x−5
0,06e(−0,15T+0,4x)
u(T)u(x)dTdx
P(x < t + 5) = 1 −
Z 5
∞
−0,4(e(0,75T−0,55x)
e(−0,4x)
)dx
P(x < t + 5) = 1 − (−0,4)((0 − 1,81e−2
) − (0 − e−2
))
P(x < t + 5) = 1 − (−0,4)2,81e−2
P(x < t + 5) = 0,96
4. Tarea 4
Resolver:
E[e2
2
] =
1
2
[
Z −1/2
1
(x + a)2
dx +
Z 1/2
−1/2
x2
dx +
Z 1
1/2
(x − a)2
dx]
Ee2
2
] =
1
2
[(
7
24
−
3a
4
+
a2
2
) +
1
12
+
12a2 − 18a + 7
24
]
E[e2
2
] =
1
2
[
6a2 − 9a + 4
6
]
E[e2
2
] =
1
2
[a2
−
3a
2
+
2
3
]
Derivamos para hallar el minimo:
d
da
1
2
[a2
−
3a
2
+
2
3
] = 0
a =
3
4
Ingenierı́a Electrónica 1

4. Estadı́stica y procesos estocásticos
5. Tarea 5
Demostrar Mx(0) = 1:
Mx(v) =
Z ∞
−∞
ejvx
Px(x)dx
Reemplazamos v = o:
Mx(0) =
Z ∞
−∞
ejv0
Px(x)dx
Mx(0) =
Z ∞
−∞
Px(x)dx
Mx(0) = 1
6. Tarea 6
Demostrar |Mx(t)| ≤ 1:
|Mx(v)| ≤ 1; v ∈ ℜ
|Mx(v)| ≤ |
Z ∞
−∞
ejvx
dFx(x)|
|Mx(v) ≤
Z ∞
−∞
|ejvx
|dFx(x)
|Mx(v) ≤
Z ∞
−∞
dFx(x)
|Mx(t)| ≤ 1
7. Tarea 7
Resolver E[x2
2] =? :
k1 = 0; k2 = 2
E[x2
2
] = (−j)2 d2Mx(v)
dv2
2
|v=0
E[x2
2
] = −1[
d
dv2
Mx(v)(−4v2 − v1)]|v=0
E[x2
2
] = −1[Mx(v)(−4) + (−4v2 − v1)Mx(v)(−4v2 − v1)]|v=0
E[x2
2
] = 4
Ingenierı́a Electrónica 2

5. Estadı́stica y procesos estocásticos
8. Tarea 8
Hallar kx y k−1
x :
Hallamos kx:
kx = BkaBT
=
t1 1
t2 1
1 1
2
1
2 b
t1 t2
1 1
kx = BkaBT
=
t1 + 1
2
t1
2 + 1
t2 + 1
2
t2
2 + 1
t1 t2
1 1
kx = BkaBT
=
t2
1 + t1 + 1 2t1t2+2t1+2t2
2 + 1
2t1t2+2t1+2t2
2 + 1 t2
2 + t2 + 1
Hallamos k−1
x :
1
|k|
= (
1
t2
1t2
2 + t2
1t2 + t1t2
2 + t2
1 + t2
2 + t1 + t2 + 1
) − (
4
4t1t2 + 4t1 + 4t2 + 8
)
Adj(kx) =
t2
2 + t2 + 1 −2t1t2−2t1−2t2−2
2
−2t1t2−2t1−2t2−2
2 t2
1 + t1 + 1
k−1
x =



t2
2+t2+1
( 1
t2
1t2
2+t2
1t2+t1t2
2+t2
1+t2
2+t1+t2+1
)−( 4
4t1t2+4t1+4t2+8
)
−t1t2−t1−t2−1
( 1
t2
1t2
2+t2
1t2+t1t2
2+t2
1+t2
2+t1+t2+1
)−( 4
4t1t2+4t1+4t2+8
)
−t1t2−t1−t2−1
( 1
t2
1t2
2+t2
1t2+t1t2
2+t2
1+t2
2+t1+t2+1
)−( 4
4t1t2+4t1+4t2+8
)
t2
1+t1+1
( 1
t2
1t2
2+t2
1t2+t1t2
2+t2
1+t2
2+t1+t2+1
)−( 4
4t1t2+4t1+4t2+8
)



9. Tarea 9
Resolver:
Ryτ =
Z ∞
−∞
F1(τ + γ)h(γ)dγ
Ry(t1, t2) = E[y(t1)y(t2)] =
Z ∞
−∞
Z ∞
−∞
Rx(α + β)h(t1 − α)h(t2 − β)dαdβ
λ = t1 − α
γ = t2 − β
Ry(t1, t2) =
Z ∞
−∞
Z ∞
−∞
Rx(t1 − t2 − λ + γ)h(λ)h(γ)dλγ = RY τ
RY τ = h(−τ) ∗ h(τ) ∗ Rx(τ)
Ingenierı́a Electrónica 3