Estimation of Lower Bound in Poisson Distribution

 MLE for Pareto distribution 

The pdf of Pareto distribution is given by:

`f(x; \alpha, \lambda) = \frac{\alpha \lambda^\alpha}{x^{\alpha+1}} \quad \text{for } x \geq \lambda.`

`f(x_i, \alpha, \lambda) = \frac{\alpha \lambda^\alpha}{x_i^{\alpha+1}} \cdot I(x_i \geq \lambda), \quad x_i \in \mathbb{R} ` 

Writing the likelihood function 

`L(\alpha, \lambda) = \prod_{i=1}^n f(x_i , \alpha, \lambda)`

`\Rightarrow L(\alpha, \lambda) = \prod_{i=1}^n \left( \frac{\alpha \lambda^\alpha}{x_i^{\alpha+1}} \cdot I(x_(i) \geq \lambda) \right)`

`= L(\alpha, \lambda) = \alpha^n \lambda^{n\alpha} \prod_{i=1}^n \frac{1}{x_i^{\alpha+1}} \cdot \prod_{i=1}^n I(x_(i)\geq \lambda)`

Finding the calculus for MLE in Pareto Distribution will be easy if we optimize it by taking log both sides because it is also a increasing function.

`= \log L(\alpha, \lambda) = n \log \alpha + n \alpha \log \lambda - (\alpha+1) \sum_{i=1}^n \log x_i + \sum_{i=1}^n \log I(x_i \geq \lambda)`

`= \log L(\alpha, \lambda) = n \log \alpha + n \alpha \log \lambda - (\alpha+1) \sum_{i=1}^n \log x_i + \log \left[ \prod_{i=1}^n I(x_(i) \geq \lambda) \right]`

1. To maximize `log(x|\alpha, \lambda)` we need to maximize 

`\alpha nlog(\lambda) \Rightarrow ` need to maximize `\lambda`.

2. Maximize `log[I(X_{(1)} > \lambda)] \Rightarrow \lambda < X_{(1)}`.

So the MLE of 

`\hat \lambda_{MLE} = X_{(1)}.`

Now differentiate the likelihood function with respect to `\alpha`

`= \frac{\partial \log L(\alpha, \lambda)}{\partial \alpha} = \frac{n}{\alpha} + n \log \lambda - \sum_{i=1}^n \log x_i`

Now putting this above equation equal to zero 

`\Rightarrow \frac{\partial \log L(\alpha, \lambda)}{\partial \alpha} =0`

`\Rightarrow \frac{n}{\alpha} + n \log \lambda - \sum_{i=1}^n \log x_i=0`

`=>\frac{n}{\alpha} = \sum_{i=1}^n \log x_i - n \log \lambda`

`= n = \alpha \left( \sum_{i=1}^n \log x_i - n \log \lambda \right)`

`= \alpha = \frac{n}{\sum_{i=1}^n \log x_i - n \log \lambda}`

`\alpha= \frac{n}{\sum_{i=1}^n \log x_i - n \log (X_{(1)})}`

` \alpha= \frac{n}{n \bar{log(x_n)} - n \log (X_{(1)})}`

=>`\hat \alpha_{MLE} = \frac{1}{ \bar{log(x_n)} - \log (X_{(1)})}`

So the joint MLE for the parameters `\alpha and \lambda` in Pareto distribution is

`(\hat \lambda, \hat \alpha) = \left( X_{(1)}, \frac{1}{\bar{\log x} - \log X_{(1)}} \right)`

MLE Estimation and Log-Likelihood Contour for Pareto Distribution

Log-Likelihood Contour for Pareto Distribution

MLE Estimation for Pareto Distribution

Python code for Python code for Pareto Distribution

# Python program to plot the Pareto distribution curve
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import pareto

# Parameters for the Pareto distribution
alpha = 3  # Shape parameter
scale = 1  # Scale parameter

# Generate x values
x = np.linspace(1, 5, 1000)

# Pareto PDF values
y = pareto.pdf(x, alpha, scale=scale)

# Plot the Pareto distribution curve
plt.plot(x, y, label='Pareto Distribution', color='orange')
plt.title('Pareto Distribution Curve')
plt.xlabel('x')
plt.ylabel('Density')
plt.legend()
plt.grid(True)

# Show the plot
plt.show()