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