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,


(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 $\lambda$.

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    $\begingroup$ This is not such an issue. In the Poisson example, the rv $Y$ cannot be defined without the knowledge of $\lambda$. In that sense, the distribution of $Y$ is conditional on $\lambda$, whether it is a constant (hence a rv with a Dirac mass) or a random variable. By the very nature of the conditional distribution, it does not make a difference. $\endgroup$ – Xi'an Dec 6 '14 at 19:46
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    $\begingroup$ That is exactly what I was thinking a minute ago. However, I was wondering if a conditional is always based on a single value: $p(y|\lambda=\lambda_{0})$. That is, when we write $p(y|\lambda)p(\lambda)$, this is actually $p(y|\lambda=\lambda_{0})p(\lambda=\lambda_{0})$ for every $\lambda_{0}$ in the distribution. $\endgroup$ – Robert Smith Dec 6 '14 at 20:22
  • $\begingroup$ No, no. no [good question!]! The definition of a conditional distribution is that, once the conditioning random variable $\lambda$ takes a realisation $\lambda_0$ say, then the distribution of $Y$ is $p(y|\lambda_0)$. But, when we write $p(y|\lambda)$, we mean that whatever the realisation of $\lambda$ is, this will be the corresponding density of $y$. So you can unreservedly write a density conditioned by a random variable. $\endgroup$ – Xi'an Dec 6 '14 at 20:29
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    $\begingroup$ Uhm, I'm not sure what is the difference between what you said and what I said. I don't mean to say that $p(y|\lambda)$ implies $p(y|\lambda=\lambda_{0})$. Since, $p(y|\lambda)$ (where $\lambda$ is a random variable) is a family of poisson distributions, each centered in a different $\lambda$. So, when we take a realization $\lambda_{0}$, there is a single poisson distribution $p(y|\lambda=\lambda_{0})$. Is this correct? $\endgroup$ – Robert Smith Dec 6 '14 at 20:55
  • $\begingroup$ (+1) this is indeed correct! $\endgroup$ – Xi'an Dec 6 '14 at 21:00

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.


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 $\lambda$
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    $\begingroup$ Right, but one also writes $p(y \mid \lambda)$ in frequentist statistics to denote a distribution depending on $\lambda$. This has not a conditional meaning, this is just another notation for $p_\lambda(y)$ for example. $\endgroup$ – Stéphane Laurent Dec 6 '14 at 18:40
  • $\begingroup$ I would write $p(y \,;\, \lambda)$ in that case to avoid ambiguity, or $p_\lambda(y)$ as you pointed out. $\endgroup$ – Adrian Dec 6 '14 at 18:59
  • $\begingroup$ Thank you, Adrian. I was about to edit my question with Stéphane's observation. Sometimes, the conditional notation is used as a way to convey dependence. $\endgroup$ – Robert Smith Dec 6 '14 at 19:05
  • $\begingroup$ @Adrian Just to clarify. In the Bayesian case, we are using a conditional notation but not conditioning on $\lambda$, correct? Otherwise, a true conditional $y|\lambda$ would be a random variate with Dirac mass instead of $F$ as Xi'an pointed out. Then, how do you describe a true conditional in a Bayesian setting? Obviously, if a prior is explicit, there is a way to identify $y|\lambda$ as $y$ depending on $\lambda$ where $\lambda$ has some distribution. However, this is not clear in all cases. $\endgroup$ – Robert Smith Dec 6 '14 at 20:32
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    $\begingroup$ no, the Bayesian approach explicitly relies on all sampling distributions being conditional on the parameter(s) since the parameter(s) are then random. So $p(y|\lambda)$ is a true conditional then. And my remark about the Dirac mass was that the prior of $\lambda$ could formally be a Dirac mass, in which case $\lambda$ would a constant, e.g. $\lambda_0$. $\endgroup$ – Xi'an Dec 6 '14 at 21:47

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