1
$\begingroup$

In this paper, the first discussion (Universal latent variable representation by C. Andrieu, A. Doucet and A. Lee) authors state that

Sampling exactly $Y \sim f(y|\theta)$ on a computer most often means that $Y=\phi(\theta,U)$ where $U$ is a random vector of probability distribution $D(\cdot)$ and $\phi(\cdot,\cdot)$ is a mapping either known analytically or available as a “black-box".

And they explain it further. I am trying to understand their point but I am lost. I think I managed to use this idea on a simple example. I did a Approximate Bayesian Computation (should work also for the exact) inference using normal likelihood (known mean, unknown precision $\tau$). So in this case $\theta=\tau$, $U\sim U[0,1]$ and $\phi(\cdot,\cdot)$ is a Box-Muller transformation. So when I simulate data from the likelihood, instead of using numpy normal distribution, I simulate $U$ from uniform distribution and use Box-Muller transformation to get my normally distributed data. However, even if I'm right with this example, I don't know how to apply it to real-world problems. In general I am working on a ABC code to infer from an Agent Based Model (that was given to me in NetLogo, I just call it from my code in Python). So instead of having to simulate from $N(\mu, \frac{1}{\tau})$, I have to run the NetLogo model with proposed parameters $\theta$ ($\theta$ is a multidimensional vector). So my only input to NetLogo model are there parameters. My question is: how can I use the idea from this paper in my case? I.e., what would $U$ and $\phi$ be? I would be grateful for any explanation and help!

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ Taken literally, the quotation is a trivial statement: it asserts that sampling on a computer requires a random number generator. $\phi$ can be understood to be any computer program, $U$ is the RNG, $\theta$ represents all non-random inputs to the program, and $Y$ is the output. Such statements, although without content, often are used to establish notation or terminology. $\endgroup$
    – whuber ♦
    Commented Aug 9, 2016 at 12:54

1 Answer 1

Reset to default
1
$\begingroup$

In the context of ABC, you want to simulate from $y$, which is a realization of your model for a fixed value of the parameter $\theta$. Then, in the step when you simulate from $y\sim f(\vert\theta)$, for a given $\theta$, the way you simulate from $f(\vert\theta)$ defines the function $\phi(U,\theta)$. This might be a complicated function, depending on your model. The specific expression of $\phi$ can only be obtained (if at all possible) in a case by case analysis. If you want to know the explicit expression of $\phi$ in a model of interest, you will need to ask a new question with all the details about the model you want to simulate from.

Improve this answer
$\endgroup$
9
  • $\begingroup$ I wrote: "I simulate U from uniform distribution and use Box-Muller transformation to get my normally distributed data." Isn't it exactly what you are saying? I think I understand what $\phi$ is, but I don't know what $U$ would be. $\endgroup$
    – Paula
    Commented Aug 9, 2016 at 13:32
  • $\begingroup$ @Paula I see, my point is that you don't need to know anything about priors in order to simulate from a normal distribution using BMT. $U$ is nothing but a uniform random variable, which is the atom of any simulation, even in complex models. $\endgroup$
    – Scorpion
    Commented Aug 9, 2016 at 13:38
  • $\begingroup$ I edited my post a bit. Now I'm a bit lost: what do you mean by "the atom of any simulation"? Sorry if it's obvious... $\endgroup$
    – Paula
    Commented Aug 9, 2016 at 13:38
  • 2
    $\begingroup$ @Paula According to the cited paper, $U$ is a latent variable with distribution $D$. This can be a normal, logistic, or what have you. However, in order to simulate from any distribution, you always start from a uniform random simulation and somehow transform it into a sample from the desired distribution. Like in your normal example. Sometimes this step is hidden since you simulate from a more complex distribution using software, but inside this software, there is a uniform random variable, which is the only thing, with known distribution, we know how to simulate from. $\endgroup$
    – Scorpion
    Commented Aug 9, 2016 at 13:43
  • 1
    $\begingroup$ It seems to me that you are using those codes as a "black box" and hence cannot find the transform function $\phi$ without opening the codes. In that sense your question has no answer. $\endgroup$
    – Xi'an
    Commented Aug 9, 2016 at 15:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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