I am not a math major, just trying to formulate my research question in a more rigorous way.
Suppose I have a probability measure $\pi_{\theta}$ and another variable $\phi=g(\theta)$. How do i formulate the conditional probability measure $\pi_{\theta\mid \phi}(A)$ for a measurable $A$?
I was trying $\pi_{\theta\mid \phi}(A)=\frac{\int_{A} \mathbf{1}\{g(\theta) =\phi\}d\pi_{\theta} (\theta)}{\int_{\Theta} \mathbf{1}\{g(\theta) =\phi\}d\pi_{\theta} (\theta)}$. (For now let's just say density of $\phi$ is non-zero on $g(\Theta)$, I would love to hear any insights about regularity condition as well.)
However, when I plug in an uniform distribution as an example, that does not seem to work. Say $\theta\sim U(-1,1)$ and $\phi=\theta^2$, then $\int_{\Theta} \mathbf{1}\{g(\theta) =\phi\}d\pi_{\theta} (\theta)$ becomes zero and that's not what i want. I was hoping to find a non-trivial conditional distribution for this scenario.
If anyone could tell me what part went wrong that will be great!
Thank you!
You want to use the dirac delta rather than an indicator. $$\pi_{\theta\mid g(\theta)}(A\mid\phi)=\mathbf 1_{\phi\in g(\Theta)}\int_A \delta_{g(\theta)-\phi}\,\mathrm d \pi_\theta(\theta)$$
For $\theta\sim\mathcal U(-1..1)$ and $g(t)=t^2$ $$\pi_{\theta\mid\theta^2}([0..1]\mid \phi)=\mathbf 1_{\phi\in[0..1^2]}\int_{0}^1\delta_{x^2-\phi}\cdot\tfrac 12\mathrm d x \\~= \frac 12\,\mathbf 1_{\phi\in[0..1^2]}$$