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?

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