Counterfactuals in the Probit Model

This post is about counterfactuals in the probit model. I wrote this while reading Pearl et al’s 2016 book Causal Inference in Statistics: A Primer.

The probit model with one normally distributed covariate can be written like this:

$\begin{eqnarray*} \epsilon & \sim & N\left(0,1\right)\\ X & \sim & N\left(0,1\right)\\ Y\mid X,\epsilon & = & 1_{X+\epsilon\geq0} \end{eqnarray*}$

Now we wish to find the density of the counterfactual $$Y_{x}$$, see Pearl et al.  (2016) for definitions. The density of $$\epsilon$$ given $$X=x$$ and $$Y=1$$ is

$\begin{eqnarray*} p\left(\epsilon\mid X=x,Y=1\right) & = & \frac{p\left(x=x,Y=1\mid\epsilon\right)p\left(\epsilon\right)}{p\left(x=x,Y=1\right)}\\ & = & \frac{p\left(Y=1\mid\epsilon,x=x\right)p\left(x=x\right)p\left(\epsilon\right)}{p\left(x=x,Y=1\right)}\\ & \propto & 1_{x+\epsilon\geq0}p\left(x=x\right)p\left(\epsilon\right) \end{eqnarray*}$ or the normal density truncated to $$\left[-x,\infty\right)$$, or $$\phi_{\left[-x,\infty\right)}\left(\epsilon\right)$$.

Probability of Necessity

Now we’ll take a look at the probability of necessity. The most obvious way to generalize the probability of necessity to continuous distributions is to allow the counterfactual $$x'$$ to be a parameter of the counterfactual $$Y_{x'}$$, like this:

$\begin{eqnarray*} P\left(Y_{x'}=0\mid X=x,Y=1\right) & = & 1-\int_{-x}^{\infty}1_{x'+\epsilon\geq0}\phi_{\left[-x,\infty\right)}\left(\epsilon\right)\\ & = & 1-\int_{\max\left\{ -x,-x'\right\} }^{\infty}\phi_{\left[-x,\infty\right)}\left(\epsilon\right)d\epsilon\\ & = & 1-\frac{\min\left(\Phi\left(x\right),\Phi\left(x'\right)\right)}{\Phi\left(x\right)} \end{eqnarray*}$ Let’s plot this function.

x <- seq(-3, 3, by = 0.01)
x_hat <- 0

ps <- function(x, x_hat) 1 - pmin(pnorm(x_hat), pnorm(x))/pnorm(x_hat)

plot(x = x, y = ps(x, x_hat),
type = "l",
xlab = "Counterfactual x'",
ylab = "Probability",
main = "Counterfactual Probabilities, x = 0")
grid()
lines(x = x, y = ps(x, x_hat), type = "l") There is nothing too strange. The probability of necessity goes to $$0$$ as $$x\to-\infty$$, which is what you would expect. For the probability of $$Y=1$$ becomes really slim as $$x\to-\infty$$. When $$x'>0$$ the probability of necessity is $$0$$, as you will always observe $$Y=1$$ counterfactually when $$X=x'$$ if you observed $$Y=1$$ with $$X=x!$$

Integrated Probability of Necessity

A more complicated question is: What rôle did the fact that $$X=x$$ have in $$Y=1$$? Or, if $$X$$ wasn’t $$x$$, what would $$Y$$ have been? With some abuse of notation, $$P\left(Y_{X'}=0\mid X=x,Y=1\right)$$ answers this question, where $$X'$$ is an independent copy of $$X$$.

$\begin{eqnarray*} P\left(Y_{X'}=0\mid X=x,Y=1\right) & = & 1-\int_{-\infty}^{\infty}\int_{\max\left\{ -x,-x'\right\} }^{\infty}\phi_{\left[-x,\infty\right)}\left(\epsilon\right)\phi\left(x'\right)dx\\ & = & 1-\int\frac{\min\left(\Phi\left(x\right),\Phi\left(x'\right)\right)}{\Phi\left(x\right)}\phi\left(x'\right)dx\\ & = & 1-\int_{x}^{\infty}\phi\left(x'\right)dx-\frac{1}{\Phi\left(x\right)}\int_{-\infty}^{x}\Phi\left(x'\right)\phi\left(x'\right)dx\\ & = & 1-\Phi\left(-x\right)-\frac{1}{2}\frac{1}{\Phi\left(x\right)}\left[\Phi\left(x\right)-\Phi\left(x\right)\Phi\left(-x\right)\right]\\ & = & \frac{1}{2}\Phi\left(x\right) \end{eqnarray*}$

Now let’s plot this.

counter = function(x) 0.5*pnorm(x)

plot(x = x,
y = counter(x),
type = "l",
xlab = "Actual x",
ylab = "Probability",
main = "Counterfactual Probabilities")
grid()
lines(x = x,
y = counter(x),
type = "l") Notice the asymptote at $$0.5$$. No matter how large $$X=x$$ we observe together with $$Y=1$$, we can never be more than $$0.5$$ certain that $$Y=0$$ if we were to draw an $$x$$ once again. The asymptote at $$0$$ if we had observed at very small value $$X=x$$ together with $$Y=1$$, and we were to draw again, we would be really certain that $$Y=1$$ would happen once again.