I'm trying to understand how to condition a probabilistic posterior distribution.

Consider the following probability density:

$$ p(\alpha, \beta | y) = \prod_{i=1}^n (\alpha+\beta t_i)^{y_i}e^{-(\alpha+\beta t_i)} $$

From this, can I say the following:

$$ p(\alpha|\beta, y) \propto \prod_{i=1}^n (\alpha+\beta t_i)^{y_i} e^{-\alpha} $$

  • $\begingroup$ Yes, accidentally removed the wrong part. Fixed that. $\endgroup$ – JC1 Sep 30 '16 at 3:51

My understanding from your equation (1) is that you mean $\mathbf{y} = (y_1, \cdots, y_n)$ is your sample, $\alpha$ and $\beta$ are parameters, and the density of the sample or likelihood is the Poisson regression: $$p(\mathbf{y}|\alpha, \beta) \propto \prod_{i=1}^n (\alpha + \beta t_i)^{y_i} e^{-(\alpha + \beta t_i)}$$

To answer the question, conditioning on a random variable, like $\beta$, means that you know or fix $\beta$ and you want to know the density $p(\alpha |\beta, \mathbf{y})$. Assuming you have a non-informative prior on $\alpha$ and $\beta$, then the posterior has the same format as the likelihood function

$$p(\alpha, \beta |\mathbf{y}) \propto \prod_{i=1}^n (\alpha + \beta t_i)^{y_i} e^{-(\alpha + \beta t_i)}$$

and the conditional density satisfies

$$p(\alpha|\beta, \mathbf{y}) \propto p(\alpha, \beta |\mathbf{y}) \propto\prod_{i=1}^n (\alpha + \beta t_i)^{y_i}e^{-\alpha}=e^{-n\alpha}\prod_{i=1}^n (\alpha + \beta t_i)^{y_i}$$

where the $\propto$ symbols are understood in terms of $\alpha$ only, given that $\beta$ and $Y$ are fixed.


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