Final answer:
The question asks to demonstrate the derivation of the maximum likelihood estimate for the variance parameter, which is the average of squared deviations from the sample mean. This involves maximizing the likelihood function, differentiating it with respect to variance, and solving for the estimate that makes the observed data most probable.
Step-by-step explanation:
The question seeks to show the derivation of the maximum likelihood estimate (MLE) for variance parameters in statistics, which is represented by the formula \(\hat \sigma^2 = \frac{1}{m} \sum_{i=1}^m (x^i - \hat \mu)^2\). In deriving the MLE, we typically assume that the data points xi come from a normally distributed population with unknown mean \(\hat \mu\) and variance \(\hat \sigma^2\). The estimate is found by maximizing the likelihood function, which involves taking the derivative of the log-likelihood with respect to the variance, setting this derivative to zero, and solving for the variance parameter to find the estimate that makes the observed data most probable.
To show that the MLE for the variance is indeed this formula, let's consider a dataset of m independent and identically distributed observations from a normal distribution with mean \(\mu\) and variance \(\sigma^2\). The likelihood function for this dataset is proportionate to the product of individual probability density functions of each observed data point. By taking the natural logarithm of the likelihood function, differentiating with respect to \(\sigma^2\), and setting the resulting expression to zero, we can solve for the MLE of the variance. It turns out that the MLE for the variance is indeed equal to the average of the squared differences between each data point and the sample mean, denoted by \(\hat \sigma^2\).