Maximum Likelihood Estimator for Log Normal Distribution

 The pdf of exponential distribution is given as 

`f(x; \theta) = \frac{1}{\theta} e^{-x / \theta}, \quad x \geq 0, \, \theta > 0.`

The MLE estimator is given by considering the likelihood function 

`L(\theta ; X) = \prod_{i=1}^n f(x_i; \theta) = \prod_{i=1}^n \frac{1}{\theta} e^{-x_i / \theta}.`

`L(\theta; X) = \frac{1}{\theta^n} exp{-\sum_{i=1}^n x_i / \theta}.`

Now taking log both sides to make calculus easy 

`\log L(\theta) = \log \left( \frac{1}{\theta^n} exp({-\frac{\sum_{i=1}^n x_i}{\theta}}) \right).`

`\log L(\theta) = -n \log \theta - \frac{\sum_{i=1}^n x_i}{\theta}.`

Differentiating partially with respect to `theta` both sides 

`\frac{\partial \log L(\theta)}{\partial \theta} = -\frac{n}{\theta} + \frac{\sum_{i=1}^n x_i}{\theta^2}.`

Putting it equal to zero 

`-\frac{n}{\theta} + \frac{\sum_{i=1}^n x_i}{\theta^2} = 0.`

`\frac{\sum_{i=1}^n x_i}{\theta^2} = \frac{n}{\theta}.`

`\sum_{i=1}^n x_i = n \theta.`

`\hat \theta_{MLE}= \frac{\sum_{i=1}^n x_i}{n}.`

R

# R code to plot an exponential distribution curve

# Define the rate parameter (lambda = 1/theta)
lambda <- class="keyword" span="">as.numeric

(readline("Enter the rate parameter (lambda): ")) # Ensure lambda is positive if (lambda <= 0) { stop("Rate parameter lambda must be greater than 0.") } # Generate x values and compute corresponding density x <- class="keyword" span="">seq(0, 10, by = 0.01) y <- class="keyword" span="">dexp(x, rate = lambda) # Plot the exponential distribution curve plot( x, y, type = 'l', col = 'blue', lwd = 2, xlab = 'x', ylab = 'Density', main = paste("Exponential Distribution (lambda =", lambda, ")") ) grid(col = 'gray')

Exponential Curve

Enter θ (θ > 0):