Document

Cálculo de Probabilidades II
Respuestas Tema 2
1. Si
Xi ∼ Ber(p) y
Y =
n
X
xi
i=1
entonces,
MY (t) = E(E ty )
= E(etx1 etx2 ...etxn )
n
Y
=
E(etxi )
i=1
t
= (e p + (1 − p)n )
Y ∼ Bin(n, p)
2. Sea
X ∼ χ2k
Y
Y = Z12 + Z22 + ... + Zn2
→ MY (t) = E(ety )
n
Y
2
=
etk(zi )
i=1
= (1 − 2t)nk/2
→ Y ∼ χ2P ni
3. Sea Y = aX + b
→ MY t = E(ety )
= E(et(aX+b) )
= etb MX (at)
4. Sea
f (x; µ, σ) =
1
√
xσ 2π
2 /2σ 2
e−(lnx−µ)
E(X) = eµ+σ
2
2 /2
V ar(X) = e2µ+2σ − e2µ+σ
1
2
5. Sea X ∼ N(0, 1), p.d.
Z = X 2 ∼ χ2(1)
Dem.
∞
Z
e−x
MZ (t) =
2 (t− 1 )
2
0
=
1
t−
1
√ dx
2π
1
2
→ Z = X 2 ∼ χ2(1)
6. Sean
Xi ∼ Ga(α, 1)
n
X
Y =
Xi
i=1
(a) Con n = 2
Z
∞
fX1 (x1 )fX2 (y − x1 )dx1
fX1 ,X2 (y) =
Z−∞
y
fX1 (x1 )fX2 (y − x1 )dx1
=
0
Z
=
0
y
xα−1
e−x1 (y − x1 )α−1 e−(y−x1 )
1
dx1
Γ(α)
Γ(α)
Sea
x1 = yt dx1 = ydt
y sustituyendo en la integral anterior,
−y α+α−1
Z
fX1 ,X2 (y) = e y
1
ytα−1 (1 − t)α−1
0
−y 2α−1
=
e y
∼ Ga(2α, 1)
Γ(2α)
(b) Sea
Y ∼ Ga(α, n) y
W = 2nY.
P (W ≤ w) = P (2nY ≤ y)
w
= P (Y ≤
)
2n
w
= FY ( )
2n
2
1
dt
Γ(α)Γ(α)
w
1
fY ( )
2n
2n
wα−1 −w/2 1
=
e
∼ χ22α
Γ(α)
2α
→ fW (w) =
(c) Sea
X ∼ Ga(α, 1) y
Y =
X
n
X
≤ y)
n
= P (X ≤ ny)
= FX (ny)
P (Y ≤ y) = P (
→ fY (y) = n ∗ fX (ny)
1 ny α−1 α
=
e y n ∼ Ga(α, n)
Γ(α)
7. Sea
Y =
n
X
ai X i
i=1
P.D.
n
n
X
X
Y ∼ N(
ai µ i ,
(ai σi )2 )
i=1
i=1
Sea
u i = ai x i
MU (t) = E(etai xi )
= exp(tai µi +
Sea
Y =
n
X
(tai xi )2
)
2
ui
i=1
Pn
MY (t) = E(e i=1 tui )
n
Y
=
ui E(etui )
i=1
n
n
X
X
(tai σi )2
= exp(
tai µi +
)
2
i=1
i=1
3
v
u n
n
X
uX
→ Y ∼ N(
ai µi , t (ai σi )2 )
i=1
i=1
8. Sea fX (x) y Y = aX + b
FY = P (Y ≤ y)
= P (aX + b ≤ y)
y−b
= P (X ≤
)
a
y−b
d
)
→ fY = F X (
dy
a
9. Sea
Y = ex
X ∼ N (µ, σ 2 )
P (Y ≤ y) = P (ex ≤ y)
= P (X ≤ ln(y))
= FX (ln(y))
ln(y) − µ
= P (Z ≤
)
σ
ln(y) − µ
= Φ(
)
σ
Derivando obtenemos
ϕY (y)
10. Sean
(a)
Y =(
x−b β
)
a
y
X ∼ W eibull(b, a, β)
x−b β
) ≤ y)
a
1
= P (X ≤ ay β + b)
Z ay β1 +b
β x β−1 −(x/a)β
=
( ) e
dx,
a a
0
= 1 − e−y
P (Y ≤ y) = P ((
→ Y ∼ Exp(1)
4
Seaβ = 1 yb = 0
(b) Sean
Y =(
x−b β
)
a
1
yX = ay β + b
P (X ≤ x) = P (ay 1/β + b ≤ x)
x−b β
= P (Y ≤ (
) )
a
x−b β
) )
= 1 − exp((
a
β x − b β−1
x−b β
(
) exp((
) )
a a
a
X ∼ W eibull(a, b, β)
→ fX =
11. Sean
X ∼ CauchyEstandar
W =
1
X
1
≤ w)
x
1
= P (X ≥ )
w
1 arctan w1
)
=1−( +
2
π
P (W ≤ w) = P (
fW =
1
π(w2 + 1)
W ∼ CauchyEstandar
12. Sean X ∼ U(-1, 1)
W = |X|
P (W ≤ w) = P (−w ≤ X ≤ w) = w
→ fW = 1,
0≤w≤1
SeaY = X 2 + 1
p
p
p
P (Y ≤ y) = P (− y − 1 ≤ X ≤ y − 1) = y − 1
1
1
fY = ( √
)
2 y−1
1
X +1
1
z−1
P (Z ≤ z) = P (X ≥ − 1) =
z
2z
SeaZ =
5
13. Sean
X ∼ exp(1)
Y = ln(X)
y
P (Y ≤ y) = P (X ≤ ey ) = 1 − e−e
fY = exp(−ey + y)
14. Sean
Y = eX
X ∼ U (0, 1)
P (Y ≤ y) = P (X ≤ ln(y)) = ln(y)
1
→ fY =
I[1,e] (y)
y
15. Sea
θ ∼ U(
−π π
, )
2 2
R = Asen(θ)
arcsen( ar )
r
P (R ≤ r) = P (θ ≤ arcsen( )) =
a
π
1
1
→ fR (r) =
(p
)
πA
1 − ( Ar )2
16. Sean
X ∼ U (0, 1)
Y = −ln(X)
P (Y ≤ y) = P (X ≥
1
1
)=1− y
y
e
e
→ fY (y) = e−y
→ Y ∼ Exp(1)
17. Sea
Long ∼ N (3.25, 0.052 ) L1 + L2 ∼ N (6.50, .005)
6.60 − 6.50
√
)
.005
= φ(1.4142)
→ P (L1 + L2 ≤ 6.60) = P (Z ≤
18. Si X1 , X2 , X3 , ..., Xn son v.a.i.i.d.
Yn = max(X1 , X2 , ..., Xn )
p.d.fYn (y) = n(FX (y))n−1 fX (y)
6
Dem.
FY = P (Yn ≤ y)
= P (max(X1 , X2 , ..., Xn ) ≤ y)
n
Y
=
P (Xi ≤ y)
i=1
= (FX (y))n
→ fY (y) = n(FX (y))n−1 fX (y)
19. Si
X ∼ U (0, 1) Y1 = minXi
FY (y) = P (Y1 ≤ y)
= P (minXi ≤ y)
= 1 − (1 − FX (y))n
1
3
→ P (Y1 ≤ ) = 1 − ( )n
4
4
20. Si X1 y X2 son v.a.i.i.d. N(0,1) y sea
Y =
(X2 − X1 )2
2
p.d.
1 1
Y ∼ Ga( , )
2 2
MY (t) = E(eyt )
Z ∞Z ∞
(x2 − x1 )2 t
1
)exp(− (x21 x22 ))dx1 dx2
=
exp(
2
2
0
0
1 1/2
1
= 1 2
para t <
2
−t
2
21. Si
X1 , X2
v.a.i.i.d. N ormal(0, 1) Y1 = X12 + X22
Y2 = X2
1
1 −y1 /2
Iy22 ,∞ (y1 )I−∞,∞ (y2 )
f (y1 , y2 ) = p
e
2 2π
2 y1 − y2
Y1 ∼ χ22
7
22. (a) Sea X1 , X2 v.a.i.i.d. Normal(0,1)
1
Y1 = (X12 + X22 ) 2
X2
Y2 = arctan( )
X1
2
y1 sec (y2 )
|J| =
1 + tan2 (y2 )
y1 −y12 /2
I< (y1 )I(0,π/2) (y2 )
e
2π
→ f (y1 , y2 ) =
No son independientes.
(b) Sea Xi ∼ U(0, 1)
Y1 = X2 + X1
X1
Y2 −
X2
y1
|J| =
(1 − y2 )2
→ f (y1 , y2 ) =
y1
(1 − y2 )2
I(0,1) (y2 )I(0,1−y2 ) (y1 )
No son independientes.
(c) Sea Xi ∼ Exp(1), i = 1, 2, 3
X1
X1 + X2
X 1 + X2
Y2 =
X1 + X2 + X3
Y3 = X1 + X2 + X3
|J| = y2 y32
Y1 =
→ f (y1 , y2 , y3 ) = y2 y32 e−y3 I(0,1) (y1 , y2 , y3 )
Si son independientes.
23. Sea
Xi ∼ N (µi , σi2 )
MW (t) = E(etw )
n
Y
=
MXi (t)
i=1
= prodni=1 expµi t + σi2 t2
X X
→ Y ∼ N(
µi ,
σi2 )
8
24. (a) Sea
X ∼ U (0, 2) Y ∼ U (0, 1)
Z =X +Y
1
1
fZ = z I(0,1) (z) + I(1,2) (z)
2
2
(b)
X, Y ∼ N (0, 1)
Y
X
W =X
→ |J| = W
Z=
w −w2 (z2 +1)
e
I(−∞,∞) (w, z)
Z2π∞
w −w2 (z2 +1)
fZ (z) =
e
dw I(−∞,∞) (z)
−∞ 2π
1
=
I< (z)
π(1 + z 2 )
f(Z,W ) =
(c) X, Y tal que
fT (t) = btb−1
I(0,1) (t)I(0,∞) (b)
Z =YX
W =Y
→ |J| =
1
W
b2 b−1
z
I(0,1) (z)I(z,1) (w)I(0,∞) (b)
w
→ fZ = b2 z b−1 ln(Z) I(0,1) (z)I(0,∞) (b)
f(Z,W ) =
25. Sea
f (xi ) = 2xi
I(0,1) (x)
Z = min(Xi )
FZ (z) = 1 − (1 − x2 )3
→ fZ = 6x(1 − x2 )2
1
mediana = √
2
√
1
1
P (Z > √ ) = 1 − Fz (1/ 2) = (1 − 1/2)3 =
8
2
9
26. (a) Sean X1 , X2 , ..., Xn v.a.i.i.d. con FX (x) y la funcion de densidad de la j-esima
estadistica de orden Yj esta dada por:
f (yj ) =
n!
fX (y)(FX (y))j−1 (1 − FX (y))n−j
(j − 1)!(n − j)!
Para variables continuas.
(b) Sea Y1 = min(Xi )
FY1 = P (Y1 ≤ y1 )
P (min(Xi ) ≤ y1 )
= 1 − P (min(Xi ) > y1 )
n
Y
=1−
P (Xi > y1 )
i=1
= 1 − (1 − FX (y1 ))n
→ fY1 (y1 ) = n(1 − FX (y1 ))n−1 fX (y1 )
(c) Sea Yn = max(Xi )
FYn = P (Yn ≤ yn )
P (max(Xi ) ≤ yn )
n
Y
=
P (Xi ≤ yn )
i=1
= (FX (yn ))n
→ fYn (yn ) = n(FX (y1 ))n−1 fX (yn )
27. (a) Sea X1 , X2 , ..., Xn una muestra aleatoria, en donde
√
√
fXi (xi ) ∼ U (µ − 3σ, µ + 3σ)
Obtener el Rango muestral:
Z = Yn − Y1
W = Y1
→ |J| = 1
w
1 z + w n−1
( √ )
I(µ−√3σ,µ+√3σ) (w)(I(0,2√3σ) (z)
f(Z,W ) (z, w) = n2 (1 − √ )n−1
12σ 2 3σ
2 3σ
Z µ−√3σ
1 z + w n−1
w
fZ (z) =
n2 (1 − √ )n−1
( √ ) dw I(0,2√3σ) (z)
√
12σ 2 3σ
2 3σ
µ− 3σ
10
Obtener el rango medio muestral:
Y1 − Yn
2
W = Yn
→ |J| = 2
Z=
2z − w n−1 1
w
√ )
( √ )n−1 I(µ−√3σ,µ) (z)I(µ−√3σ,2z−µ+√3σ) (w)
12σ 2 3σ
2 3σ
√
√
+ I(2z−µ− 3σ,µ+ 3σ) (w)I(µ,µ+√3σ) (z)
fZ,W (z, w) =2n2 (1 −
Z
→ fZ (z) =
√
2z−µ+ 3σ
√
µ− 3σ
Z
+
2n2 (1 −
√
µ+ 3σ
√
2z−µ− 3σ
w
2z − w n−1 1
√ )
( √ )n−1 dw
12σ 2 3σ
2 3σ
w
2z − w n−1 1
√ )
( √ )n−1 dw
12σ 2 3σ
2 3σ
2n2 (1 −
I(µ−√3σ,µ) (z)
I(µ,µ+√3σ) (z)
(b) Sean X1 , X2 , ..., Xn ∼ Exp(θ). Obtener el rando muestral:
Z = Yn − Y1
W = Y1
→ |J| = 1
fZ,W (z, w) = θ2 n2 (1 − eθ(z+w) )n−1 eθw(n+1) eθz I(0,∞) (z)I(z,∞) (w)
Z ∞
fZ (z) =
θ2 n2 (1 − eθ(z+w) )n−1 eθw(n+1) eθz dw I(0,∞) (z)
z
Obtener el Rango medio muestral:
Y1 − Yn
2
W = Yn
→ |J| = 2
Z=
fZ,W (z, w) = 2θ2 n2 (1 − eθ(z+w) )n−1 eθw(n+1) eθz I(0,∞) (z)I(0,2z) (w)
Z 2z
fZ (z) =
2θ2 n2 (1 − eθ(z+w) )n−1 eθw(n+1) eθz dw I(0,∞) (z)
0
28. Sea
L ∼ N (µ, 1)
(a)
P (Ŷ > 8) = 1 − φ(4.5)
11
(b)
P (6.2 ≤ Ŷ ≤ 6.8) = 2φ(0.9) − 1
(c)
P (µ − 0.3 ≤ Ŷ ≤ µ + 0.3) = 0.95
√
→ 2φ(0.3 n) − 1 = 0.95
→ n = 42.6844
29. Sea N = 10, σ = 1
P (a ≤ S 2 ≤ b) = 0.90
n
1 X
n−1 n−1
(xi − x̄)2 ∼ Ga(
,
)
n − 1 i=1
2
2σ 2
Z
Z
( 92 )9/2 a 7 − 7x
( 29 )9/2 b 7 − 7x
x 2 e 2 dx −
x 2 e 2 dx
→ 0.90 =
Γ( 92 ) 0
Γ( 29 ) 0
en donde S 2 =
30. Sea
N1 = 6 N2 = 10
ambas con la misma varianza de poblacional.
P(
S12
≤ b) = 0.90
S22
sabemos que
S12
∼ F (5, 9)
S22
→ b = 2.611
31. Sea
P (X ≤ Q1 ) = 0.25,
X ∼ N (0, 1)
Qx1 = 1 − φ(0.68)
Qx2 = 0
Qx3 = φ(0.68)
Sea
Y ∼ t − student(10)
Qy1 = 1 − FX (.700)
Qy2 = 0.129
Qy3 = 0.700
W ∼ χ2 (20)
12
Qw1 = 15425
Qw2 = 19337
Qw3 = 23828
32. Sean
X ∼ U (0, 1)
Y = ln(
x
)
1−x
x
≤ y)
1−x
ey
= P (x ≤
)
1 + ey
P (Y ≤ y) = P (ln(
→ fY (y) =
ey
(1 + ey )2
33. Sea
Xi ∼ χ22
X 1 − X2
2
Z = X1
→ |J| = 2
W =
→ fZ,W (z, w) = e−z−w I(0,∞) (w)I(2w,∞) (z)
→ fW (w) = e−3w I(0,∞) (w)
34. Sean
X ∼ N (µ, σ 2 ) Y =
x−µ
∼ N (0, 1) W = Y 2
σ
1
1
1
→ fW (w) = √ w 2 e− 2w
2π
1 21
( )
1
1
= 2 1 w 2 e− 2w
Γ( 2 )
√
1
I(0,∞) (w) como π = Γ( )
2
I(0,∞) (w)
1 1
→ W ∼ Ga( , ) = χ21
2 2
35. Sea
X ∼ Ga(α, 1)
xi
yi =
Yk+1
Yk+1 =
k+1
X
i=1
13
xi
sea
x1 = y1 ∗ Yk+1 , x2 = y2 ∗ Yk+1 , ..., xk = yk ∗ Yk+1
Y
xk+1 = (1 −
k
X
yi ) ∗ yk+1
i=1
k
k
→ |J| = |yk+1
− 2(yk+1
∗
1 k+1
→ fY (y) = (
)
Γ(α)
Sea
∞
Z
k
k
∗
−2(yk+1
(yk+1
X
X
yi )|
X
2(α−1)
yi ))∗e−yk+1 yk+1 (1−
yi )
α−1
0
k
Y
i=1
k
Y
X
X
1 k+1
α−1
γ=(
yiα−1
) (1 − 2
yi )(1 −
yi )
Γ(α)
i=1
Z ∞
2(α−1)
k
yk+1
→γ
e−yk+1 dyk+1 yk+1 = γ ∗ Γ(2α + k − 1)
0
k
X
X
Y
1 k+1
α−1
→ fY (y) = ( ) (1 − 2
yi )(1 −
yi )
yiα−1 ∗ Γ(2α + k − 1)
Γα
i=1
Con K = 1 se tiene una Beta
36. (a) Sean
SeaX1 , X2 , ..., Xn v.a.i.N (µ, σ)
n
1 X
2
YS =
(xi − x̄)2
n − 1 i=1
→M =
X
(xi − µ)2 (t) = M
X
(xi − x̄)2 (t) ∗ M (nx̄ − nµ)2
Si son independientes,
P
M (xi − µ)2 (t)
→M
(xi − x̄) (t) =
M (nx̄ − nµ)2
P
X xi − x̄
M ( xiσ−µ )2 (t)
2
→M
(
) (t) =
σ
)2
M ( nx̄−nµ
σ
X
2
En donde
X xi − µ
(
)2 (t) ∼ χ2n
σ
nx̄ − nµ 2
M(
) ∼ χ21
σ
M
14
yiα−1 dyk+1
1
n
( 1 2−t ) 2
X xi − x̄
2
→M
(
) (t) = 2 1 1
σ
( 1 2−t ) 2
2
= (1
2
1
2
−t
)
n−1
2
Por Teorema de unicidad.
X xi − x̄
(n − 1)S 2
(
)2 =
σ
σ2
∼ χ2n−1
(b) Sabemos que si
X ∼ N (0, 1) Y ∼ χ2n
entonces
X
T = q ∼ t(n)
Y
n
√
→
x̄−µ
∼ N (0, 1)
√σ
n(x̄ − µ)
n
q 2 ∼ t(n − 1)
=q
χn−1
SX
(n−1)S 2
∼
σ 2 (n−1)
n−1
37. Sea n = 3
(a)
1
P (Y1 > median) median = √
2
Z
P (Y1 > median) =
√1
2
6(1 − y 2 )2 y1 dy1
0
= 0.125
(b) Sean
Y1
Y2
Y2
Z2 =
Y3
Z3 = Y3
→ |J| = z2 z32
→ f (z1 , z2 , z3 ) = z2 z32 fY1 (z1 z2 z3 )fY2 (z2 z3 )fY3 (z3 )
Z1 =
Son independientes
15
38. Sea
12
x(x + y)I(0,1) (x)I(0,1) (y)
7
U = min(X, Y ) V = min(X, Y )
fX,Y (x, y) =
FU,V = P (U ≤ u, V ≤ v)
= P (V ≤ v) − P (U > u, V ≤ v)
v
Z
Z
P (V ≤ v) =
=v
0
0
4
Z
v
Z
v
12
x(x + y)dxdy
7
v
12
x(x + y)dydx
u
u 7
6
4
4
= v 4 + u4 − u2 v 2 − v 3 u − u3 v
7
7
7
→ P (U > u, V ≤ v) =
1
fU,V (u, v) = (4v 3 u + 4u3 v − 7u4 + 6v 2 u2 ) I(0 < u < v < 1)
7
39. Si X1 , X2 , X3 , ..., Xn es una muestra aleatoria.
E(Xi ) = µ V ar(Xi ) = σ 2
n
X
σ2
1
xi ) =
V (x̄) = V (
n
n
i=1
n
1 X
E(S ) = E(
(xi − x̄)2 )
n − 1 i=1
2
n
1 X
=
(
V ar(xi ) − nV ar(x̄))
n − 1 i=1
→ E(S 2 ) = σ 2
40. Sea
Y = a + bx
Cov(X, Y )
σY σX
bV (X)
=
bσX σX
=1
ρX,Y =
16
41. Si X1 , X2 , X3 , ..., Xn es una muestra con V(Xi ) = σ 2 . P.D. Cov(xi − x̄, x̄) = 0.
Cov(xi − x̄, x̄) = Cov(xi , x̄) − V ar(x̄)
σ2 σ2
=
−
n
n
=0
42. Sea X1 , X2 , X3 , ..., Xn una sucesion de v.a.i.i.d. con media µ y varianza σ 2 .
(a)
N m ∼ P o(γ)
E(SN ) = E(E(N |SN ))
nµ
=
γ
V (SN ) = V (E(N |SN )) + E(V (N |SN ))
n
= 2 (nσ 2 + µ)
γ
(b)
1
1−p
N ∼ Geo( ) γ =
γ
p
E(SN ) = E(E(N |SN ))
1
= nµ
γ
V (SN ) = V (E(N |SN )) + E(V (N |SN ))
1
1
1
= (nσ )2 + (1 − )nµ
γ
γ
γ
43. Sea X una v.a. con segundo momento finito.
E((X − a)2 ) = E(X 2 ) − 2aE(X) + a2 + 2aE(X) − a2
= E(X 2 ) − E 2 (X)
= V ar(X)
44. Sean
Z ∼ N (0, 1) Y = a + bz + cz 2
Cov(Z, Y )
σZ σY
b
=√
2
b + 2c2
→ ρY,Z =
17
45.
V(
k
X
k
k
X
X
αi xi ) = E[(
αi xi )2 ] − E 2 [
α i xi ]
i=1
i=1
k
X
i=1
αi xi −
= E[
i=1
k
X
= E[
i=1
k
X
= E[
n
X
αi µi ]
i=1
αi (xi − µi )2 ]
αi2 V ar(xi ) + 2
X
i=1
αi αj cov(xi , xj )]
i<j
46. Sean
Z ∼ N (0, 1) Y = a + bz + cz 2
ρ=
Cov(Z, Y )
σY σZ
Cov(Z, Y ) = E(ZY ) − E(Z)E(Y )
= E(ZY )
= E(az + bz 2 + cz 3 )
= b + cE(Z 3 )
= b por independencia
→ σY =
√
V ar(Z) = V ar(a + bz + cz 2 )
= b2 + 2c2
b2 + 2c2
b
ρX,Y = √
2
b + 2c2
47. Sea X ∼ N(0, 1) y sea I otra v.a.i. de X tal que P(I = 1) = P(I = 0) = 12 . Y sea
tambien Y = X si I = 1 y Y = -X si I = 0.
Y ∼ N (0, 1) c.p. 21
fY =
Y ∼ N (0, 1) c.p. 12
1 −x2
e 2 ∼ N (0, 1)
2π
Cov(X, Y ) = 0
→ fY =
18
48. Sean
f (Y |X) = Bin(n, x) X ∼ U (0, 1)
n
fX,Y (x, y) = (y )xy (1 − x)n−y
n
fX,Y (x, y) = (y )
Z
I(0,1) (x)I0,1,2,...,n (y)
1
xy (1 − x)n−y dx I0,1,2,...,n (y)
0
Γ(y + 1)Γ(n − y + 1)
= (y )
Γ(n + 2)
n
I0,1,2,...,n (y)
* Kernel de una beta.
49. Sea
ECM = E[(y − h(x))2 |X]
P.D.
minf ECM (f ) = E(Y |X = x)
minfX E[(y − h(x))2 |X] = minc E[(y − c)2 |X]
= minc [E(y 2 |X) − 2E(yc|X = x) + E(c2 |X = x)]
→ −2E(Y |X) + 2c = 0
→ fX (x) = E(Y |X)
50. Sean
u=
f (y2 |y1 ) =
y1 − µ1
σ1
v=
y 2 − µ2
σ2
1√
−1
2
exp[ 1−ρ
2 (u −
2πσ1 σ2 1−ρ2
1
e−u2 /2
2πσ1
=√
2ρuv + v 2 )]
−1 v − ρu 2
1
p
exp[ ( p
)]
2
2πσ2 1 − ρ2
1 − ρ2
σ
−1 y2 − (µ2 + ρ σ12 (y1 − µ1 ))
p
→ f (y2 |y1 ) = √
exp[
]
2
2πσ2 1 − ρ2
σ2 1 − ρ2
p
σ2
→ f (y2 |y1 ) ∼ N (µ2 + ρ (y1 − µ1 ), (σ2 1 − ρ2 )2 )
σ1
1
p
51. Sea
X ∼ N (0, 1)
1
φX (t) = √
2π
Z
2 /2
= e−t
∞
eixt e−x
Z−∞
∞
−∞
−t2 /2
=e
19
2 /2
dx
1
2
√ e−(x−it) /2 dx
2π
En general, sea
W = µ + σx, W ∼ N (µ, σ 2 )
φW (t) = E(ei(µ∗σx)t )
= eiµt E(eitσx )
= eiµt φX (σt)
= eiµt e−σ
52.
fp|n =
1
Z
2 t2 /2
fp fn|p
fn
n+m
( n )pn (1 − p)m dp
fn =
0
n+m
=( n )
Γ(n + 1)Γ(m + 1)
Γ(n + m + 2)
fp|n =
pn (1 − p)m
Γ(n+1)Γ(m+1
Γ(n+m+2)
53.
fX,Y = e−(x+y) I[0,∞) (x)I[0,∞) (y)
Sea
X
Y
W =Y
↔ X = WZ
Z=
↔Y =W
J=
Z W
1 0
→ |J| = W
f(W,Z) = w ∗ f(X,Y ) (wz, w)
= w ∗ e−w(z+1) I[0,∞) (w)I[0,∞) (z)
Z
fZ =
∞
we−w(z+1) dwI[0,∞) (z)
0
=
1
I[0,∞) (z)
z+1
20
54. Sea
X ∼ P o(θ),
Y ∼ P o(λ),
Z = X + Y ∼ P o(θ + λ).
p.d.
fX|Z ∼ Bin(Z,
θ
)
θ+λ
fX,Z = f(X,Y ) (x, z − x)
θx −(θ) λz−x −λ
e I[0,1,2,..,z] (x)I[0,1,2,..) (y)
= e
x!
(z − x)!
fZ =
fX,Z
=
fZ
(λ + θ)z −(λ+θ)
e
I[0,1,2,..) (z)
z!
θx −(θ) λz−x −λ
e
e I(0,1,2,..,z) (x)I(0,1,2,..) (y)
x!
(z−x)!
z
(λ+θ) −(λ+θ)
e
I(0,1,2,..) (z)
z!
z!
θx λz−x (λ + θ)−z I(0,1,2,..,z) (x)
(x − z)!x!
z
θ x λ z−x
=
(
) (
) I[0,1,2,..,z) (x)
x θ+λ θ+λ
=
f(X|Z) ∼ Bin(z,
θ
)
θ+λ
55. Sean X1 , X2 , X3 ,.., Xn v.a.i.i.d
P (Xk = 1) = P (Xk = −1) =
1
2
N ∼ Geo(α)
P (N = n) = α(1 − α)n I1,2,3,4,... (n)
Y =
n
X
xi
xi ∼ Geo(α)
i=1
→ Y ∼ BinN eg(n, α)
r+y−1 r
→ fY =
α (1 − α)y I1,2,3,4,... (y)
y
21
56. Sea
(a)
fS = k(s2 + 1) I(1,4) (s)
(b)
fT = ct I(0,600) (t)
(c)
Z
4
(s2 + 1)ds
1=k
→k=
1
1
24
(d)
Z
6
00tdt
1=c
→c=
0
1
180000
(e)


100 c.p.0.11112,
fX = 200 c.p.0.33336,


300 c.p.0.5556.
(f)


1.5 c.p.0.1388,
fY = 2.5 c.p.0.3055,


3.5 c.p.0.5555
(g)
fY =



























x = 100
x = 100
x = 100
x = 200
x = 200
x = 200
x = 300
x = 300
x = 300
y
y
y
y
y
y
y
y
y
= 1.5
= 2.5
= 3.5
= 1.5
= 2.5
= 3.5
= 1.5
= 2.5
= 3.5
c.p. 0.022224,
c.p. 0.044448,
c.p. 0.044448,
c.p. 0.181963,
c.p. 0.060654,
c.p. 0.090981,
c.p. 0.06486,
c.p. 0.20039,
c.p. 0.42007
57. Sean X1 , X2 , X3 , ...,Xn v.a.i.i.d.
fXi = 2Xi
I(0,1) (x)
W = Yn − Y1
Z = Yn
→ |J| = 1
fW,Z = 4n2 (1 − (z − w)2 )n−1 (z − w)z 2n−1 I(0,1) (w)I(w,1) (z)
Z 1
fW =
4n2 (1 − (z − w)2 )n−1 (z − w)z 2n−1 dz I(0,1) (x)
w
22
Z
1
Z
1
4n2 (1 − (z − w)2 )n−1 (z − w)z 2n−1 dzdw
E(W ) =
w
w
58. Sean
1
4
fX =
I(−2,−1,1,2) (x)
Y sea
Y = X2
X/Y
-2
-1
(a)
1
2
1
4
0
0.25
0.25
0
0.25
0
0
0.25
(b)
CovXY = E(XY ) − E(X)E(Y )
X
X
XX
yf (y)
xf (x)
xyf (x, y) −
=
x
x
y
y
=0
(c)
ρXY = 0
(d) Pero no son independientes porque Y siempre va a depender de X
59. Sea
X ∼ U (0, 1)
Y
Y = log
P (Y ≤ log
X
1−X
ey
X
) = P (X ≤
)
1−X
1 + ey
→ fY = (1 + ey )ey − e2y (1 + ey )−2 I(−∞,∞) (Y )
23