11 Day 11 (June 17)

11.1 Announcements

  • Questions about assignment 3
  • Update on assignment 2
  • Read Ch. 2 in linear models with R (estimation)
  • Read Ch. 3 in linear models with R (inference)

11.2 Maximum Likelihood Estimation

  • Maximum likelihood estimation for the linear model
    • Assume that \(\mathbf{y}\sim \text{N}(\mathbf{X}\boldsymbol{\beta},\sigma^2\mathbf{I})\)
    • Write out the likelihood function \[\mathcal{L}(\boldsymbol{\beta},\sigma^2)=\prod_{i=1}^n\frac{1}{\sqrt{2\pi\sigma^2}}\textit{e}^{-\frac{1}{2\sigma^2}(y_{i} - \mathbf{x}_{i}^{\prime}\boldsymbol{\beta})^2}.\]
    • Maximize the likelihood function
      1. First take the natural log of \(\mathcal{L}(\boldsymbol{\beta},\sigma^2)\), which gives \[\mathcal{l}(\boldsymbol{\beta},\sigma^2)=-\frac{n}{2}\text{ log}(2\pi\sigma^2) - \frac{1}{2\sigma^2}\sum_{i=1}^{n}(y_{i} - \mathbf{x}_{i}^{\prime}\boldsymbol{\beta})^2.\]
      2. Recall that \(\sum_{i=1}^{n}(y_{i} - \mathbf{x}_{i}^{\prime}\boldsymbol{\beta})^2 = (\mathbf{y} - \mathbf{X}\boldsymbol{\beta})^{\prime}(\mathbf{y} - \mathbf{X}\boldsymbol{\beta})\)
      3. Maximize the log-likelihood function \[\underset{\boldsymbol{\beta}, \sigma^2}{\operatorname{argmax}} -\frac{n}{2}\text{ log}(2\pi\sigma^2) - \frac{1}{2\sigma^2}(\mathbf{y} - \mathbf{X}\boldsymbol{\beta})^{\prime}(\mathbf{y} - \mathbf{X}\boldsymbol{\beta})\]
    • f- Three ways to do it in program R
    • Using matrix calculus and algebra
    # Data 
y <- c(0.16,2.82,2.24)
X <- matrix(c(1,1,1,1,2,3),nrow=3,ncol=2,byrow=FALSE)
n <- length(y)

# Maximum likelihood estimation using closed from estimators
beta <- solve(t(X)%*%X)%*%t(X)%*%y 
sigma2 <- (1/n)*t(y-X%*%beta)%*%(y-X%*%beta)

beta
    ##       [,1]
## [1,] -0.34
## [2,]  1.04
    sigma2
    ##        [,1]
## [1,] 0.5832
    • Using modern (circa 1970’s) optimization techniques
    # Maximum likelihood estimation estimation using the Nelder-Mead algorithm
y <- c(0.16,2.82,2.24)
X <- matrix(c(1,1,1,1,2,3),nrow=3,ncol=2,byrow=FALSE)

negloglik <- function(par){
beta <- par[1:2]
sigma2 <- par[3]
-sum(dnorm(y,X%*%beta,sqrt(sigma2),log=TRUE))
}
optim(par=c(0,0,1),fn=negloglik,method = "Nelder-Mead")
    ## $par
## [1] -0.3397162  1.0398552  0.5831210
## 
## $value
## [1] 3.447978
## 
## $counts
## function gradient 
##      150       NA 
## 
## $convergence
## [1] 0
## 
## $message
## NULL
    • Using modern and user friendly statistical computing software
    # Maximum likelihood estimation using lm function
# Note: estimate of sigma2 is not the MLE! 
df <- data.frame(y = c(0.16,2.82,2.24),x = c(1,2,3))
m1 <- lm(y~x,data=df)

coef(m1)
    ## (Intercept)           x 
##       -0.34        1.04
    summary(m1)$sigma^2
    ## [1] 1.7496
    summary(m1)
    ## 
## Call:
## lm(formula = y ~ x, data = df)
## 
## Residuals:
##     1     2     3 
## -0.54  1.08 -0.54 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept)  -0.3400     2.0205  -0.168    0.894
## x             1.0400     0.9353   1.112    0.466
## 
## Residual standard error: 1.323 on 1 degrees of freedom
## Multiple R-squared:  0.5529, Adjusted R-squared:  0.1057 
## F-statistic: 1.236 on 1 and 1 DF,  p-value: 0.4663
  • Live example

11.3 Confidence intervals for paramters

  • Example 
    y <- c(63, 68, 61, 44, 103, 90, 107, 105, 76, 46, 60, 66, 58, 39, 64, 29, 37,
27, 38, 14, 38, 52, 84, 112, 112, 97, 131, 168, 70, 91, 52, 33, 33, 27,
18, 14, 5, 22, 31, 23, 14, 18, 23, 27, 44, 18, 19)
year <- 1965:2011
df <- data.frame(y = y, year = year)
plot(x = df$year, y = df$y, xlab = "Year", ylab = "Annual count", main = "",
col = "brown", pch = 20, xlim = c(1965, 2040))
    • Is the population size really decreasing?
    • Write out the model we should use answer this question
    • How can we assess the uncertainty in our estimate of \(\beta_1\)?