Formal definition of random assignment

I am looking for a formal definition of random assignment.

Let $\mathbf{Z}$ be a vector of treatment assignments in which each element is 0 (unit not assigned to treatment) or 1 (unit assigned to treatment). In a JASA article, Angrist, Imbens, and Rubin (1996, 446-47) say that treatment assignment $Z_i$ is random if $\Pr(\mathbf{Z} = \mathbf{c}) = \Pr(\mathbf{Z} = \mathbf{c'})$ for all $\mathbf{c}$ and $\mathbf{c'}$ such that $\iota^T\mathbf{c} = \iota^T\mathbf{c'}$, where $\iota$ is a column vector with all elements equal to 1.

In words, the definition seems to be this: assignment $Z_i$ is random if any vector of assignments that includes $m$ assignments to treatment is as likely as any other vector that includes $m$ assignments to treatment.

This definition seems unsatisfactory. What if I decide a priori that I want to rule out a particular vector of assignments, and choose one of the remaining vectors at random? This practice would not satisfy the AIR definition, but it would still be random assignment.

Here is an example. Imagine a binary assignment to treatment for each of two subjects. Let $\mathbf{Z}$ be the vector of treatment assignments. Then $\mathbf{Z}$ has four possible values: {0, 0}, {0, 1}, {1, 0}, and {1, 1}. By the AIR definition, assignment is random only if $\Pr(\mathbf{Z} = \{1, 0\}) = \Pr(\mathbf{Z} = \{0, 1\})$. But why should this be the definition of random assignment, or even a necessary condition for it? What if I simply decide that I want to rule out {0, 1} and choose at random from the three remaining vectors? It seems that this practice is consistent with conventional understanding of random assignment but inconsistent with the AIR definition.

So: is there a formal definition of random assignment that encompasses the idea that the experimenter may rule out some assignment vectors a priori?

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No. If you rule out some assignment vectors a priori you no longer have random assignment, strictly speaking. –  Peter Flom Sep 17 '12 at 12:38
Please explain the definitions in the second paragraph more clearly, i.e. whats is $\mathbf{z}$, $\mathbf{c}$ and $\mathbf{c}'$? –  mpiktas Sep 17 '12 at 12:39
@user697473, please give a link to the full text of the paper you cite. Respect the time of the people answering this; or at least assume they will be too lazy even to Google it (I certainly was). –  StasK Sep 17 '12 at 17:03
@StasK, thanks for the suggestion. I added a link to the original post. –  user697473 Sep 17 '12 at 17:28
@mpiktas, $Z_i$ is the $i$th element of $\mathbf{Z}$. $\mathbf{Z}, \mathbf{c},$ and $\mathbf{c'}$ are vectors of zeroes and ones; all of these vectors have the same length. $\iota$ is a "counting vector" -- all ones -- so $\iota^T\mathbf{c}$ is just the number (i.e., the sum) of ones in $\mathbf{c}$. And a similar definition holds for $\iota^T\mathbf{c'}$. I hope that this helps. –  user697473 Sep 17 '12 at 17:31

While Michael Chernick gave a good answer, I do not think that the people who are involved in treatment effect estimation think in terms of finite populations and randomization-based inference. Economists (Angrist and Imbens are well-known econometricians) usually don't; if the OP comes from the same tradition, that is the central issue of this question.

Economists have a model perspective instead, where there's a conceptual population from which units are taken, and there's some sort of an implicit permutation invariance, or "labels do not matter", assumption being made for these sampled units. It is this permutation invariance that is being characterized and quantified in the definition given by the OP. In finite populations, though, every unit is assumed to be unique, and denying a certain treatment for it in the randomization mechanism would produce an unestimable treatment effect. Switching from the model-based inference to randomization-based inference is very difficult; this may have been done in the cited paper, but not very clearly.

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Thank you. Your answer gets at my confusion. My deeper concern is: if I want to use a difference of means between control and treatment groups to estimate an ATE (w/o bias), must my assignment mechanism satisfy the AIR "random assignment" condition? Formally, let $\{Z_1, ..., Z_n\}$ be treatment status (0 or 1) and $\{Y_1, ..., Y_n\}$ be the outcomes of interest. If I have random assignment in the AIR sense, then $\sum Y_iZ_i - \sum Y_i(1-Z_i)$ is an unbiased estimator of the ATE. But if I don't satisfy the AIR condition, is my difference-of-means estimator of the ATE necessarily biased? –  user697473 Sep 17 '12 at 17:51
In randomization inference, the concept of bias is whether the expected value of that difference over all possible assignments, weighted with their probabilities, equals the population value. You may be able to create some special assignments (analogous to balanced sampling) such that the expected value is equal to the population value even though some combinations have zero probability of occurring. For instance, with 5 subjects and equally likely assignments of a pair {1,2}, {3,4}, {1,5}, {2,4}, {3,5} to the treatment, you can still get an unbiased estimate of the difference (I believe). –  StasK Sep 17 '12 at 23:03
Thanks again. Your example checks out, and I am not even sure that it is "special" -- which raises a deeper question of why we need random assignment at all if unbiased inference is our aim. Perhaps the question boils down to differences between the model-based and randomization-inference perspectives, but I am not sure of that. In any case, I'll write the question up in a separate post. –  user697473 Sep 21 '12 at 14:07
@user697473 I am not an expert on causal modeling. i think the Rubin paper deals with causal modeling and I think it is possible that there confusion of terminology. I feel skeptical like gung seems to on the other post. My intuition tells me that if I do a weighted selection of all possible conditions that that lead to the distribution of treatment As effect and treatment Bs effect then an unweighted average would possibly be biased but the bias could be corrected for by taking a particular weighted average. –  Michael Chernick Sep 22 '12 at 2:45
But if an assignment is possible its effect on the estimate is unknown so there is no way to compensate to guarantee unbiasedness. The population is unknown what you have is the sample and an assumed model for the population. What I don't see in your example or StasK's is how you prove that you have a way to guarantee unbiasedness of the estimate of the effect of treatment differences. Can you explain that to me? Is there something wrong with my intuition? –  Michael Chernick Sep 22 '12 at 2:53

This definition of random assignment seems to be assigning with equal probability. To assign 0 weight any of the possible assignments could create bias and should be considered a nonrandom assignemnt by any definition. However sampling with unequal nonzero weights can be an acceptable procedure (e.g. sampling randomly proportional to size or stratified random sampling with unequal samples per stratum are survey sampling examples). They fit into a more general definition of random sampling. If one is estimating a mean a weighted average can be used to get a unbiased estimate of the population mean. By excluding a possible outcome you change the population and it is not apprpriate to draw an inference to the large population that you chose not to sample from. Also because potential samples were excluded it is impossible to make a weighted adjustment to guarantee an unbiased estimation for the mean of the unrestricted population.

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Thank you. This response helped me to see that there was a simple error in my thinking. I was conflating (a) the idea of unequal probabilities attached to different assignment vectors, and (b) the possibility that, in an experiment with multiple treatments, subjects are assigned to the different treatment conditions with unequal probability. In the sorts of experiments that I see, (b) is very common. But the Angrist, Imbens, and Rubin definition speaks only to (a), not (b). Your answer helped me to see this. –  user697473 Sep 17 '12 at 18:26
Michael, one other thing: you write that "To assign 0 weight [to] any of the possible assignments could create bias," but this doesn't seem right. It seems possible to systematically exclude certain vectors of assignments in a way that preserves the possibility of unbiased estimation. I've given an example at stats.stackexchange.com/questions/37775/…. Of course, it's possible that I am overlooking some important flaw in my example. –  user697473 Sep 22 '12 at 1:31
One thing that you'll notice in the AIR paper is that they do not condition on covariates $X$. You can generalize the AIR exposition by doing so.
Let $X$ be an indicator for whether a subject is male. Also suppose that you want men to be more likely to receive treatment than women. You can have $$\begin{equation*}\Pr(z = c \mid X=1) = \Pr(z = c^\prime \mid X=1) \end{equation*}$$ and $$\begin{equation*}\Pr(z = c \mid X=0) = \Pr(z = c^\prime \mid X=0),\end{equation*}$$ but $$\begin{equation*}\Pr(z = c \mid X=1) > \Pr(z = c \mid X=0)\end{equation*}$$ and still satisfy random assignment in this context. This would justify stratified sampling, for example.