$$\hat{\lambda} = \bar{x}$$
$$L(\mu, \sigma^2) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(x_i-\mu)^2}{2\sigma^2}\right)$$
The theory of point estimation is a fundamental concept in statistics, which deals with the estimation of a population parameter using a sample of data. The goal of point estimation is to find a single value, known as an estimator, that is used to estimate the population parameter. In this essay, we will discuss the theory of point estimation, its importance, and provide a solution manual for some common problems.
There are two main approaches to point estimation: the classical approach and the Bayesian approach. The classical approach, also known as the frequentist approach, assumes that the population parameter is a fixed value and that the sample is randomly drawn from the population. The Bayesian approach, on the other hand, assumes that the population parameter is a random variable and uses prior information to update the estimate. theory of point estimation solution manual
$$\frac{\partial \log L}{\partial \sigma^2} = -\frac{n}{2\sigma^2} + \sum_{i=1}^{n} \frac{(x_i-\mu)^2}{2\sigma^4} = 0$$
$$\frac{\partial \log L}{\partial \mu} = \sum_{i=1}^{n} \frac{x_i-\mu}{\sigma^2} = 0$$
$$\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i-\bar{x})^2$$ There are two main approaches to point estimation:
Solving these equations, we get:
Solving this equation, we get:
Taking the logarithm and differentiating with respect to $\mu$ and $\sigma^2$, we get: we get: Solving this equation
The likelihood function is given by:
Suppose we have a sample of size $n$ from a Poisson distribution with parameter $\lambda$. Find the MLE of $\lambda$.
Suppose we have a sample of size $n$ from a normal distribution with mean $\mu$ and variance $\sigma^2$. Find the MLE of $\mu$ and $\sigma^2$.