Standard deviation of filtered noise In an other post i've found that the variance of filtered noise can be found by

$$\min_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|^2\sigma_x^2\le
\sigma_y^2\le\max_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|^2
\sigma_x^2\tag{1}$$

The derivation is based on spectral density functions and the relationship $$S_y(e^{j\theta})=S_x(e^{j\theta})\left|H(e^{j\theta})\right|^2\tag{2}$$ with $$\sigma_x^2=\frac{1}{2\pi}\int_{-\pi}^{\pi}S_x(e^{j\theta})d\theta=
\frac{1}{\pi}\int_{\theta_1}^{\theta_2}S_x(e^{j\theta})d\theta\tag{3}$$

The important point for me is that the variances $\sigma_x^2$ and $\sigma_y^2$ are related by a quadratic relationship.

I'm interested in the relation of the standard deviations $\sigma_y$ and $\sigma_x$, not the variances. So the question is: How is the standard devation $\sigma_y$ related to $\sigma_x$?

My naive idea (I'm yet not too experience with random variables in the frequency domain) is to take the square root and find  $\sigma_y= |H(\omega)|\sigma_x$. So the variances should have a linear relationship.

Is this correct or am I missing some point?
Thank's to MattL's comment I found the 'answer' myself... I was mixing up equations (1) and (2).

As MattL. pointed out, it makes sense to take the square root of equation (1), resulting in something like $$\min_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|\sigma_x\le
\sigma_y\le\max_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|
\sigma_x\tag{1}$$ if all terms are positive.

If one takes equation (2), applies the integral and the same scaling as in (3) the result is $$\frac{1}{2\pi}\int_{-\pi}^{\pi}S_y(e^{j\theta})d\theta=\frac{1}{2\pi}\int_{-\pi}^{\pi}S_x(e^{j\theta}) \left|H(e^{j\theta})\right|^2 d\theta \tag{2}$$

The left hand side equals the variance of y, so taking the square root gives the standard deviation. However the right hand side is different, as $\left|H(e^{j\theta})\right|$ is a function of frequency.

Hence in general  $\sigma_y\ne |H(e^{j\theta})|\sigma_x$.