If $X\mid V=v \overset{id}\sim F$ is identically distributed only conditioning on $V$, is it biased to sample $X$ without relying on $V$? Suppose $X\mid V=v \overset{id}{\sim} F$ where $X$ is identically distributed only on levels of a discrete $V$. Now suppose we want to estimate the mean and variance of $X$. The associated procedure with $E(X)$ is normally to sample $X$ many times and to average those samples.
However, in this case I feel we should first draw a value of $v$, then draw a value of $X$ conditional on $v$. (i.e., the available values of $X$ corresponding to a given $v$. Then, we repeat it many times, before averaging it over.
What is the difference for $E(X)$ in both procedures? What about for $Var(X)$?
 A: I am not sure I understand this question.
First of all, what does it mean $X \mid V = v \sim F$?  Do you mean $F(v)$ instead of $F$?  I.e., among all cases where $V = v$, the variable $X \sim F(v)$, and among all cases where $V = v' \neq v$, the variable $X \sim F(v')$ but not $\sim F(v)$?  One example I have in mind is e.g. you roll a die to get $V \in \{1,2,3,4,5,6\}$ and then $X \sim Gaussian(V, 1)$, i.e. $V$ determines which Gaussian $X$ is.  Is this similar to what you have in mind?
Assuming the above, or perhaps even more generally, $E[X]$ and $Var[X]$ are well defined regardless of whether you know $V$'s values or not.  I.e., if you know $V=v$, that might e.g. enable you to estimate $E[X\mid V=v]$, but that in no way affects $E[X]$.
Or think of it this way: What does it mean to "draw $V$ first" vs not?  How do you "draw $X$ without drawing $V$ first"?  In the die-roll example above, you might think that implementationally you cannot "draw $X$" by itself, but in reality $X$ is well defined, and in practice you can easily ask a friend to "draw $X$" for you without telling you the "intermediate" or "hidden" value $v$.  Clearly, whether your friend told you $v$ or not has no bearing on $E[X]$ (the "true" mean) nor the sample mean of $X$.

UPDATED to address comments:  If you disregard $V=v$ (e.g. if its value is hidden from you), the different draws of $X$ will certainly still be i.i.d..  Lets simplify and consider this example:

*

*Roll a fair die and let $V \in \{1,2,3,4,5,6\}$ be the result.


*Flip a fair coin.  If Heads, $X= V$, and if Tails, $X = V + 0.1$.


*So $X$ is uniformly distributed between $12$ distinct values $\{1, 1.1, 2, 2.1, 3, 3.1, 4, 4.1, 5, 5.1, 6, 6.1 \}$.  All draws of $X$ will be i.i.d. and uniform like this.


*$E[X] = 3.55 =$ average of those $12$ equally likely values.  It is well defined, and a constant, and has nothing to do with $V$ or $v$.


*$E[X \mid V = v] = v + 0.05 =$ average of those $2$ equally likely values $\{v, v+ 0.1\}$.  This is not a number -- it is a function of $v$.


*Similarly $Var[X]$ is a number, and $Var[X \mid V = v]$ is a function of $v$.


*Lets stress the "function-ness" and write $f(v) = v + 0.05 = E[X \mid V = v]$.  Then we have $E[X] = E_V [f(V)] =$ the average of those $6$ equally likely values $\{f(1), f(2), f(3), f(4), f(5), f(6)\}$.  This is an intuitive fact with the fancy name Law of Total Expectation, and is often written in the fancy form $E[X] = E_V[E[X|V]]$.
So to conclude, if you are only interested in $E[X]$, e.g. estimating it from samples, then it does not matter whether during your data gathering you also record the underlying values of $V$ or not.
Hope this clears things up, as opposed to adding more confusion?
