Composition of probability density

Question

I know probability distribution for parameter $\phi$. I have the empirical distribution/statistical distribution of $X$ that is dependent on parameter $\phi$ for $\phi \in [0,1]$. I assimilate this empirical distribution to the probability distribution for X.

1/ Can I do so? Are notions similar?

Then, knowing distribution for $\phi$, and 'empirical distribution' for $X$, I would like to compute the distribution for $X(\phi)$.

I first thought of using the inverse density function method. This gives a random number generator for my unknown distribution, if I can compute inverse cdf $F^{-1}$.

However, I cannot always compute $F^{-1}$. I thought then at some rejection method.

2/ I wonder if I can deal with this problem as a composition of probability density problem, and what solutions are at hand. I had a quick look and saw some optimization approaches for this class of problems (ex here).

3/ Finally, why not find the distribution of X multiplying each value of $X(\phi)$ by the probability density for this value of $\phi$? (simple product of underlying probability density by X values)

Thanks!

#

EDIT

I try to reformulate the problem in term of Bayesian statistics.

I have a prior $\phi$ with uniform distribution. I then know the distribution of X conditional to this prior for $\phi$, $P(X|\phi)$. From Bayes rule, I can deduce $P(\phi|X) = \frac{P(X|\phi)*P(\phi)}{P(X)}$.

Now, my prior is no more uniformly distributed. In other words, I have same parametric model for the distribution of X, but it is now parametrized with a non uniform RV $\theta$ following some know distribution with density $f_{\theta}$. I would like to know new $P(X|\theta)$. From Bayes rule, $P(X|\theta) = \frac{P(\theta|X)*P(X)}{P(\theta)}$.

My problem is solved if I have a relationship between $P(\theta|X)$ and $P(\phi|X)$. Should I plug in $P(\phi|X)$, which I know, in the equality $P(X|\theta) = \frac{P(\theta|X)*P(X)}{P(\theta)}$?

I hope that tentative explanations sound clearer. If not, could you guide me towards could formulation and solution?

#

EDIT - I try to clarify first sentence after Zen's comment, and to reformulate

With 'I have the empirical distribution/statistical distribution of $X$ that is dependent on parameter $\phi$ for $\phi \in [0,1]$.', I wanted to say: I know the distribution of data in a situation where some parameter, $\phi$, is uniformly distributed. I assume a model for the empirical distribution, that is parametrized with $\phi$.

Now, I am in another situation, where this parameter, which I consider a random variable, is having another distribution, with some known density $f_{\phi}$. I also assume that empirical distribution's model holds with the new underlying distribution.

Data can be produced in this model where the parameter distribution is no more uniform, but is $f_{\phi}$. I want to find the distribution for these data.

Thanks, apologize for naiv question and ackward stats lexicon.

Best regards.

Answer 1

The problem posed in language that is familiar to me is that you want to determine the posterior distribution of phi given a prior distribution on phi that is

uniform on [0, 1].

X is distributed according to a parametric distribution say with density f$_φ$(x).

Take a sample of size n iid from f$_φ$(x).

The posterior distribution for φ is gotten by Bayes rule

g(φ|x) = c f$_φ$(x$_1$)f$_φ$(x$_2$)....f$_φ$(x$_n$) for 0<=φ<=1

g(φ|x) = 0 otherwise. c is the normalization constant that makes ∫g(φ|x)dx=1 where the integration is over the interval [0,1]. The uniform prior density for φ appears as the constant 1 when 0<=φ<=1 and is 0 otherwise. The product of the f$_φ$s is the likelihood function given X$_1$=x$_1$, X$_2$=x$_2$..., X$_n$=x$_n$.

Based on your edit it sounds like you want to infer the likelihood from the posterior and the prior. But I don't understand why you would want to do that. You are normally given data and you pick a prior and a model and then compute the posterior. But if by P(X) you mean ∫p(X|θ)p(θ)dθ then the formula you have is correct.