T

#### PREGUNTA N°1
rm(list = ls())
X=vector()
Y=vector()
U=vector()
U1=vector()
U2=vector()
n=1000

for (i in 1:n) {
  U1[i]=runif(1,0,1)
  U2[i]=runif(1,0,1)
  U[i]=runif(1,0,1)
  Y[i]=(exp(U[i]^2)+(exp((1-U[i])^2)))/2
  X[i]=(exp(U1[i]^2)+(exp(U2[i]^2)))/2
}

Antiteticas1=mean(Y)
Antiteticas1
var(Y)/n

Diferentes1=mean(X)
Diferentes1
var(X)/n

#### PREGUNTA N°2
rm(list = ls())
X2=vector()
Y2=vector()
U=vector()
U1=vector()
U2=vector()
n=1000

for (i in 1:n) {
  U1[i]=runif(1,0,1)
  U2[i]=runif(1,0,1)
  U[i]=runif(1,0,1)
  Y2[i]=(exp((2*U[i])^2)+(exp((1-2*U[i])^2)))/2
  X2[i]=(exp((U1[i]+U2[i])^2)+(exp((U1[i]+U2[i])^2)))/2
}

Antiteticas2=mean(Y2)
Antiteticas2
var(Y2)/n

Diferentes2=mean(X2)
Diferentes2
var(X2)/n

######pregunta 3

rm(list = ls())
## Parte a)

n=1000
x=vector()
y=vector()
cont=0
for (j in 1:n)
{
  S=0
  for(i in 1:5)
  {
    u=runif(1,0,1)
    y[i]=-logb(u)
    S=S+i*y[i]
  }
    if(S>=21.6)
    {cont=cont+1}
  }
cont

p=cont/1000;p
var=p*(1-p);var

## Parte b)
rm(list = ls())
n=1000
x=vector()
u=vector()
y1=vector()
y2=vector()
cont1=0
cont2=0

for (j in 1:n)
{
  s1=0
  s2=0
  for(i in 1:5)
  {
    u[i]=runif(1,0,1)
    y1[i]=-logb(u[i])
    y2[i]=-logb(1-u[i])

    s1=s1+(i*y1[i])
    s2=s2+(i*y2[i])
  }
    if(s1>=21.6)
    { cont1=cont1+1}
    if(s2>=21.6)
    {cont2=cont2+1}
  }

p=(cont1+cont2)/1000;p
var2=p*(1-p);var2

## Parte c)
No siempre por que no logra en todas las iteraiones reducir la varianza.

#### PREGUNTA 5

## parte a)

rm(list = ls())
x=vector()
y=vector()
U=vector()
n=1000
for (i in 1:n) {
  z=0
  for (j in 1:12) {
    U[j]=runif(1,0,1)
    z[j]=sum(U)-6
  }
  theta=(z^3)*(exp(z))
  }
theta
med=mean(theta);med
var1=(var(theta)/n);var1

##Parte b)
Longitud <- (2*(1.96)*sqrt(var1))/(sqrt(n));Longitud
ILI=med-((1.96)*sqrt(var1)/sqrt(n));ILI
ILS=med+((1.96)*sqrt(var1)/sqrt(n));ILS

### Pregunta 6:
rm(list = ls())
x1=vector()
x2=vector()
n=500
for(i in 1: n)
{
  U=runif(1,0,1)
  x1[i]=-logb(U)
  x2[i]=-logb(1-U)
}
cor(x1,x2)
X=c(x1,x2)
length(X)
med=mean(X);med
var1=var(X)/n;var1

rm(list = ls())
### Pregunta 7:
# Caso a: X~U(0,1)
n=1000
X=vector() ## Variable de control
I=vector()
med_X=0.5  ## Media de la Uniforme

for(i in 1:n)
{X[i]=runif(1)
 if( X[i]<0.8)
 { I[i]=1 }
 else {I[i]=0 }}

p=cor(I,X);p
c=-cov(X,I)/var(X);c
EstSes=mean(I)+(c*(mean(X)-med_X));EstSes
Var_Est=var(I)*(1-p^2);Var_Est
Reduc<- (1-(Var_Est/var(I)));Reduc

rm(list = ls())
# Caso b: X~exp(1)
n=1000
X=vector()  ## Variable de control
u=vector()
I=vector()
med_X=1     ## Media de la expoenecial

for(i in 1:n)
{u[i]=runif(1,0,1)
 X[i]=-logb(u[i])
  if( X[i]<0.8)
  { I[i]=1 }
  else {I[i]=0 }}

p1=cor(I,X); p1
c1=-cov(X,I)/var(X); c1
EstSes1=mean(I)+(c1*(mean(X)-med_X));EstSes1
Var_Est1=var(I)*(1-p1^2); Var_Est1
Reduc1=(1-(Var_Est1/var(I)));Reduc1

##La varianza utilizando la variable de control se reduce a 0.11
##con respecto a la varianza de la exponencial que es 0.24

## PREGUNTA 9
## b) Simulación con variable de control
## Y=U (variable de control)
rm(list = ls())
n <- 100
X <- vector()
U <- vector()
med_Y=0.5

for(i in 1:n)
{
  U[i]=runif(1,0,1) #variable  de control
  X[i]=exp((U[i])^2)
}
c=-cov(X,U)/var(X)
p=cor(U,X)
Var_IC=var(U)*(1-p^2)
Var_IC

## c) Variable antiteticas

rm(list = ls())
X=vector()
U=vector()
n=100

for (i in 1:n) {
  U[i]=runif(1,0,1)
  X[i]=(exp(U[i]^2)+(exp((1-U[i])^2)))/2
}
var(X)/n
#### PREGUNTA N°10
## a)
rm(list = ls())
Y2=vector()
U=vector()
n=100

for (i in 1:n) {
 U[i]=runif(1,0,1) #Variable de control
Y2[i]=(exp((U[i]+U[i])^2))
}
c=-cov(Y2,U)/var(U)
p=cor(U,Y2);p
Var_IC=var(U)*(1-p^2)
Var_IC

#### b)antiteticas
rm(list = ls())
Y2=vector()
U=vector()
n=100
for (i in 1:n) {
  U[i]=runif(1,0,1)
  Y2[i]=(exp((2*U[i])^2)+(exp((1-2*U[i])^2)))/2
}
var(Y2)/n

#### Pregunta 12
## a)

n <- 10000
x <- runif(n, -1, 1)
y <- runif(n, -1, 1)
indice <- (x+y < 0)
# Aproximación por simulación
sum(indice)/n
mean(indice)