Exercise 6.4 - Estimation of $\sigma^{2}$ when $\mu$ is known

Answers

The likelihood is:

p (x | σ^{2}) = \frac{1}{\sqrt{2 π σ^{2}}} \cdot \exp (- \frac{{(x - μ)}^{2}}{2 σ^{2}}) .

Taking logarithm and gradient:

\frac{d}{d σ^{2}} \ln p (𝒟 | σ^{2}) = - \frac{N}{2 σ^{2}} + \frac{1}{2 σ^{4}} \sum_{n = 1}^{N} {(x_{n} - μ)}^{2} .

Setting it to zero yields:

σ_{MLE}^{2} = \frac{\sum_{n = 1}^{N} {(x_{n} - μ)}^{2}}{N} .

In this case the MLE is unbiased since:

𝔼 (x_{n} - μ^{2}) = 𝔼 (x_{n}^{2}) + μ^{2} - 2 𝜇𝔼 (x_{n}) = σ^{2} .

solour_lfq

2021-03-24 13:42