Math Problem Statement

solve this

Solution

To solve the problem, we'll follow these steps:

Problem Overview:

  • We have a random sample X1,X2,,XnX_1, X_2, \dots, X_n from a normal distribution N(μ,σ2)N(\mu, \sigma^2).
  • The goal is to show that the maximum likelihood estimator (MLE) of the variance σ2\sigma^2 is given by: σ^2=1ni=1n(XiXˉ)2\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2 where Xˉ\bar{X} is the sample mean: Xˉ=1ni=1nXi\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i

Step-by-Step Solution:

  1. Log-Likelihood Function: The likelihood function for the normal distribution N(μ,σ2)N(\mu, \sigma^2) based on the sample X1,X2,,XnX_1, X_2, \dots, X_n is: L(μ,σ2)=i=1n12πσ2exp((Xiμ)22σ2)L(\mu, \sigma^2) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(X_i - \mu)^2}{2\sigma^2}\right) Taking the natural logarithm of the likelihood function (the log-likelihood), we have: (μ,σ2)=n2log(2πσ2)12σ2i=1n(Xiμ)2\ell(\mu, \sigma^2) = -\frac{n}{2} \log(2\pi\sigma^2) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i - \mu)^2

  2. Without Differentiation: To find the MLE for σ2\sigma^2 without differentiating the log-likelihood function:

    • Notice that the log-likelihood function is maximized when the sum of squared differences i=1n(Xiμ)2\sum_{i=1}^n (X_i - \mu)^2 is minimized.
    • The minimum value of this sum occurs when μ=Xˉ\mu = \bar{X}, the sample mean. Thus, the MLE for μ\mu is μ^=Xˉ\hat{\mu} = \bar{X}.
  3. Substituting μ=Xˉ\mu = \bar{X}: Substitute μ^=Xˉ\hat{\mu} = \bar{X} into the expression for the log-likelihood: (Xˉ,σ2)=n2log(2πσ2)12σ2i=1n(XiXˉ)2\ell(\bar{X}, \sigma^2) = -\frac{n}{2} \log(2\pi\sigma^2) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i - \bar{X})^2 The term i=1n(XiXˉ)2\sum_{i=1}^n (X_i - \bar{X})^2 represents the sum of squared deviations from the mean, also known as the total variance.

  4. MLE for σ2\sigma^2: The MLE for σ2\sigma^2 maximizes the log-likelihood. It is obtained by minimizing the denominator in the second term, which happens when: σ^2=1ni=1n(XiXˉ)2\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2 This is the desired result.

Thus, the MLE for σ2\sigma^2 is σ^2=1ni=1n(XiXˉ)2\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2.

Would you like a more detailed explanation of any part of this solution? Here are 8 related questions to deepen your understanding:

  1. How is the log-likelihood function for a normal distribution derived?
  2. Why is the sample mean Xˉ\bar{X} the MLE for μ\mu?
  3. What is the significance of the sum of squared deviations in statistics?
  4. Can the MLE for σ2\sigma^2 be biased or unbiased? How can it be adjusted if biased?
  5. How would the MLE change if the distribution were not normal?
  6. What role does the Central Limit Theorem play in this context?
  7. How would the estimation change for small sample sizes?
  8. What assumptions are critical for the validity of this MLE?

Tip: When working with MLE, understanding the properties of the likelihood function is key to finding estimators that maximize the likelihood without necessarily requiring complex calculus.

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Math Problem Analysis

Mathematical Concepts

Statistics
Maximum Likelihood Estimation
Normal Distribution

Formulas

MLE for variance in normal distribution

Theorems

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Suitable Grade Level

Advanced Undergraduate