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Maximum Likelihood Estimator for Log Normal Distribution

Published by Sumit Kumar On December 11, 2024

If $X_{1}, X_{2},......X_{n} ~ LogN(\mu, \sigma^2) \tag{1}$ The pdf of log normal distribution is given by `f(x; \mu, \sigma^2) = \frac{1}{x \sigma \sqrt{2 \pi}} \exp \left( -\frac{1}{2} \left( \frac{\log(x) - \mu}{\sigma} \right)^2 \right), \quad x > 0 \tag{2}` The likelihood function of log Normal pdf is given by `L(\mu, \sigma^2) = \prod_{i=1}^{n} f_X (x_i) \tag{3}` `L(\mu, \sigma^2) = \prod_{i=1}^{n} \frac{1}{x_i \sigma \sqrt{2 \pi}} \exp \left[ -\frac{1}{2} \left( \frac{\log(x_i) - \mu}{\sigma} \right)^2 \right] \tag{4}` `L(\mu, \sigma^2) = \frac{1}{(\sigma \sqrt{2 \pi})^n \prod_{i=1}^{n} x_i} \exp \left( -\frac{1}{2} \sum_{i=1}^{n} \left( \frac{\log(x_i) - \mu}{\sigma} \right)^2 \right) \tag{5}` Now taking the log both sides `\log L(\mu, \sigma^2) = -n \log(\sigma \sqrt{2 \pi}) - \sum_{i=1}^{n} \log(x_i) - \frac{1}{2} \sum_{i=1}^{n} \left( \frac{\log(x_i) - \mu}{\sigma} \right)^2 \tag{6}` Now...

Methods of Estimation in Statistical Inference

Published by Sumit Kumar On November 29, 2024

Statistical estimation is the cornerstone of statistical inference, enabling us to derive insights about population characteristics using sample data. The population is described by a probability density function (pdf) $f(x;θ)$, which depends on unknown parameters θ. These parameters are often not directly observable but play a vital role in defining the population's behavior and structure. To obtain a good such estimators following method can be used: Method of Maximum Likelihood Estimator Method of Minimum Variance Method of Moments Method of Least Squares Method of Minimum Chi-square Method of Inverse Probability Method of Maximum Likelihood Estimator For each sample point $x$ let $\hat\theta(x)$ be a estimator value at which $L(\theta|x)$ attains it maximum value as a function of $\theta$ with $x$ held fixed. A maximum likelihood estimator based on sample points $x_{i}$ is $\hat\theta_{MLE}$. In practical scenarios, θ is estimated using a random s...

Maximum likelihood estimation for Pareto observations

Published by Sumit Kumar On December 03, 2024

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, \l...

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