# Show that the function is symmetric.

Consider the following proposal function $$q(\vec{\theta'}|\vec{\theta})$$:

For a given vector $$\vec{\theta}=(\theta_1,\theta_2,\theta_3)$$ generate $$\vec{\theta'}=(\theta_1',\theta_2',\theta_3')$$ as

$$\theta_1'=|\theta_1+\delta_1|$$ $$\theta_2'=|\theta_2+\delta_2|$$ $$\theta_3'=|\theta_3+\delta_3|$$

where $$\delta_i\sim N(0,0.1^2).$$ Show that the proposal function is symmetric, i.e, that $$q(\vec{\theta'}|\vec{\theta}) = q(\vec{\theta}|\vec{\theta'}).$$

I have that $$q(\vec{\theta'}|\vec{\theta}) = q(|\vec{\theta} + \vec{\delta}|)$$

but when I try to calculate $$q(\vec{\theta}|\vec{\theta'})$$ I find that I need to solve $$\vec{\theta}$$ from $$\vec{\theta'}=|\vec{\theta}+\vec{\delta}|.$$ Am I missing something here?