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I am working on trying to recover correct group means in a causal inference model (I'm pretty new to causal inference) that I'm running on a simulated dataset. I believe that the issue I'm running into is the method and order in which I am controlling for different effects in the dataset.

Broadly speaking, here's the model that I'm interested in fitting.

Model DAG

We have data divided into 4 groups. For each observed value of obs, we get four pieces of information: which individual individual 1 was, which individual individual 2 was, and the value of the covariate effect. Each individual in the collections of individual 1 and individual 2 has some unknown, unobservable effect mean; the values of effect are observed for each observation, as is which group it came from. Additionally, note that the means of the effect values are associated with an individual 1 value, but have some random fluctuation.

The means of the unobserved values of individual 1 and individual 2 vary by group, as does the mean of the values of effect. We're interested in recovering the group means of individual 1 after controling for individual 2 and effect.

Here is the data generating process in R:

set.seed(20)
group <- c(rep(1, 100), rep(2, 100), rep(3, 100), rep(4, 100))
individual1 <- 1:400
individual2 <- 1:400
rand_effect1 <- c(rnorm(100, mean = 0, sd = 1),
                  rnorm(100, mean = 0.5, sd = 1),
                  rnorm(100, mean = 1, sd = 1),
                  rnorm(100, mean = 1.5, sd = 1))
rand_effect2 <- c(rnorm(100, mean = 0, sd = 1),
                  rnorm(100, mean = -1, sd = 1),
                  rnorm(100, mean = -2, sd = 1),
                  rnorm(100, mean = -3, sd = 1))
fix_effect <- c(rnorm(100, mean = 0, sd = 1),
                rnorm(100, mean = 0.25, sd = 1),
                rnorm(100, mean = 0.5, sd = 1),
                rnorm(100, mean = 0.75, sd = 1))
ind1 <- sample(individual1, size = 5000, replace = TRUE)
re1 <- rand_effect1[ind]
fe <- fix_effect[ind] + rnorm(5000, mean = 0, sd = 0.5)
grp <- group[ind]
f <- (floor((ind1-1) / 100) * 100) + 1
c <- (floor((ind1-1) / 100) + 1) * 100
ind2 <- c()
for(i in 1:5000) {
  v <- floor(i / 100)
  ind2 <- c(ind2, 
            sample(f[i]:c[i], 1))
}
re2 <- rand_effect2[ind2]
obs <- re1 + re2 + fe + rnorm(5000, mean = 0, sd = 4)
training_data <- data.frame(
  group = as.character(grp),
  individual1 = as.character(ind1),
  individual2 = as.character(ind2),
  effect = fe,
  obs = obs
)

I could certainly throw all of this into a random effects model with individual 1 and individual 2 handled appropriately. However, it would be ideal to include group means in the model; this model should be able to handle individuals with small sample sizes from different groups appropriately. Knowing nothing about an individual 1 value except for the fact that they came from group 3 should be able to tell us, eventually, that their individual random effect is in the ballpark of 1 (as is clear in the data generating process). However, it's clearly not as easy as throwing in a group factor into the model, since there are different group effects on all three inputs.

Any suggestions about how to appropriately control for each component in this model would be very appreciated!

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  • $\begingroup$ "Additionally, note that the means of the effect values are associated with an individual 1 value, but have some random fluctuation." This is not very clear because it contradicts with the image. Do you mean that there should also be an arrow from individual 1 to effect. The individual 1 has a causal influence on the effect. $\endgroup$ Jun 24 at 7:08
  • $\begingroup$ "Each individual ... has some unknown, unobservable effect mean" What does this mean? Are you talking about a more specific model, like a linear model. The phrase 'each individual has a mean', or 'Alice and Bob have a mean' makes little sense without context. $\endgroup$ Jun 24 at 7:12
  • $\begingroup$ Your code does explain your model a bit better. ------ One remark, the part where you sample the ind2 is a bit confusing. You use large functions including floor operations, in order to get a subset of 100 individuals, but it seems like you could just use the grp variable to reach the same goal. Addition, it seems like individual 1 and individual 2 can be the same id, is that supposed to be correct? $\endgroup$ Jun 24 at 7:25

1 Answer 1

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However, it's clearly not as easy as throwing in a group factor into the model, since there are different group effects on all three inputs.

I see only one way how group has an effect and that is in the different means of the normal distributions.

rand_effect1 <- c(rnorm(100, mean = 0, sd = 1),
                  rnorm(100, mean = 0.5, sd = 1),
                  rnorm(100, mean = 1, sd = 1),
                  rnorm(100, mean = 1.5, sd = 1))
rand_effect2 <- c(rnorm(100, mean = 0, sd = 1),
                  rnorm(100, mean = -1, sd = 1),
                  rnorm(100, mean = -2, sd = 1),
                  rnorm(100, mean = -3, sd = 1))
fix_effect <- c(rnorm(100, mean = 0, sd = 1),
                rnorm(100, mean = 0.25, sd = 1),
                rnorm(100, mean = 0.5, sd = 1),
                rnorm(100, mean = 0.75, sd = 1))

You can just as well consider normal distributions with zero mean and a separate fixed effect for group.

rand_effect1 <- c(rnorm(100, mean = 0, sd = 1),
                  rnorm(100, mean = 0, sd = 1) + 0.5,
                  rnorm(100, mean = 0, sd = 1) + 1,
                  rnorm(100, mean = 0, sd = 1) + 1.5)
rand_effect2 <- c(rnorm(100, mean = 0, sd = 1),
                  rnorm(100, mean = 0, sd = 1) -1 ,
                  rnorm(100, mean = 0, sd = 1) - 2,
                  rnorm(100, mean = 0, sd = 1) - 3)
fix_effect <- c(rnorm(100, mean = 0, sd = 1),
                rnorm(100, mean = 0, sd = 1) + 0.25,
                rnorm(100, mean = 0, sd = 1) + 0.5,
                rnorm(100, mean = 0, sd = 1) + 0.75)

Or is this just a simplification of your true problem and do you have more complicated ways how group has an effect?


In addition, because the three different effects of group all result in effectively a single effect, you can not obtain the 12 individual means. You can only discover the sum of three means and obtain 4 parameters. Or if you mix individuals of different groups then you should be able to figure out 8 parameters.

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  • $\begingroup$ This is a simplification of the true problem. In reality, while knowing the 'true group' of an individual should tell you about their mean, we may observe individuals in groups that aren't their true groups. For example, suppose ind1 represents teacher, ind2 represents student, and group represents school. Even if a teacher primarily teaches at school 1, they may make a guest appearance in school 2, and may interact with students from school 3 in school 2. Could you explain about how we should be able to figure out 8 (but not 12) parameters if we mix individuals of different groups? $\endgroup$
    – drum75mph
    Jun 24 at 14:04
  • $\begingroup$ @drum75mph the reason why you are never able to figure out all 12 variables is because the expression obs <- re1 + re2 + fe + rnorm(5000, mean = 0, sd = 4)contains the sum/term re1 + fe which completely depends on a single parameter, the group. If all rand_effect1 would be one unit less and all fix_effect would be one unit more, then it would give the same obs. That is, the following would be the same: obs <- (re1+1) + re2 + (fe-1) + rnorm(5000, mean = 0, sd = 4) The grouphas only an effect on the intercept, giving 4 parameters one intercept for each grouo. $\endgroup$ Jun 24 at 14:57
  • $\begingroup$ That makes sense. In the true problem, we can observe the values of fe, and know the model to be obs <- re1 + re2 + fe + rnorm(...). Could we instead directly model obs2 := obs - fe instead with a model along the lines of obs2 <- re1 + re2 + rnorm(...) and find ways to discern the different group means for re1 and re2? $\endgroup$
    – drum75mph
    Jun 24 at 16:48

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