Calculo de Probabilidades IIRespuestas Tema 2
1. Si
Xi ∼ Ber(p) y Y =n∑i=1
xi
entonces,
MY (t) = E(Ety)
= E(etx1etx2 ...etxn)
=n∏i=1
E(etxi)
= (etp+ (1− p)n)
Y ∼ Bin(n, p)
2. SeaX ∼ χ2
k
YY = Z2
1 + Z22 + ...+ Z2
n
→MY (t) = E(ety)
=n∏i=1
etk(zi)2
= (1− 2t)nk/2
→ Y ∼ χ2∑ni
3. Sea Y = aX + b
→MY t = E(ety)
= E(et(aX+b))
= etbMX(at)
4. Sea
f(x;µ, σ) =1
xσ√
2πe−(lnx−µ)2/2σ2
E(X) = eµ+σ2/2
V ar(X) = e2µ+2σ2 − e2µ+σ2
1
5. Sea X ∼ N(0, 1), p.d.Z = X2 ∼ χ2
(1)
Dem.
MZ(t) =
∫ ∞0
e−x2(t− 1
2) 1√
2πdx
=1
t− 12
→ Z = X2 ∼ χ2(1)
6. Sean
Xi ∼ Ga(α, 1)
Y =n∑i=1
Xi
(a) Con n = 2
fX1,X2(y) =
∫ ∞−∞
fX1(x1)fX2(y − x1)dx1
=
∫ y
0
fX1(x1)fX2(y − x1)dx1
=
∫ y
0
xα−11 e−x1
Γ(α)
(y − x1)α−1e−(y−x1)
Γ(α)dx1
Seax1 = yt dx1 = ydt
y sustituyendo en la integral anterior,
fX1,X2(y) = e−yyα+α−1
∫ 1
0
ytα−1(1− t)α−1 1
Γ(α)Γ(α)dt
=e−yy2α−1
Γ(2α)∼ Ga(2α, 1)
(b) SeaY ∼ Ga(α, n) y W = 2nY.
P (W ≤ w) = P (2nY ≤ y)
= P (Y ≤ w
2n)
= FY (w
2n)
2
→ fW (w) =1
2nfY (
w
2n)
=wα−1
Γ(α)e−w/2
1
2α∼ χ2
2α
(c) Sea
X ∼ Ga(α, 1) y Y =X
n
P (Y ≤ y) = P (X
n≤ y)
= P (X ≤ ny)
= FX(ny)
→ fY (y) = n ∗ fX(ny)
=1
Γ(α)enyyα−1nα ∼ Ga(α, n)
7. Sea
Y =n∑i=1
aiXi
P.D.
Y ∼ N(n∑i=1
aiµi,n∑i=1
(aiσi)2)
Seaui = aixi
MU(t) = E(etaixi)
= exp(taiµi +(taixi)
2
2)
Sea
Y =n∑i=1
ui
MY (t) = E(e∑ni=1 tui)
=n∏i=1
uiE(etui)
= exp(n∑i=1
taiµi +n∑i=1
(taiσi)2
2)
3
→ Y ∼ N(n∑i=1
aiµi,
√√√√ n∑i=1
(aiσi)2)
8. Sea fX(x) y Y = aX + b
FY = P (Y ≤ y)
= P (aX + b ≤ y)
= P (X ≤ y − ba
)
→ fY =d
dyFX(
y − ba
)
9. SeaY = ex X ∼ N(µ, σ2)
P (Y ≤ y) = P (ex ≤ y)
= P (X ≤ ln(y))
= FX(ln(y))
= P (Z ≤ ln(y)− µσ
)
= Φ(ln(y)− µ
σ)
Derivando obtenemos
ϕY (y)
10. Sean
(a)
Y = (x− ba
)β y X ∼ Weibull(b, a, β)
P (Y ≤ y) = P ((x− ba
)β ≤ y)
= P (X ≤ ay1β + b)
=
∫ ay1β +b
0
β
a(x
a)β−1e−(x/a)βdx, Seaβ = 1 yb = 0
= 1− e−y
→ Y ∼ Exp(1)
4
(b) Sean
Y = (x− ba
)β yX = ay1β + b
P (X ≤ x) = P (ay1/β + b ≤ x)
= P (Y ≤ (x− ba
)β)
= 1− exp((x− ba
)β)
→ fX =β
a(x− ba
)β−1exp((x− ba
)β)
X ∼ Weibull(a, b, β)
11. Sean
X ∼ CauchyEstandar W =1
X
P (W ≤ w) = P (1
x≤ w)
= P (X ≥ 1
w)
= 1− (1
2+arctan 1
w
π)
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 = X2 + 1
P (Y ≤ y) = P (−√y − 1 ≤ X ≤
√y − 1) =
√y − 1
fY =1
2(
1√y − 1
)
SeaZ =1
X + 1
P (Z ≤ z) = P (X ≥ 1
z− 1) =
z − 1
2z
5
13. SeanX ∼ exp(1) Y = ln(X)
P (Y ≤ y) = P (X ≤ ey) = 1− e−ey
fY = exp(−ey + y)
14. SeanX ∼ U(0, 1) Y = eX
P (Y ≤ y) = P (X ≤ ln(y)) = ln(y)
→ fY =1
yI[1,e](y)
15. Sea
θ ∼ U(−π2,π
2) R = Asen(θ)
P (R ≤ r) = P (θ ≤ arcsen(r
a)) =
arcsen( ra)
π
→ fR(r) =1
πA(
1√1− ( r
A)2
)
16. SeanX ∼ U(0, 1) Y = −ln(X)
P (Y ≤ y) = P (X ≥ 1
ey) = 1− 1
ey
→ fY (y) = e−y
→ Y ∼ Exp(1)
17. SeaLong ∼ N(3.25, 0.052) L1 + L2 ∼ N(6.50, .005)
→ P (L1 + L2 ≤ 6.60) = P (Z ≤ 6.60− 6.50√.005
)
= φ(1.4142)
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−1fX(y)
6
Dem.
FY = P (Yn ≤ y)
= P (max(X1, X2, ..., Xn) ≤ y)
=n∏i=1
P (Xi ≤ y)
= (FX(y))n
→ fY (y) = n(FX(y))n−1fX(y)
19. SiX ∼ U(0, 1) Y1 = minXi
FY (y) = P (Y1 ≤ y)
= P (minXi ≤ y)
= 1− (1− FX(y))n
→ P (Y1 ≤1
4) = 1− (
3
4)n
20. Si X1 y X2 son v.a.i.i.d. N(0,1) y sea
Y =(X2 −X1)2
2
p.d.
Y ∼ Ga(1
2,1
2)
MY (t) = E(eyt)
=
∫ ∞0
∫ ∞0
exp((x2 − x1)2t
2)exp(−1
2(x2
1x22))dx1dx2
=
( 12
12− t
)1/2
para t <1
2
21. SiX1, X2 v.a.i.i.d. Normal(0, 1) Y1 = X2
1 +X22 Y2 = X2
f(y1, y2) =1
2√y1 − y2
2
1
2πe−y1/2Iy22 ,∞(y1)I−∞,∞(y2)
Y1 ∼ χ22
7
22. (a) Sea X1, X2 v.a.i.i.d. Normal(0,1)
Y1 = (X21 +X2
2 )12
Y2 = arctan(X2
X1
)
|J| = y1sec2(y2)
1 + tan2(y2)
→ f(y1, y2) =y1
2πe−y
21/2I<(y1)I(0,π/2)(y2)
No son independientes.
(b) Sea Xi ∼ U(0, 1)
Y1 = X2 +X1
Y2 −X1
X2
|J| = y1
(1− y2)2
→ f(y1, y2) =y1
(1− y2)2I(0,1)(y2)I(0,1−y2)(y1)
No son independientes.
(c) Sea Xi ∼ Exp(1), i = 1, 2, 3
Y1 =X1
X1 +X2
Y2 =X1 +X2
X1 +X2 +X3
Y3 = X1 +X2 +X3
|J| = y2y23
→ f(y1, y2, y3) = y2y23e−y3I(0,1)(y1, y2, y3)
Si son independientes.
23. SeaXi ∼ N(µi, σ
2i )
MW (t) = E(etw)
=n∏i=1
MXi(t)
= prodni=1expµit+ σ2i t
2
→ Y ∼ N(∑
µi,∑
σ2i )
8
24. (a) SeaX ∼ U(0, 2) Y ∼ U(0, 1)
Z = X + Y
fZ =1
2z I(0,1)(z) +
1
2I(1,2)(z)
(b)X, Y ∼ N(0, 1)
Z =Y
XW = X
→ |J| = W
f(Z,W ) =w
2πe−w
2(z2+1) I(−∞,∞)(w, z)
fZ(z) =
∫ ∞−∞
w
2πe−w
2(z2+1)dw I(−∞,∞)(z)
=1
π(1 + z2)I<(z)
(c) X, Y tal quefT (t) = btb−1 I(0,1)(t)I(0,∞)(b)
Z = Y X
W = Y
→ |J| = 1
W
f(Z,W ) =b2
wzb−1 I(0,1)(z)I(z,1)(w)I(0,∞)(b)
→ fZ = b2zb−1ln(Z) I(0,1)(z)I(0,∞)(b)
25. Seaf(xi) = 2xi I(0,1)(x) Z = min(Xi)
FZ(z) = 1− (1− x2)3
→ fZ = 6x(1− x2)2
mediana =1√2
P (Z >1√2
) = 1− Fz(1/√
2) = (1− 1/2)3 =1
8
9
26. (a) Sean X1, X2, ..., Xn v.a.i.i.d. con FX(x) y la funcion de densidad de la j-esimaestadistica de orden Yj esta dada por:
f(yj) =n!
(j − 1)!(n− j)!fX(y)(FX(y))j−1(1− FX(y))n−j
Para variables continuas.
(b) Sea Y1 = min(Xi)
FY1 = P (Y1 ≤ y1)
P (min(Xi) ≤ y1)
= 1− P (min(Xi) > y1)
= 1−n∏i=1
P (Xi > y1)
= 1− (1− FX(y1))n
→ fY1(y1) = n(1− FX(y1))n−1fX(y1)
(c) Sea Yn = max(Xi)
FYn = P (Yn ≤ yn)
P (max(Xi) ≤ yn)
=n∏i=1
P (Xi ≤ yn)
= (FX(yn))n
→ fYn(yn) = n(FX(y1))n−1fX(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
f(Z,W )(z, w) = n2(1− w
2√
3σ)n−1 1
12σ(z + w
2√
3σ)n−1 I(µ−
√3σ,µ+
√3σ)(w)(I(0,2
√3σ)(z)
fZ(z) =
∫ µ−√
3σ
µ−√
3σ
n2(1− w
2√
3σ)n−1 1
12σ(z + w
2√
3σ)n−1dw I(0,2
√3σ)(z)
10
Obtener el rango medio muestral:
Z =Y1 − Yn
2W = Yn
→ |J| = 2
fZ,W (z, w) =2n2(1− 2z − w2√
3σ)n−1 1
12σ(
w
2√
3σ)n−1 I(µ−
√3σ,µ)(z)I(µ−
√3σ,2z−µ+
√3σ)(w)
+ I(2z−µ−√
3σ,µ+√
3σ)(w)I(µ,µ+√
3σ)(z)
→ fZ(z) =
∫ 2z−µ+√
3σ
µ−√
3σ
2n2(1− 2z − w2√
3σ)n−1 1
12σ(
w
2√
3σ)n−1dw I(µ−
√3σ,µ)(z)
+
∫ µ+√
3σ
2z−µ−√
3σ
2n2(1− 2z − w2√
3σ)n−1 1
12σ(
w
2√
3σ)n−1dw 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) = θ2n2(1− eθ(z+w))n−1eθw(n+1)eθz I(0,∞)(z)I(z,∞)(w)
fZ(z) =
∫ ∞z
θ2n2(1− eθ(z+w))n−1eθw(n+1)eθzdw I(0,∞)(z)
Obtener el Rango medio muestral:
Z =Y1 − Yn
2W = Yn
→ |J| = 2
fZ,W (z, w) = 2θ2n2(1− eθ(z+w))n−1eθw(n+1)eθz I(0,∞)(z)I(0,2z)(w)
fZ(z) =
∫ 2z
0
2θ2n2(1− eθ(z+w))n−1eθw(n+1)eθzdw I(0,∞)(z)
28. SeaL ∼ N(µ, 1)
(a)P (Y > 8) = 1− φ(4.5)
11
(b)P (6.2 ≤ Y ≤ 6.8) = 2φ(0.9)− 1
(c)
P (µ− 0.3 ≤ Y ≤ µ+ 0.3) = 0.95
→ 2φ(0.3√n)− 1 = 0.95
→ n = 42.6844
29. Sea N = 10, σ = 1
P (a ≤ S2 ≤ b) = 0.90
en donde S2 =1
n− 1
n∑i=1
(xi − x)2 ∼ Ga(n− 1
2,n− 1
2σ2)
→ 0.90 =(9
2)9/2
Γ(92)
∫ b
0
x72 e−
7x2 dx−
(92)9/2
Γ(92)
∫ a
0
x72 e−
7x2 dx
30. SeaN1 = 6 N2 = 10
ambas con la misma varianza de poblacional.
P (S2
1
S22
≤ b) = 0.90
sabemos queS2
1
S22
∼ F (5, 9)
→ b = 2.611
31. SeaP (X ≤ Q1) = 0.25, X ∼ N(0, 1)
Qx1 = 1− φ(0.68)
Qx2 = 0
Qx3 = φ(0.68)
SeaY ∼ t− student(10)
Qy1 = 1− FX(.700)
Qy2 = 0.129
Qy3 = 0.700
W ∼ χ2(20)
12
Qw1 = 15425
Qw2 = 19337
Qw3 = 23828
32. SeanX ∼ U(0, 1) Y = ln(
x
1− x)
P (Y ≤ y) = P (ln(x
1− x≤ y)
= P (x ≤ ey
1 + ey)
→ fY (y) =ey
(1 + ey)2
33. SeaXi ∼ χ2
2
W =X1 −X2
2Z = X1
→ |J| = 2
→ 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, σ2) Y =x− µσ∼ N(0, 1) W = Y 2
→ fW (w) =1√2πw
12 e−
12w I(0,∞)(w) como
√π = Γ(
1
2)
=(1
2)12
Γ(12)w
12 e−
12w I(0,∞)(w)
→ W ∼ Ga(1
2,1
2) = χ2
1
35. SeaX ∼ Ga(α, 1)
yi =xiYk+1
Yk+1 =k+1∑i=1
xi
13
seax1 = y1 ∗ Yk+1, x2 = y2 ∗ Yk+1, ..., xk = yk ∗ Yk+1
Y
xk+1 = (1−k∑i=1
yi) ∗ yk+1
→ |J| = |ykk+1 − 2(ykk+1 ∗∑
yi)|
→ fY (y) = (1
Γ(α))k+1
∫ ∞0
(ykk+1−2(ykk+1∗∑
yi))∗e−yk+1y2(α−1)k+1 (1−
∑yi)
α−1
k∏i=1
yα−1i dyk+1
Sea
γ = (1
Γ(α))k+1(1− 2
∑yi)(1−
∑yi)
α−1
k∏i=1
yα−1i
→ γ
∫ ∞0
ykk+1e−yk+1dyk+1y
2(α−1)k+1 = γ ∗ Γ(2α + k − 1)
→ fY (y) = (1
Γα)k+1(1− 2
∑yi)(1−
∑yi)
α−1
k∏i=1
yα−1i ∗ Γ(2α + k − 1)
Con K = 1 se tiene una Beta
36. (a) Sean
SeaX1, X2, ..., Xnv.a.i.N(µ, σ)
Y S2 =1
n− 1
n∑i=1
(xi − x)2
→M =∑
(xi − µ)2(t) = M∑
(xi − x)2(t) ∗M(nx− nµ)2
Si son independientes,
→M∑
(xi − x)2(t) =M∑
(xi − µ)2(t)
M(nx− nµ)2
→M∑
(xi − xσ
)2(t) =M∑
(xi−µσ
)2(t)
M(nx−nµσ
)2
En donde
M∑
(xi − µσ
)2(t) ∼ χ2n
M(nx− nµ
σ)2 ∼ χ2
1
14
→M∑
(xi − xσ
)2(t) =(
12
12−t)
n2
(12
12−t)
12
= (12
12− t
)n−12
Por Teorema de unicidad.∑(xi − xσ
)2 =(n− 1)S2
σ2∼ χ2
n−1
(b) Sabemos que siX ∼ N(0, 1) Y ∼ χ2
n
entonces
T =X√Yn
∼ t(n)
→√n(x− µ)
SX=
x−µσ√n
∼ N(0, 1)√(n−1)S2
σ2(n−1)∼√
χ2n−1
n−1
∼ t(n− 1)
37. Sea n = 3
(a)
P (Y1 > median) median =1√2
P (Y1 > median) =
∫ 1√2
0
6(1− y2)2y1dy1
= 0.125
(b) Sean
Z1 =Y1
Y2
Z2 =Y2
Y3
Z3 = Y3
→ |J | = z2z23
→ f(z1, z2, z3) = z2z23fY1(z1z2z3)fY2(z2z3)fY3(z3)
Son independientes
15
38. Sea
fX,Y (x, y) =12
7x(x+ y)I(0,1)(x)I(0,1)(y)
U = min(X, Y ) V = min(X, Y )
FU,V = P (U ≤ u, V ≤ v)
= P (V ≤ v)− P (U > u, V ≤ v)
P (V ≤ v) =
∫ v
0
∫ v
0
12
7x(x+ y)dxdy
= v4
→ P (U > u, V ≤ v) =
∫ v
u
∫ v
u
12
7x(x+ y)dydx
= v4 + u4 − 6
7u2v2 − 4
7v3u− 4
7u3v
fU,V (u, v) =1
7(4v3u+ 4u3v − 7u4 + 6v2u2) I(0 < u < v < 1)
39. Si X1, X2, X3, ..., Xn es una muestra aleatoria.
E(Xi) = µ V ar(Xi) = σ2
V (x) = V (n∑i=1
1
nxi) =
σ2
n
E(S2) = E(1
n− 1
n∑i=1
(xi − x)2)
=1
n− 1(n∑i=1
V ar(xi)− nV ar(x))
→ E(S2) = σ2
40. SeaY = a+ bx
ρX,Y =Cov(X, Y )
σY σX
=bV (X)
bσXσX= 1
16
41. Si X1, X2, X3, ..., Xn es una muestra con V(Xi) = σ2. P.D. Cov(xi − x, x) = 0.
Cov(xi − x, x) = Cov(xi, x)− V ar(x)
=σ2
n− σ2
n= 0
42. Sea X1, X2, X3, ..., Xn una sucesion de v.a.i.i.d. con media µ y varianza σ2.
(a)Nm ∼ Po(γ)
E(SN) = E(E(N |SN))
=nµ
γ
V (SN) = V (E(N |SN)) + E(V (N |SN))
=n
γ2(nσ2 + µ)
(b)
N ∼ Geo(1
γ) γ =
1− pp
E(SN) = E(E(N |SN))
= nµ1
γ
V (SN) = V (E(N |SN)) + E(V (N |SN))
= (nσ1
γ)2 +
1
γ(1− 1
γ)nµ
43. Sea X una v.a. con segundo momento finito.
E((X − a)2) = E(X2)− 2aE(X) + a2 + 2aE(X)− a2
= E(X2)− E2(X)
= V ar(X)
44. SeanZ ∼ N(0, 1) Y = a+ bz + cz2
→ ρY,Z =Cov(Z, Y )
σZσY
=b√
b2 + 2c2
17
45.
V (k∑i=1
αixi) = E[(k∑i=1
αixi)2]− E2[
k∑i=1
αixi]
= E[k∑i=1
αixi −n∑i=1
αiµi]
= E[k∑i=1
αi(xi − µi)2]
= E[k∑i=1
α2iV ar(xi) + 2
∑i<j
αiαjcov(xi, xj)]
46. SeanZ ∼ N(0, 1) Y = a+ bz + cz2
ρ =Cov(Z, Y )
σY σZ
Cov(Z, Y ) = E(ZY )− E(Z)E(Y )
= E(ZY )
= E(az + bz2 + cz3)
= b+ cE(Z3)
= b por independencia
V ar(Z) = V ar(a+ bz + cz2)
= b2 + 2c2
→ σY =√b2 + 2c2
ρX,Y =b√
b2 + 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.
fY =
{Y ∼ N(0, 1) c.p. 1
2
Y ∼ N(0, 1) c.p. 12
→ fY =1
2πe−x22 ∼ N(0, 1)
Cov(X, Y ) = 0
18
48. Seanf(Y |X) = Bin(n, x) X ∼ U(0, 1)
fX,Y (x, y) = (ny)xy(1− x)n−y I(0,1)(x)I0,1,2,...,n(y)
fX,Y (x, y) = (ny)
∫ 1
0
xy(1− x)n−ydx I0,1,2,...,n(y)
= (ny)
Γ(y + 1)Γ(n− y + 1)
Γ(n+ 2)I0,1,2,...,n(y)
* Kernel de una beta.
49. SeaECM = E[(y − h(x))2|X]
P.D.minfECM(f) = E(Y |X = x)
minfXE[(y − h(x))2|X] = mincE[(y − c)2|X]
= minc[E(y2|X)− 2E(yc|X = x) + E(c2|X = x)]
→ −2E(Y |X) + 2c = 0
→ fX(x) = E(Y |X)
50. Sean
u =y1 − µ1
σ1
v =y2 − µ2
σ2
f(y2|y1) =
1
2πσ1σ2√
1−ρ2exp[ −1
1−ρ2 (u2 − 2ρuv + v2)]
12πσ1
e−u2/2
=1
√2πσ2
√1− ρ2
exp[−1
2(v − ρu√
1− ρ2)2]
→ f(y2|y1) =1
√2πσ2
√1− ρ2
exp[−1
2
y2 − (µ2 + ρσ2σ1
(y1 − µ1))
σ2
√1− ρ2
]
→ f(y2|y1) ∼ N(µ2 + ρσ2
σ1
(y1 − µ1), (σ2
√1− ρ2)2)
51. SeaX ∼ N(0, 1)
φX(t) =1√2π
∫ ∞−∞
eixte−x2/2dx
= e−t2/2
∫ ∞−∞
1√2πe−(x−it)2/2dx
= e−t2/2
19
En general, seaW = µ+ σx,W ∼ N(µ, σ2)
φW (t) = E(ei(µ∗σx)t)
= eiµtE(eitσx)
= eiµtφX(σt)
= eiµte−σ2t2/2
52.
fp|n =fpfn|pfn
fn =
∫ 1
0
(n+mn )pn(1− p)mdp
= (n+mn )
Γ(n+ 1)Γ(m+ 1)
Γ(n+m+ 2)
fp|n =pn(1− p)mΓ(n+1)Γ(m+1
Γ(n+m+2)
53.fX,Y = e−(x+y)I[0,∞)(x)I[0,∞)(y)
Sea
Z =X
Y↔ X = WZ
W = Y ↔ Y = W
J =
(Z W1 0
)→ |J| = W
f(W,Z) = w ∗ f(X,Y )(wz,w)
= w ∗ e−w(z+1)I[0,∞)(w)I[0,∞)(z)
fZ =
∫ ∞0
we−w(z+1)dwI[0,∞)(z)
=1
z + 1I[0,∞)(z)
20
54. Sea
X ∼ Po(θ),
Y ∼ Po(λ),
Z = X + Y ∼ Po(θ + λ).
p.d.
fX|Z ∼ Bin(Z,θ
θ + λ)
fX,Z = f(X,Y )(x, z − x)
=θx
x!e−(θ) λz−x
(z − x)!e−λI[0,1,2,..,z](x)I[0,1,2,..)(y)
fZ =(λ+ θ)z
z!e−(λ+θ)I[0,1,2,..)(z)
fX,ZfZ
=
θx
x!e−(θ) λz−x
(z−x)!e−λI(0,1,2,..,z)(x)I(0,1,2,..)(y)
(λ+θ)z
z!e−(λ+θ)I(0,1,2,..)(z)
=z!
(x− z)!x!θxλz−x(λ+ θ)−zI(0,1,2,..,z)(x)
=
(z
x
)(
θ
θ + λ)x(
λ
θ + λ)z−xI[0,1,2,..,z)(x)
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
2N ∼ Geo(α)
P (N = n) = α(1− α)nI1,2,3,4,...(n)
Y =n∑i=1
xi xi ∼ Geo(α)
→ Y ∼ BinNeg(n, α)
→ fY =
(r + y − 1
y
)αr(1− α)yI1,2,3,4,...(y)
21
56. Sea
(a)fS = k(s2 + 1) I(1,4)(s)
(b)fT = ct I(0,600)(t)
(c)
1 = k
∫ 4
1
(s2 + 1)ds → k =1
24
(d)
1 = c
∫ 6
0
00tdt → c =1
180000
(e)
fX =
100 c.p.0.11112,
200 c.p.0.33336,
300 c.p.0.5556.
(f)
fY =
1.5 c.p.0.1388,
2.5 c.p.0.3055,
3.5 c.p.0.5555
(g)
fY =
x = 100 y = 1.5 c.p. 0.022224,x = 100 y = 2.5 c.p. 0.044448,x = 100 y = 3.5 c.p. 0.044448,x = 200 y = 1.5 c.p. 0.181963,x = 200 y = 2.5 c.p. 0.060654,x = 200 y = 3.5 c.p. 0.090981,x = 300 y = 1.5 c.p. 0.06486,x = 300 y = 2.5 c.p. 0.20039,x = 300 y = 3.5 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)z2n−1 I(0,1)(w)I(w,1)(z)
fW =
∫ 1
w
4n2(1− (z − w)2)n−1(z − w)z2n−1dz I(0,1)(x)
22
E(W ) =
∫ 1
w
∫ 1
w
4n2(1− (z − w)2)n−1(z − w)z2n−1dzdw
58. Sean
fX =1
4I(−2,−1,1,2)(x)
Y seaY = X2
(a)
X/Y 1 4-2 0 0.25-1 0.25 01 0.25 02 0 0.25
(b)
CovXY = E(XY )− E(X)E(Y )
=∑x
∑y
xyf(x, y)−∑x
xf(x)∑y
yf(y)
= 0
(c)ρXY = 0
(d) Pero no son independientes porque Y siempre va a depender de X
59. SeaX ∼ U(0, 1)
Y
Y = logX
1−X
P (Y ≤ logX
1−X) = P (X ≤ ey
1 + ey)
→ fY = (1 + ey)ey − e2y(1 + ey)−2I(−∞,∞)(Y )
23