Enunciado: Let \(X_{ij}, j=1,\ldots,r>1\), \(i=1,\ldots,n\), be independently distributed as \(N(\mu_i,\sigma^2)\). Find the MLE of \((\mu_1,\ldots,\mu_n,\sigma^2)\).

Resposta: Vetor de estimativa das médias populacionais

\[ \hat{\mu}_i=\sum_{j=1}^{r}\frac{X_{ij}}{r}, \quad \forall\; i \in \{1,2,...,n\}. \]

set.seed(20/11/2021)
n <- 10 ; r <- 1e6
sigma <- 2

mu<-seq(-3,3,length.out=n)

mat<-matrix(0, nrow = n, ncol = r)
for (i in 1:n){
  mat[i,] <- rnorm(r, mean = mu[i], sd = sigma)
}

mean.est <- apply(mat,1,mean)
mean.par <- mu

info <- matrix(c(mean.par,mean.est),nrow=2,byrow=TRUE)
rownames(info)<-c("vet.par","vet.est")
knitr::kable(info)
vet.par -3.000000 -2.333333 -1.666667 -1.0000000 -0.3333333 0.3333333 1.0000000 1.666667 2.333333 3.000000
vet.est -3.000075 -2.334066 -1.668646 -0.9995568 -0.3320011 0.3291641 0.9986403 1.664645 2.332346 3.000019 
plot(mean.par,mean.est,main="Estimativa de MV x Parâmetro",xlab=bquote(mu),ylab=bquote(hat(mu)))
abline(a=0,b=1,col="red")

Resposta: Estimativa da variância populacional

\[ \hat{\sigma^2}=\frac{1}{nr}\sum\limits_{i=1}^{n}\sum\limits_{j=}^{r}(X_{ij}-\bar{X}_{i.})^2, \quad \bar{X}_{i.}=\hat{\mu}_i. \]

mat.dif<-matrix(0,nrow = n, ncol = r)

for (j in 1:r){
  mat.dif[,j] <- mat[,j] - mean.est
}

sigma2.est <- sum ( mat.dif^2 )  / (n*r)

setNames( c(sigma^2,sigma2.est), c("variância ", " est. da variância"))
##         variância   est. da variância 
##           4.000000           4.000835