What is the meaning of the conditional $y|b$ I think I'm confused about a very simple thing. When we say that some variable is distributed as a Poisson distribution and we write $y \sim \text{Pois}(\lambda)$, is this the same that saying $y|\lambda\sim \text{Pois}(\lambda)$? 
I remember that at first, I thought the conditional was meant to describe things that we already know, so if I write $y|\lambda$, I assumed that $\lambda$ was a known constant. Then I encountered cases where $\lambda$ was a distribution, so I retroactively thought that $y\sim \text{Pois}(\lambda)$  was used with $\lambda$ as an implicit parameter because it is obvious that $y$ depends on $\lambda$, and many books routinely make some parameters implicit. Furthermore, it was unusual to think of a distribution $\lambda$ as a known value. In any case, it could be a possible realization of $\lambda$. However, using $y$ (without conditional), I think, is a bit a misleading since this implies a marginalized $y$.
So, what is the correct interpretation of the conditional in distributions such as $y \sim \text{Pois}(\lambda)$? 
As a way to describe more clearly where both interpretations of conditional can be confusing, take this example: Suppose a Poisson distribution $y_{t}\sim \text{Pois}(\mu_{t})$ where $\log(\mu_{t}) = X_{t}\beta + b$. Then,
$$cov(y_{t},y_{s})=E[cov(y_{t},y_{s}|b)]+cov[E(y_{t}|b),E[y_{s}|b)]$$
(using the law of total covariance)
Therefore, $y_{t}$ should be understood as the Poisson distribution without conditional, because in the right hand side of this identity, we are using the conditional on $b$.
As you can see, I'm confused about what it really means $y|\lambda$ as opposed to a marginalized $y$ or a distribution $y$ with parameter $\lambda$.
 A: It seems ambiguous to me whether writing $y$ or $y \; | \; \lambda$ is correct.  I'd say the answer depends on whether you're in a bayesian or a frequentist setting:


*

*If you're being bayesian, you'll often model parameters as random. In that case, you'd write $y \;|\; \lambda \; \sim \; F(\cdot\,; \lambda)$, and you would have some prior over $\lambda$ (which you could use to calculate the marginal distribution of $y$, if desired)

*If you're being frequentist, you think of $\lambda$ as something unknown but fixed, so you just write $y \; \sim \; F(\cdot\,; \lambda)$ -- it wouldn't make sense to condition on $\lambda$

A: As a Bayesian, it does not make much of a difference to me to think of $\mathcal{P}(\lambda)$ as a given distribution, indexed by a parameter $\lambda$, or as a conditional distribution, conditional on the realisation of a random variable $\lambda$. Indeed, in either case, when I observe 
$$
y_1,\ldots,y_n\stackrel{\text{iid}}{\sim}\mathcal{P}(\lambda)
$$
the data is indexed by a given if unknown value of $\lambda$. Whether or not this $\lambda$ is the realisation of a random variable does not change the behaviour of the data. Remember, there is only one realisation of $\lambda$ for a given dataset, no matter its size. So, even when making the assumption that $\lambda\sim\pi(\lambda)$ (a certain prior distribution chosen by me), I do not get observations from the marginal
$$
m(y_i) = \int_0^\infty f(y_i|\lambda)\pi(\lambda)\,\text{d}\lambda
$$
but observations from the conditional. Hence, for the likelihood function and related inference, conditioning or not does not make a difference as the parameter is assumed fixed for the data at hand.
Conditioning only makes a difference when running a Bayesian analysis, since the posterior
$$
\pi(\lambda|y_1,\ldots,y_n) \propto f(y_1|\lambda)\cdots f(y_n|\lambda)\pi(\lambda)
$$
only makes sense as a conditional distribution from the joint distribution
$$
\pi(\lambda,y_1,\ldots,y_n) = f(y_1|\lambda)\cdots f(y_n|\lambda)\pi(\lambda)
$$
(meaning that $\lambda$ has to be a random variable for this derivation to make sense).
At a probabilistic level, assuming the $\sigma$-algebra on $\mathcal{Y}\times\Theta$ is induced by the products of the measurable sets on $\mathcal{Y}$ and on $\Theta$, writing $p_\lambda(y)$ or $p(y|\lambda)$ does not make a difference either.
