A hierarchical Bayesian model in pymc3 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 ...  
 A: 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!
A: 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 $\sigma.$
