Solved – Exponential distribution: Log-Likelihood and Maximum Likelihood estimator

The likelihood is given as

$$L(lambda,x) = L(lambda,x_1,…,x_N) = prod_{i=1}^N f(x_i,lambda)$$

where the second identity use the IID assumption and with $x = (x_1,…,x_N)$. The log-likelikelihood is given as

$$l(lambda,x) := log L(lambda,x) = sum_{i=1}^N log f(x_i, lambda),$$

where $log f(x_i,lambda) = log lambda – lambda x_i$. This implies that

$$l(lambda,x) = sum_{i=1}^N log lambda – lambda x_i = N log lambda – lambda sum_{i=1}^N x_i.$$ Since we are interested in maximum a positive monotone transformation such as dividing with $N$ is fine. This gets us to

$$frac{1}{N} l(lambda , x) = log lambda – lambda bar x$$

differentiate and set to zero to get first order condition

$$frac{1}{lambda} – bar x = 0 Leftrightarrow lambda = frac{1}{bar x}$$

Small simulation in R

lambda_0 <- 0.5 N <- 10000 x <- rexp(N, rate = lambda_0)  loglik <- function(theta)     {         ll <- N * log(theta) - theta*sum(x)         return(ll)     }  # Calculate estimate m_x <- 1/mean(x)  # Create vector for plot of loglikelihood t <- seq(0.5*m_x,1.5*m_x,length.out=100)  plot(t,loglik(t),type="l") abline(v=m_x,col="red") 

This will genrate this plot of loglikelihood function to see maximum …