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I've seen Bayesian models specified as \begin{align*} Y_i|v_i &\overset{ind}{\sim} f_i(y_i|v_i),\\ v_i & \overset{ind}{\sim} g_i(v_i). \end{align*} My question is about the top line $Y_i|v_i\overset{ind}{\sim}f_i(y_i;v_i)$. It seems like, strictly speaking, this isn't mathematically precise because $Y_i$ and $Y_j$ are being conditioned on two separate variables. I can think of two ways to interpret this notation: $Y_i\perp Y_j|v_i,v_j$, for all $i,j$, or $\{Y_i\}_i$ is an independent family of random variables given the vector $\{v_i\}_i$. I can't tell whether or not they're the same. So are these two interpretations equivalent, and if not, which is correct?

Edit: To be clear, I'm confused about the "ind" above the "distributed as" sign $\sim$ on the first line, given that the conditioning variable has the subscript $i$.

Another possibility also occurred to me, which is that the "ind" is redundant and that the joint distribution of all $Y_i$ and $v_i$ would be determined by just writing \begin{align*} Y_i|v_i & \sim f_i(y_i;v_i),\\ v_i & \overset{ind}{\sim} g_i(v_i), \end{align*} but I'm not sure how to show that either. Edit: I realized this can't be true, because \begin{align*} v_1, v_2 &\text{ iid } N(0, 1),\\ Y_i | v_i & \sim N(v_i, 2),\quad \text{for } i =1, 2, \end{align*} is true for both $Y_i|v_1,v_2 \text{ iid } N(v_1+v_2,1)$, and $Y_i|v_1,v_2 \overset{ind}{\sim} N(v_i,2)$.

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    $\begingroup$ Technically you should condition on all the $v$’s in the first line. Or drop the conditioning entirely and just say that each line is implicitly conditional on all lines below. $\endgroup$
    – guy
    Jun 13 '18 at 20:59
  • $\begingroup$ @guy thanks, so you mean that technically in these models, the $Y_i$ are conditionally independent given the entire vector of parameters $\{v_i\}_i$? $\endgroup$
    – user209990
    Jun 13 '18 at 21:18
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    $\begingroup$ Yes, that is what is usually intended. If you take the statements at face value then you don’t have a fully-specified model since you are being ambiguous about what dependencies there are if you condition on the whole $v$ vector. I usually either say that these independencies are implicit, or else I describe things in terms of a generative scheme. $\endgroup$
    – guy
    Jun 13 '18 at 23:01
  • $\begingroup$ Wait, you said "you are being ambiguous about what dependencies there are if you condition on the whole $v$ vector." Isn't that less ambiguous? If you say, $v$ has independent components, and $y$ has conditionally independent components given the vector $v$, and $v_i\sim g$ and $y_i|v_i\sim f_{v_i}$ doesn't that specify the joint distribution of $v$ and $y$? $\endgroup$
    – user209990
    Jun 13 '18 at 23:31
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    $\begingroup$ I mean, if you only report the independencies in the OP, then you are being ambiguous about what happens if you condition on the whole vector; that’s what I said but it is a bit difficult to parse. $\endgroup$
    – guy
    Jun 14 '18 at 0:10
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You are correct to say that the statement is not mathematically precise; it is a shorthand that is sometimes used to set out a hierarchical model. I'm not a fan of this notation personally, since it is an unnecessary abuse of notation. (I prefer to specify independence separately.)

Presumably the intention of such statements is to set up mutually independent random variables with specified conditional distributions. Implicitly, the argument variable is taken to be independent of any other iteration of the conditioning variable not mentioned in the conditioning statement. So I would think that the shorthand statement,

$$Y_i|v_i \overset{ind}{\sim} f_i(y_i|v_i),$$

would formally mean that $\{ Y_k \}$ are mutually independent conditional on $\{ v_k \}$, and each element of the former has conditional distribution $Y_i | \{ v_k \} \sim f_i(v_i)$. So your second interpretation is the one I would use. The reason I would think that this is the interpretation is simply that you cannot proceed with your analysis with weaker assumptions. If what was intended by the notation was certain kinds of pairwise independence, or independence conditional only on subsets of $\{ v_k \}$ then there is not enough specification to set up a hierarchical model with a fixed likelihood function.

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Taking the simpler case

$$ v_i \overset{ind}{\sim} g_i(v_i) $$

presumably means that "$v_i$'s are independently distributed according to $g_i$ distributions each". The notation is a shortcut for describing each $i$-th variable as separate case. There's often trade-off between simplicity and formality.

But the notation you quote is strange, e.g. it puts $v_i$ on both sides of $\sim$, while usually you'd see variable on left hand side and distribution with parameters on right hand side, so it is redundant if used like this.

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  • $\begingroup$ Thanks for the response. Maybe my notation was unclear, I was trying to abstract from a specific model that I saw. What I meant was that $f(y_i; v_i)$ is a distribution with parameter $v_i$. So for example $v_i \sim N(0, \sigma^2)$ could be a random effect inducing overdispersion in a Poisson distribution, and $Y_i|v_i\sim Poisson(exp(v_i))$. $\endgroup$
    – user209990
    Jun 13 '18 at 18:55

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