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# CIS Primer Question 3.4.1

Here are my solutions to question 3.4.1 of Causal Inference in Statistics: a Primer (CISP). \(\DeclareMathOperator{\do}{do}\)

If we can only measure one additional variable to estimate the causal effect of \(X\) on \(Y\) in figure 3.8, then we should measure \(W\). From question 3.3.1 we see that no single variable satisfies the backdoor criteria. Moreover, visual inspection of the graph verifies that \(W\) satisfies the frontdoor criteria:

- it intercepts all (the only) directed paths from \(X\) to \(Y\);
- there is no unblocked path from \(X\) to \(W\); and
- all backdoor paths from \(W\) to \(Y\) are blocked by \(X\).

To illustrate this, lets simulate the causal effect in 3 separate ways:

- by intervention,
- via the backdoor, and
- via the frontdoor.

Here are the data. Note that we have created functions for \(W\) and \(Y\) for use later.

```
N <- 100000
W <- function(x) {
N <- length(x)
rbinom(N, 1, inv_logit(-x))
}
Y <- function(d, w, z) {
N <- length(d)
rbinom(N, 1, inv_logit(-d - w + 3*z))
}
df <- tibble(id = 1:N) %>%
mutate(
b = rnorm(N, 0, 1),
a = b + rnorm(N, 0, 0.1),
c = rnorm(N, 0, 1),
d = rbinom(N, 1, inv_logit(-1 + c)),
z = rbinom(N, 1, inv_logit(-2 + 2*b + c)),
x = rbinom(N, 1, inv_logit(a + z)),
w = W(x),
y = Y(d, w, z)
)
```

id | b | a | c | d | z | x | w | y |
---|---|---|---|---|---|---|---|---|

1 | 0.3641297 | 0.3917626 | 1.0369530 | 1 | 1 | 0 | 0 | 1 |

2 | 0.0287563 | 0.0397299 | 0.5736271 | 0 | 0 | 1 | 1 | 0 |

3 | -0.7727052 | -0.5993870 | -0.5179657 | 0 | 1 | 0 | 1 | 1 |

4 | 0.4107888 | 0.5737898 | 1.2586840 | 0 | 1 | 1 | 0 | 1 |

5 | 2.3512417 | 2.1631719 | 0.6746523 | 1 | 1 | 1 | 0 | 1 |

## Intervention

In order to simulate an intervention, we assign values to \(X\) randomly, then assign new values for all its descendents. After intervention, the causal effect of \(X\) on \(Y\) is simply \(\mathbb P(Y \mid X)\).

```
intervention <- df %>%
# intervene on x
mutate(
x = rbinom(n(), 1, 0.5),
w = W(x),
y = Y(d, w, z)
) %>%
# model P(y | do(x))
glm(
formula = y ~ x,
family = binomial(),
data = .
) %>%
# predict
augment(
newdata = tibble(x = 0:1),
type.predict = 'response'
)
```

x | .fitted | .se.fit |
---|---|---|

0 | 0.4637566 | 0.0022239 |

1 | 0.5072714 | 0.0022422 |

We can compare this causal effect to the simple statistical effect to see the difference.

```
noncausal <- df %>%
# model P(y | x)
glm(
formula = y ~ x,
family = binomial(),
data = .
) %>%
# predict
augment(
newdata = tibble(x = 0:1),
type.predict = 'response'
)
```

x | .fitted | .se.fit |
---|---|---|

0 | 0.3797282 | 0.0022595 |

1 | 0.5759927 | 0.0021293 |

## Backdoor

Since \(\{X, Z\}\) satisfies the backdoor criteria, we can use it to apply the backdoor adjustment. First we’ll need \(\mathbb P(D, Z)\).

```
# P(d, z)
p_d_z <- df %>%
group_by(d, z) %>%
count() %>%
ungroup() %>%
mutate(p_d_z = n / sum(n))
```

Now we model \(\mathbb P(Y \mid X, D, Z)\), multiply it by \(\mathbb P(D, Z)\), then take the sum for each value of \(X\).

```
backdoor <- formula(y ~ 1 + x + z + d) %>%
# model P(y | x, d, z)
glm(
family = binomial(),
data = df
) %>%
# predict
augment(
type.predict = 'response',
newdata =
crossing(
d = c(0, 1),
x = c(0, 1),
z = c(0, 1)
)
) %>%
# get P(d, z)
mutate(p_y_given_d_x_z = .fitted) %>%
inner_join(p_d_z, by = c('d', 'z')) %>%
# backdoor adjustment over d, z
group_by(x) %>%
summarise(p_y_given_do_x = sum(p_y_given_d_x_z * p_d_z))
```

x | p_y_given_do_x |
---|---|

0 | 0.4681530 |

1 | 0.5033398 |

Note that the backdoor adjusted estimates are similar to the estimates from intervention.

## Frontdoor

To apply the frontdoor adjustment with \(W\), we’ll need \(\mathbb P(W \mid X)\), \(\mathbb P(X^\prime)\), and \(\mathbb P(Y \mid X, W)\).

```
p_w_given_x <- df %>%
group_by(x, w) %>%
count() %>%
group_by(x) %>%
mutate(p_w_given_x = n / sum(n)) %>%
ungroup()
p_xprime <- df %>%
group_by(xprime = x) %>%
count() %>%
ungroup() %>%
mutate(p_xprime = n / sum(n))
p_y_given_xprime_w <- formula(y ~ 1 + x + w) %>%
glm(
family = binomial(),
data = df
) %>%
augment(
newdata = crossing(x = 0:1, w = 0:1),
type.predict = 'response'
) %>%
transmute(
xprime = x,
w,
p_y_given_xprime_w = .fitted
)
```

Now we apply the frontdoor adjustment:

\[

\mathbb P (Y \mid \do(X))

=

\sum_{x^\prime, w}

\mathbb P(x^\prime)

\cdot

\mathbb P(w \mid x)

\cdot

\mathbb P (y \mid x^\prime, w)

.

\]

```
frontdoor <- p_w_given_x %>%
inner_join(p_y_given_xprime_w, by = 'w') %>%
inner_join(p_xprime, by = 'xprime') %>%
group_by(x) %>%
summarise(sum(p_w_given_x * p_y_given_xprime_w * p_xprime))
```

x | sum(p_w_given_x * p_y_given_xprime_w * p_xprime) |
---|---|

0 | 0.4623710 |

1 | 0.5041105 |

Our frontdoor estimates of \(\mathbb P(Y \mid \do(X))\) are very similar to the intervention and backdoor estimates.

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**Brian Callander**.

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