I raised this question because of an argument I am having with a question from user697473 here. The title of his question is "Formal definition of random assignment." In the post, he makes a claim that one can get unbiased estimates of the difference in treatment effect even if one group that is represented in the population is not sampled. At least that is the way I interpret the claim. He points to a paper coauthored by Don Rubin on causal modeling here. Both he and StasK argue through examples that this claim is true. This issue was also raised in another question that he posted here. In that post titled "Random assignment: why bother?" gung expresses skeptism.

In the first question, my answer included a statement to the effect that without sampling all possible assignments of a potential confounding factor you cannot construct an estimate of the difference in treatment effect without assigning samples to each stratum for the confounding factor.

The Issue

I am confused about the claim. As I interpret the claim I am sure it is false. The claim is supported by two examples one given by user697473 and the other by StasK. The examples are confusing to me and in neither case is a proof or demonstration given to show that the claim is true.

I could be misinterpreting the claim but based on my interpretation it is false. To illustrate my point let's look at a famous example from the design of experiments book of the Godfather of randomization Sir Ronald Aylmer Fisher: "The Lady Tasting Tea".

In case you don't know the example I will describe it and provide a modification to clarify the example. The lady claims that she has the ability to determine just by tasting a cup of tea whether the tea was poured first or the milk was poured first. Fisher poses an experiment to test her claim by providing some cups with tea poured first and others with the milk poured first. If the Lady has the expertise that she claims she should be able to correctly identify the types better than a person who guesses at random. So the experiment is designed to see if her probability of correct classification is better than 0.5. So he randomly provides the Lady with cups from both groups (in the example there were 6 cups 3 with milk poured first and 3 with tea poured first.

Now let me modify the experiment slightly. Suppose I have three brands of tea and for my population and these brands are served at a tea parties equally. The question I want to ask is: when given a cup of tea at a tea party can the Lady predict whether milk was poured first or the tea was poured first? I want to test whether her prediction accuracy is greater than 0.5 in the setting of tea parties. So I will apply Fisher's experiment to estimate this probability when the Lady is at one of these tea parties.
Note: I do not care whether or not she can differentiate the brands.

Suppose that in the population of tea parties the Lady's ability is:

Brand A Can predict tea first with probability 0.7 and can predict milk first with probability 0.7.

Brand B Can predict tea first with probability 0.75 and can predict milk first with probability 0.75.

Brand C Can predict tea first with probability 0.5 and can predict milk first with probability 0.6.

So she does no better than chance with Brand C but can do better than chance with Brands A and B.

For the experiment I only provide Brands A and B for her to taste. All I want to know is her ability to predict correctly in the normal tea party situation.

My Questions

  1. Can I get an unbiased estimate of her prediction capabilities with this experiment?

  2. If the answer to (1) is no, is user697473's claim false or have I misinterpreted it?

Randomization is used to avoid bias that is produced by confounding variables. This is important when trying to draw inference from a sample to a population. In my example the Brands are confounders. The treatment is tasting the cups. In the population she would get Brand A, Brand B and Brand C each 1/3rd of the time. If I know this and I sample Brands with unequal probabilities I claim that I can get an unbiased estimate of her prediction capabilities by taking a specific weighted average of the estimates of the proportions if I use a stratified random sample with strata totals n$_1$, n$_2$ and n$_3$ all greater than 0. But I cannot if n$_3$ = 0. Furthermore, there is no other estimate I can calculate from a sample where allocation to Brand C equals 0 that will be an unbiased estimate.


2 Answers 2


You have not correctly interpreted user697473's claim. He is not talking about failing to include any data from brand C. He was talking about giving a particular vector of assignemnts $0$ probability. He was not saying that you can magically determine the value of some variable while never testing it. He wants to be able to use a balanced random subset, so that each point is included in the random subset with the right probability, but not a uniformly random one.

For example, if the set is $\lbrace x_1,x_2,x_3,x_4 \rbrace$, then the following random subsets of uniform size $2$ all have the property that the probability that $x_i$ is included is $1/2$:

$S_1 = 1/6\lbrace x_1,x_2\rbrace + 1/6\lbrace x_1,x_3\rbrace + 1/6\lbrace x_1,x_4\rbrace + 1/6\lbrace x_2,x_3\rbrace + 1/6\lbrace x_2,x_4\rbrace + 1/6\lbrace x_3,x_4\rbrace $

$S_2 = 1/4\lbrace x_1,x_3\rbrace + 1/4\lbrace x_1,x_4\rbrace + 1/4\lbrace x_2,x_3\rbrace + 1/4\lbrace x_2,x_4\rbrace $

$S_3 = 1/2\lbrace x_1,x_2\rbrace + 1/2\lbrace x_3,x_4\rbrace $

These are all balanced in the sense that if you compute the average value of some function $f$ over the random set, the expected value is $1/4(f(x_1) + f(x_2) + f(x_3)+f(x_4))$. In the third random subset, the probability of the subset $\lbrace x_1,x_3 \rbrace$ is $0$.

That said, the point of the experiment should not be to produce an unbiased estimate. That is just one consideration. Another goal is to provide useful information. If you know that you may want to estimate $f$ on a subset $T$ (say $\lbrace x_1,x_2 \rbrace$) and its complement and to subtract the two, the quality of your estimate depends on $\#(T \cap S)$ and $\#(T^c \cap S)$. Then not all balanced subsets have the same quality. For that task, $S_3$ is worse than random assignment ($S_1$), while $S_2$ is better than random assignment.

  • 3
    $\begingroup$ +1 The point about good experiments needing more than just unbiased estimates seems to get to the core of the controversy. $\endgroup$
    – whuber
    Commented Sep 22, 2012 at 19:03
  • 1
    $\begingroup$ That is more understandable. I understand about things like fractional factorial experiments and other designs like Latin squares where you can do inference and get unbiased estimates when there are certain assumptions suchas certain interaction terms being zero. Somehow that wasn't clear to me in his posts. My comments were always based on the assumptions illustrated in my tea example. $\endgroup$ Commented Sep 22, 2012 at 19:54
  • 1
    $\begingroup$ Apologies for off-topic ping: there is a suggestion on Meta to make [randomized-experiment] a synonym of [random-allocation] tag (stats.meta.stackexchange.com/a/4651). You have enough reputation in this tag in order to vote for this suggestion here: stats.stackexchange.com/tags/random-allocation/synonyms - it now needs 4 upvotes to go through. If you disagree with the proposal, consider commenting on Meta to explain why. I will delete this comment soon. Cheers. $\endgroup$
    – amoeba
    Commented May 2, 2017 at 12:04

Michael, I did not say that when you exclude a group from randomization you can still get unbiased estimates, and I would never argue in favor of this idea; this is certainly not true. What is true is that you don't have to have the same sampling rates, and that's the imbalance that you can correct with weights. You illustrate this point with your Lady-Tasting-Tea example, which works unless one of the sample sizes is zero.

The example I gave is aimed at a different subtle point. It uses all units in the population, but it's not a full design (which would be ${5 \choose 2}=10$ possible assignments); yet every unit is assigned to treatment twice, and to control, three times. So with respect to randomization to these assignments, the difference of the treatment group means is an unbiased estimator of the mean population difference (in potential outcomes). This example shows that you don't have to fully randomize, in the sense of tossing a coin independently for every unit (which may put you to an awkward situation of having no controls or no treatments with probability 1/16 for sample this small, by the way). However, for this example to work out, you need to have a sample size that was fixed in advance, and you had to toss a five-sided coin in advance before the very first unit enters the study. So each approach has its pros and cons; I doubt that the balanced sampling example that I gave is hugely practical, although cube sampling appears to be picking up in survey statistics.

I am not a causal inference statistician, nor am I terribly good with experimental designs. But I know my randomization inference from survey statistics, so the way I've been viewing this problem is from the perspective of Horvitz-Thompson estimator.

  • $\begingroup$ Thanks StasK. I think your example and the OPs example both addressed the point he was making. I didn't understand the examples initially and the OP was suggesting that it might indicate that randomization is not necessary. Certainly the tea example shows the importance of randomization as did Fishers simpler form of the lady tasting tea scenario. I suspected that I could be misinterpreting and that was one of the reasons I asked the question to find out what I was misinterpretation. I think the example I gave showed clearly what my interpretation was and help you and Doug to correct me. $\endgroup$ Commented Sep 24, 2012 at 1:50
  • $\begingroup$ OK, I am glad things are clearer now! $\endgroup$
    – StasK
    Commented Sep 24, 2012 at 3:42
  • $\begingroup$ Apologies for off-topic ping: there is a suggestion on Meta to make [randomized-experiment] a synonym of [random-allocation] tag (stats.meta.stackexchange.com/a/4651). You have enough reputation in this tag in order to vote for this suggestion here: stats.stackexchange.com/tags/random-allocation/synonyms - it now needs 4 upvotes to go through. If you disagree with the proposal, consider commenting on Meta to explain why. I will delete this comment soon. Cheers. $\endgroup$
    – amoeba
    Commented May 2, 2017 at 12:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.