# CIS Primer Question 1.5.2

February 9, 2019
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# CIS Primer Question 1.5.2

Here are my solutions to question 1.5.2 of Causal Inference in Statistics: a Primer (CISP).

I’ll use different indexing to make the notation clearer. In particular, the indices will match the values of the conditioning variables.

## Part a

The full joint probability is

$\mathbb P(x, y, z) = \mathbb P (z) \cdot \mathbb P (x \mid z) \cdot \mathbb P (y \mid x, z)$

using the decomposition formula. Each factor is given by

\begin{align} \mathbb P (z) &= z r + (1 – z) (1 – r) \\ \mathbb P (x \mid z) &= xq_z + (1 – x)(1 – q_z) \\ \mathbb P (y \mid x, z) &= yp_{x, z} + (1 – y)(1 – p_{x, z}) \end{align}

where each parameter is assumed to have support on $$\{0, 1\}$$.

The marginal distributions are given by

\begin{align} \mathbb P(x, z) &= \mathbb P(x \mid z) \cdot \mathbb P (z) \\ \mathbb P(y, z) &= \mathbb P(0, y, z) + \mathbb P(1, y, z) \\ \mathbb P(x, y) &= \mathbb P(x, y, 0) + \mathbb P(x, y, 1) \\ &= yp_{x, 0} + (1 – y)(1 – p_{x, 0}) + yp_{x, 1} + (1 – y)(1 – p_{x, 1}) \\ &= y (p_{x, 0} + p_{x, 1}) + (1 – y)(2 – p_{x, 0} – p_{x, 1}) . \end{align}

Furthermore,

\begin{align} \mathbb P (x) &= \sum_z \mathbb P(x \mid z) \mathbb P (z) \\ &= \sum_z (xq_z + (1 – x)(1 – q_z))(zr + (1 – z)(1 – r)) \end{align}

so that

\begin{align} \mathbb P(X = 0) &= (1 – q_0)(1 – r) + (1 – q_1)r \\ \mathbb P(X = 1) &= q_0(1 – r) + q_1r \end{align}

## Part b

The increase in probability from taking the drug in each sub-population is:

• $$\mathbb P(y = 1 \mid x = 1, z = 0) – \mathbb P(y = 1 \mid x = 0, z = 0) = p_{1, 0} – p_{0, 0}$$; and
• $$\mathbb P(y = 1 \mid x = 1, z = 1) – \mathbb P(y = 1 \mid x = 0, z = 1) = p_{1, 1} – p_{0, 1}$$.

In the whole population, the increase is $$\mathbb P(Y = 1 \mid X = 1) – \mathbb P(Y = 1 \mid X = 0)$$, calcualted via

\begin{align} & \sum_{z = 0}^1 \mathbb P(Y = 1, Z = z \mid X = 1) – \mathbb P(Y = 1, Z = z \mid X = 0) \\ &= \sum_{z = 0}^1 \frac{\mathbb P(X = 1, Y = 1, Z = z)}{\mathbb P(X = 1)} – \frac{\mathbb P(X = 0, Y = 1, Z = z)}{\mathbb P(X = 0)} \\ &= \frac{(1 – r)q_0p_{1, 0} + rq_1p_{1, 1}}{q_0(1 – r) + q_1r} \frac{(1 – r)(1 – q_0)p_{0, 0} + r(1 – q_1)p_{0, 1}}{(1 – q_0)(1 – r) + (1 – q_1)r} \end{align}

## Part c

There’s no need to be smart about this. Let’s just simulate lots of values and find some combination with a Simpson’s reversal. We’ll generate a dataset with a positive probability difference in each sub-population, then filter out anything that also has a non-negative population difference.

set.seed(8168)

N <- 10000

part_c <- tibble(
id = 1:N %>% as.integer(),

r = rbeta(N, 2, 2),   # P(Z = 1)

q0 = rbeta(N, 2, 2),  # P(X = 1 | Z = 0)
q1 = rbeta(N, 2, 2),  # P(X = 1 | Z = 1)

p00 = rbeta(N, 2, 2), # P(Y = 1 | X = 0, Z = 0)
p10 = rbeta(N, 2, 2) * (p00 - 1) + 1, # P(Y = 1 | X = 1, Z = 0)
p01 = rbeta(N, 2, 2), # P(Y = 1 | X = 0, Z = 1)
p11 = rbeta(N, 2, 2) * (p01 - 1) + 1, # P(Y = 1 | X = 1, Z = 1)

diff_pop = (p10 * q0 * (1 - r) + p11 * q1 * r) / (q0 * (1 - r) + q1 * r) - (p00 * (1 - q0) * (1 - r) + p01 * (1 - q1) * r) / ((1 - q0) * (1 - r) + (1 - q1) * r),
diff_z0 = p10 - p00,
diff_z1 = p11 - p01
) 

As a check, there should be no rows with a non-positive difference.

check <- part_c %>%
filter(diff_z0 <= 0 | diff_z1 <= 0) %>%
nrow()

# throw error if there are rows
stopifnot(check == 0)

check
[1] 0

Now we simply throw away any rows with a non-negative population difference. Here is one combination of parameters exhibiting Simpson’s reversal.

simpsons_reversal <- part_c %>%
filter(diff_pop < -0.05) %>%
gather(term, value)
term value
id 109.0000000
r 0.2837123
q0 0.0664811
q1 0.8468126
p00 0.8441892
p10 0.8827558
p01 0.5273831
p11 0.5816885
diff_pop -0.1933634
diff_z0 0.0385666
diff_z1 0.0543054

As a final check, let’s generate a dataset for this set of parameters.

df <- simpsons_reversal %>%
crossing(unit = 1:N) %>%
mutate(
z = rbinom(N, 1, r),
x = rbinom(N, 1, if_else(z == 0, q0, q1)),
p_y_given_x_z = case_when(
x == 0 & z == 0 ~ p00,
x == 0 & z == 1 ~ p01,
x == 1 & z == 0 ~ p10,
x == 1 & z == 1 ~ p11
),
y = rbinom(N, 1, p_y_given_x_z)
) %>%
select(unit, x, y, z)

The empirical joint probability distribution is as follows.

p_x_y_z <- df %>%
group_by(x, y, z) %>%
count() %>%
ungroup() %>%
mutate(p = n / sum(n))
x y z n p
0 0 0 1068 0.1068
0 0 1 197 0.0197
0 1 0 5609 0.5609
0 1 1 224 0.0224
1 0 0 52 0.0052
1 0 1 1016 0.1016
1 1 0 400 0.0400
1 1 1 1434 0.1434

The population-level probability difference is given by:

diff_pop <- p_x_y_z %>%
group_by(x) %>%
summarise(p = sum(n * y) / sum(n)) %>%
mutate(diff = 1 - 0)
0 1 diff
0.8217808 0.6319779 -0.1898028

which is close to the theoretical value.

Similarly, the sub-population differences are

diff_z <- p_x_y_z %>%
group_by(x, z) %>%
summarise(p = sum(n * y) / sum(n)) %>%
mutate(diff = 1 - 0)
z 0 1 diff
0 0.8400479 0.8849558 0.0449078
1 0.5320665 0.5853061 0.0532396

which are also close to the theoretical values we calculated. More importantly, they have a different sign to the population difference, confiming that we have case of Simpson’s reversal.

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