Consider the following simple linear regression equation:

$$y_i=\beta_0+\beta_1 x_i+\epsilon_i,$$

where $\epsilon_i\sim N(0,\sigma^2)$.

Suppose I have confidence interval of $\sigma$ which is (.561, .972).

And I know the parameter value of $\sigma^2=.356$.

I want to check whether true value of standard deviation lies in the confidence interval of $\sigma$.

Since I have already computed the CI for $\sigma$ and knows the parameter value of $\sigma^2$, can I simply take square root of $\sigma^2$, i.e., $\sigma=\sqrt\sigma^2=\sqrt(.356)$, and check whether it falls in the confidence interval of $\sigma$, $$.561\le \sqrt(.356) \le .972$$???


1 Answer 1


Yes you can, the square root is a monotonic transformation, so if:

$$a^2 \leq b^2 \leq c^2$$

then, if all a, b, c are positive reals, then it is also true that:

$$a \leq b \leq c$$

Note that the transform has to be monotonic for the inequalities not to switch.

  • $\begingroup$ For a complex model, $a$ in $a^2\le b^2 \le c^2 $ is not positive real. And the output of a software package just gives me the confidence interval $(a,c)$ of standard deviation. But I know the value of $b^2$ , not the $b$. So, still can I check $a\le\sqrt( b^2) \le c $? $\endgroup$
    – time
    Sep 20, 2018 at 21:49
  • $\begingroup$ The standard deviation cannot be negative (by a complex model, I'm assuming you mean complicated), so $a$ will be a positive real. Yes, you can apply a square root to the value of $b^2$ without messing up the ordering between a, b and c. $\endgroup$
    – adityar
    Sep 20, 2018 at 21:55

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