1
$\begingroup$

Suppose we have the following model:

$X$ unobserved

$Y$ such that $Y|X \sim \mathcal{N}(X,\sigma^2)$, observed

$Z$ such that $Z|X \sim \mathcal{B}(1,X)$, observed

and suppose, given observed data $(y_i,z_i)_{i}$, we want to estimate $\sigma^2$

To do this in a Bayesian fashion, we can put some priors on $X$ and $\sigma^2$ and run a MCMC. Consider the following pymc3 implementation (uniform prior on $X$ and half normal on $\sigma^2$)

with pm.Model() as my_model:

    x = pm.Uniform('x', lower=0, upper=1)
    my_sd = pm.HalfNormal('my_sd', sd=1)
    y = pm.Normal('y', mu=x, sd=my_sd,observed=observed_y)
    z = pm.Bernoulli('z',p=x,observed=observed_z)

with my_model:
    step = pm.Metropolis()
    trace = pm.sample(100000, step=step)

If I remove the line z = pm.Bernoulli('z',p=x,observed=observed_z) I get the same results. Is this expected? One could expect that providing more data that are dependent on $X$ would make the difference ...

$\endgroup$
0
$\begingroup$

This does not seem to be necessarily wrong. If the process that generates Y given X is the same underlying physical phenomenon that generates Z given X - in other words, if Z is another view on Y when both are caused by the same X - then observing more data through Z wouldn't necessarily change what the model knows about X.

Of course, it is difficult to say anything with certainly without being able to tinker with the actual data!

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

You can see that \begin{align} p(\sigma \vert Y) &\propto p(\sigma)p(Y\vert\sigma) %\\&= %p(\sigma) \int_{0}^1\frac{1}{\sigma}\phi\left(\frac{x - y}{\sigma}\right)dx\\ %&= p(\sigma)\left[\Phi\left(\frac{y}{\sigma}\right) - \Phi\left(\frac{y-1}{\sigma}\right)\right], \end{align}

and that

\begin{align*} p(\sigma \vert Y, Z) &\propto p(\sigma)p(Y, Z\vert \sigma) \\ &= p(\sigma)p(Y\vert\sigma)p(Z\vert\sigma). \end{align*}

But, the distribution of $Z$ does not depend on $\sigma$, so \begin{align*} p(\sigma \vert Y, Z) &\propto p(\sigma\vert Y), \end{align*} Therefore, $Z$ carries no additional information about $\sigma.$

| cite | improve this answer | |
$\endgroup$

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.