4
$\begingroup$

Let's say that the likelihood of observation $x$ given a random latent variable $z$ and a model parameter $\theta$ is defined as $p(x|\theta, z)$.

As far as I know, if I want to obtain $p(x| \theta)$, I would have to compute the expectation of $p(x|\theta, z)$ with respect to $z\sim p(z)$:

$p(x|\theta)=\mathbb E_{z\sim p(z)}[p(x|\theta, z)]=\int p(x|\theta, z)p(z)dz$

(assuming that the prior $p(z)$ has nothing to do with $\theta$)

However, I know only $p(z|x)$, the posterior of $z$ given $x$, not the prior $p(z)$.

Then can I use $\mathbb E_{z\sim p(z|x)}[p(x|\theta, z)]$ as an approximate for $p(x|\theta)$, or is this just nonsense?

Thanks

$\endgroup$
  • $\begingroup$ nuance: I'm not sure if it's wrong to say mean, but I think it would sound more standard to me to say the expectation, in this case. $\endgroup$ – Hugh Perkins Oct 12 '17 at 12:15
  • $\begingroup$ @HughPerkins thanks for pointing out! Just changed them $\endgroup$ – whkang Oct 12 '17 at 14:28
3
$\begingroup$

This is a strange and unusual setting and the question would benefit from an explanation of how the integrated posterior $p(z|x)$ is available. The question never mentions the prior distribution on $\theta$, $p(\theta)$, which matters in all subsequent calculations.

If both $p(x|z)$ and $p(z|x)$ are available [or approximated by converging estimators], Chib's formula [also known, as an earlier occurence, as the candidate's formula] provides the prior $p(z)$ through Bayes' formula [or the dual decomposition of a joint distribution] as $$p(z) \overbrace{\propto}^{\text{as function of $z$}} \frac{p(z|x)}{p(x|z)}$$ which can be approximated by $$\frac{n\,p(z|x)}{\sum_{i=1}^n p(x|z,\theta_i)}\qquad\text{when}\qquad\theta_1,\ldots,\theta_n\sim p(\theta|x,z)$$ and when only the numerator is available in closed form. (This is a special case of Rao-Blackwellisation.)

Using instead $p(z|x)$ in the integral as you propose, i.e. $$p(x|\theta)\approx\int p(x|\theta, z)p(z|x)dz$$is not coherent from a probabilistic view point since $x$ appears on both sides, i.e. as conditioned and as conditional. For instance this approximation does not integrate to one. The distinction between $p(z)$ and $p(z|x)$ is non-negligible, as shown for instance by the identity $$p(z|x) = \int p(z,\theta|x)\text{d}\theta \overbrace{\propto}^{\text{as function of $z$}} \int p(z)p(\theta)p(x|\theta,z)\text{d}\theta = p(z) \int p(\theta)p(x|\theta,z)\text{d}\theta$$which involves a second function of $z$, depending on the choice of the prior distribution on $\theta$.

$\endgroup$
2
$\begingroup$

I think you are confusing two x in your mind possibly:$\def\x{\mathbf{x}}\def\z{\mathbf{z}}$

  • there is the single vector $\mathbf{x}$ associated with a new example for which you want the probability distribution, given the parameters $\theta$. Let's denote this $\x'$
    • with no additional information, this will simply reflect your priors
  • there is the evidence, a set of training data, $\mathcal{X}$
    • by taking into account this evidence, you can improve the estimate for the distribution of $\z'$, $\theta$, $\x'$, and so on, conditioned on various combinations of the other variables
    • these will be posterior distributions (posterior, updated with the provided evidence, $\mathcal{X}$)
    • we can see they are posterior, since they will be conditioned on $\mathcal{X}$

So, what you are looking for, I reckon is $\def\X{\mathcal{X}}p(\x' \mid \theta, \X)$, ie conditioned on the training data, the evidence, $\X$.

Here's a diagram of one way of depicting these variables:

enter image description here

Looking at this diagram, and thinking about your question, we probably want to do in sequence:

  • obtain the posterior $p(\z \mid \mathcal{X}, \theta)$
  • use this posterior to obtain $p(\x' \mid \theta, \X)$

However... thinking this through a bit, I'm not sure we can just 'infer' a distribution: I think what we want to do is postulate a parametric family of distributions for $p(\z)$, eg using $\phi$ as the parameters, this would be:

$$ p(\z \mid \phi) = f_\phi(\z) $$

... where $f(\cdot)$ is some parametric probability distribution we choose. We'd then learn $\phi$ using Bayesian inference.

Our diagram will become:

enter image description here

Now things start to become fairly standard I think. We need to use Bayesian inference to estimate a distribution for $p(\phi \mid \X)$, and then slot this into standard likelihood-type formulae, to obtain the final expression for $p(\x' \mid \theta, \X)$, which we now see is something more like the marginal over $p(x', \phi \mid \theta, \X)$.

Let's start by writing down expressions we know. We have:

  • a prior $p(\phi)$, eg some uniformed prior, standard Gaussian etc
  • distribution of $\z$ given $\phi$, $p(\z \mid \phi)$
  • prior over $\theta$, $p(\theta)$
  • likelihood function $p(\x \mid \z, \theta)$

Let's express the likelihood function in terms of $\phi$. Working our way forwards we have:

  • prior $p(\phi)$
  • $p(\z, \phi) = p(\z \mid \phi)p(\phi)$
  • $p(\x, \z, \theta, \phi) = p(\x \mid \z, \phi) p(\z \mid \phi)p(\phi)$

Let's write down the joint, and start to factorize:

$$ p(\x, \z, \theta, \phi) = p(\x, \z, \phi \mid \theta)p(\theta) $$

$$ = p(\x, \z \mid \theta, \phi)p(\phi \mid \theta) p(\theta) $$

$\phi$ is independent of $\theta$, so we have:

$$ p(\x, \z, \theta, \phi) = p(\x, \z \mid \theta, \phi) p(\phi)p(\theta) $$

$$ = p(\x \mid \z, \theta, \phi)p(\z \mid \theta, \phi) p(\phi)p(\theta) $$

$\z$ is independent of $\theta$. $\x$ is conditionally independent of $\phi$, given $\z$. So we have:

$$ p(\x, \z, \theta, \phi) = p(\x \mid \z, \theta) p(\z \mid \phi) p(\phi) p(\theta) $$

We have closed-form expressions for each of the terms on the right-hand side, ie the likelihood, the priors over the parameters, and the distribution of $\z$ given $\phi$.

Let's start to consider the evidence, $\X$. We can decompose the evidence into individual examples:

  • $\X = \{ \mathbf{X}_1, \mathbf{X}_2, \dots, \mathbf{X}_n \}$
  • $\mathcal{Z} = \{ \mathbf{Z}_1, \mathbf{Z}_2, \dots, \mathbf{Z}_n \}$

So we have for example:

$$ p(\X) = \prod_{i=1}^n p(\mathbf{X}_i) $$

We want to obtain:

$$ p(\phi \mid \X) = p(\phi \mid \mathbf{X}_1, \mathbf{X}_2, \dots, \mathbf{X}_n) $$

By Bayes Rule we have:

$$ p(\phi \mid \X) = \frac{p(\X \mid \phi)p(\phi)} {p(\X)} $$

$p(\phi)$ is just the prior, so that easy. The other expressions will take some more work. For the numerator, we'll need to marginalize. But before we do that, we note that in the ultimate target expression, for the original question, we are conditioning on $\phi$, ie we want $p(\x' \mid \theta, \X)$. So, let's condition on $\theta$ here too:

$$ p(\phi \mid \X, \theta) = \frac{p(\X \mid \phi, \theta)p(\phi \mid \theta)} {p(\X \mid \theta)} $$

$\phi$ is independent of $\theta$, so we have:

$$ p(\phi \mid \X, \theta) = \frac{p(\X \mid \phi, \theta)p(\phi)} {p(\X \mid \theta)} $$

where $p(\phi)$ is just the prior over $\phi$, as before.

The denominator is the marginal:

$$ E_1 = p(\X \mid \theta) = \prod_{i=1}^n p(\mathbf{X}_i \mid \theta) = \prod_{i=1}^n \int_\z \int_\phi p(\mathbf{X}_i, \z, \phi \mid \theta)\,d\phi\,d\z $$

The first term on the numerator is another marginal:

$$ E_2 = p(\X \mid \phi, \theta) = \prod_{i=1}^n p(\mathbf{X}_i \mid \phi, \theta) = \prod_{i=1}^n \int_\z p(\mathbf{X}_i, \z \mid \phi, \theta)\,d\z $$

As for how to solve these marginalizations, I think I will leave that to you for now. However, I think the answer to your original question is:

  • no, we cant just form an expectation in the way you are doing
  • we need to introduce a parametric distribution over $\z$, eg parameterized by $\phi$, and perform Bayesian inference to estimate $\phi$, or a distribution over $\phi$
  • in addition, I think it's important to distinguish between the $\x'$ of a new example for which you want a probability distribution, and the $\X$ that forms your evidence, based on which you will form posterior distributions, eg for $\phi$
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.