Is $\hat{\sigma^2}=\hat{\sigma}^2$? In simple linear regression $$Y_i=b_0+b_1X_i+\varepsilon_i$$ where $$\varepsilon_i\sim N(0, \sigma^2)$$ is it true to say that the estimator for variance and the estimator for SD squared are equal? I mean$$\hat{\sigma^2}=\hat{\sigma}^2$$
Also is there a generalized answer to this question (not in linear regression specifically)?
Thank you.
 A: This depends on the estimation procedure. It is generally not true that $\hat{\sigma^2}=\hat{\sigma}^2$. However, maximum likelihood estimates are an important exception.
A: This is not true if you want unbiased estimators.
For the model $Y_i = \mu + \epsilon_i$, the sample standard deviation $s = \sqrt{\frac{\sum (y_i-\bar{y})^2}{n-1}}$ (square root of the unbiased estimate of $\sigma^2$) is a biased estimator of $\sigma$. Namely, $\mathbb{E}[s] = c_n \sigma$ with
$$
c_n = \sqrt{\frac{2}{n-1}}\frac{\Gamma\bigl(\frac{n}{2}\bigr)}{\Gamma\bigl(\frac{n-1}{2}\bigr)}.
$$
Thus, an unbiased estimate of $\sigma$ is $\hat\sigma = \frac{s}{c_n}$, and one does not have the equality $(\hat\sigma)^2 = \widehat{\sigma^2}$ when $\widehat{\sigma^2} = s^2$ is the well-known unbiased estimate of $\sigma^2$.
I don't know the result for the simple linear regression, but it should be similar.
If you want maximum likelihood estimators, then it is true that $\widehat{f(\theta)} = f(\hat\theta)$ for any monotonic function $f$.
