Closed-Form Probit-Normal Dropout Probabilities

Why marginalize a probit dropout equation?

In a current-outcome selection model, dropout at a visit can depend on the outcome at that same visit. The conditional dropout probability describes this dependence for a specified outcome value. Often, however, the quantity of interest is the dropout probability averaged over the distribution of the current outcome, given the history available before the dropout decision. When that outcome distribution is normal and the conditional model uses a probit link, this average has a closed form.

The distinction between conditional and marginal probabilities matters. The conditional equation is evaluated at a particular outcome; the marginal probability averages its values over the possible outcomes. In general, averaging a nonlinear probability function is not the same as evaluating it at the mean outcome.

Definition: Let \(D_t=1\) indicate dropout at visit \(t\), and let \(H_{t-1}\) denote the history observed before the dropout decision. Suppose, conditional on that history, the current outcome satisfies \(Y_t\mid H_{t-1}\sim N(\mu_t,\sigma_t^2)\), and the conditional dropout probability is \(P(D_t=1\mid Y_t=y,H_{t-1})=\Phi(a_t+b_ty)\). Here \(\Phi\) is the standard normal cumulative distribution function, \(a_t\) is the probit intercept, and \(b_t\) is the slope per unit of outcome. Then the dropout probability averaged over the current outcome, conditional on \(H_{t-1}\), is \(\Phi\left((a_t+b_t\mu_t)/\sqrt{1+b_t^2\sigma_t^2}\right)\).

The formula conditions on a fixed pre-decision history. Its normality assumption concerns the distribution of the current outcome given that history, not necessarily the outcome distribution across all participants or histories. The conditional probit equation and the normal outcome distribution must refer to the same decision time and conditioning information.

Deriving the closed form

The marginal dropout probability is the conditional probability averaged over the normal distribution of \(Y_t\). For a fixed history, suppress the time subscripts temporarily and write \(Y\sim N(\mu,\sigma^2)\) and \(P(D=1\mid Y=y)=\Phi(a+by)\). The desired probability is therefore an expectation:

P(D=1)=E\{\Phi(a+bY)\} =\int_{-\infty}^{\infty}\Phi(a+by) \frac{1}{\sigma}\phi\left(\frac{y-\mu}{\sigma}\right)\,dy,

where \(\phi\) is the standard normal density. To evaluate the expectation, represent the conditional probit probability with a latent standard normal variable \(U\), independent of \(Y\): dropout occurs when \(U\leq a+bY\). This representation gives the specified conditional probability because \(P(U\leq a+by)=\Phi(a+by)\).

Center and scale the outcome.
Write \(Y=\mu+\sigma Z\), where \(Z\sim N(0,1)\) and \(Z\) is independent of \(U\).

Express dropout as a normal event.
The latent condition becomes \(U-b\sigma Z\leq a+b\mu\).

Use the distribution of the left-hand side.
Because \(U\) and \(Z\) are independent standard normal variables, \(U-b\sigma Z\) is normal with mean \(0\) and variance \(1+b^2\sigma^2\).

Standardize the event.
Dividing the threshold by the standard deviation yields the marginal probability \(\Phi\left((a+b\mu)/\sqrt{1+b^2\sigma^2}\right)\).

This derivation also explains the denominator. There is variability in the latent probit error, with variance \(1\), as well as variability in the outcome contribution \(bY\), with variance \(b^2\sigma^2\). Their independent variances add. If \(b=0\), outcome level does not affect dropout and the result reduces to \(\Phi(a)\). If \(\sigma^2=0\), the outcome is fixed at its mean and the result reduces to \(\Phi(a+b\mu)\).

Worked calculations

Worked Example: A Positive Outcome Effect

Suppose that for a specified pre-decision history, a hypothetical current outcome is normally distributed as \(Y\mid H\sim N(10,4)\), in outcome units. The conditional dropout equation is \(P(D=1\mid Y=y,H)=\Phi(-1+0.20y)\). The outcome standard deviation is \(2\) units, and the slope is \(0.20\) per outcome unit. We assume the stated normal distribution and probit equation apply to this same history and visit.

The marginal probit index in the numerator is the conditional index at the outcome mean: \(-1+0.20(10)=1\). The variance adjustment is \(1+(0.20)^2(4)=1.16\). Therefore,

P(D=1\mid H)= \Phi\left(\frac{-1+0.20(10)}{\sqrt{1+(0.20)^2(4)}}\right) =\Phi\left(\frac{1}{\sqrt{1.16}}\right) =\Phi(0.9285)\approx 0.8234.

For this history, the marginal dropout probability is about \(0.8234\), or \(82.34\%\). By contrast, at exactly the mean outcome \(y=10\), the conditional probability is \(\Phi(1)\approx0.8413\). These are different quantities: the first averages over outcome variability, while the second conditions on one outcome value.

Worked Example: Outcome Variability Changes the Average

Consider a second hypothetical setting with \(Y\mid H\sim N(4,9)\) and \(P(D=1\mid Y=y,H)=\Phi(-0.30+0.20y)\). The mean is \(4\) outcome units and the variance is \(9\) squared units. Assume again that both models condition on the same pre-decision history.

The index at the mean is \(-0.30+0.20(4)=0.50\), and the denominator is \(\sqrt{1+(0.20)^2(9)}=\sqrt{1.36}\). Substitution gives:

P(D=1\mid H)= \Phi\left(\frac{0.50}{\sqrt{1.36}}\right) =\Phi(0.4287)\approx 0.6659.

Thus the marginal dropout probability is approximately \(66.59\%\). The conditional dropout probability at the mean outcome is \(\Phi(0.50)\approx0.6915\), not \(0.6659\). Outcome variation changes the average because the probit curve is nonlinear; the denominator in the closed form accounts for that variation.

Worked Example: A Negative Outcome Effect and Response Probability

Suppose \(Y\mid H\sim N(50,100)\), measured in hypothetical score points, and dropout follows \(P(D=1\mid Y=y,H)=\Phi(1-0.04y)\). A larger score lowers the conditional dropout probability because the slope is negative. The normal model and probit equation are taken to apply conditional on the same history.

The index at the mean is \(1-0.04(50)=-1\), and the adjusted standard deviation is \(\sqrt{1+(0.04)^2(100)}=\sqrt{1.16}\). Hence:

P(D=1\mid H)= \Phi\left(\frac{-1}{\sqrt{1.16}}\right) =\Phi(-0.9285)\approx0.1766.

The marginal dropout probability is about \(17.66\%\). If response at this decision is defined as \(R=1-D\), then \(P(R=1\mid H)=1-0.1766=0.8234\), or \(82.34\%\). The sign of the slope and the definition of the indicator must be checked before interpreting either probability.

Standardized slopes and interpretation

Sometimes the outcome enters the probit equation after standardization. If the conditional model is \(P(D=1\mid Y,H)=\Phi\left(\alpha+\delta (Y-\mu)/\sigma\right)\), then its average over \(Y\mid H\sim N(\mu,\sigma^2)\) is \(\Phi\left(\alpha/\sqrt{1+\delta^2}\right)\). Here \(\delta\) is the change in the probit index for a one-standard-deviation increase in the outcome. This parameterization makes clear that the adjustment depends on the standardized slope, rather than on the outcome’s measurement units.

In the unstandardized equation, \(b\) must be paired with an outcome measured in the units used to fit the model. If the outcome is rescaled, the numerical slope changes while the product \(bY\), and therefore the model, can remain unchanged. The variance term \(b^2\sigma^2\) is also unchanged when the coefficient and outcome scale are transformed consistently.

Quantity	Meaning	Role in the marginal probability
\(a+b\mu\)	Conditional probit index at the outcome mean	Sets the numerator of the adjusted index
\(b^2\sigma^2\)	Variance in the outcome contribution to the index	Adds to the latent probit-error variance
\(\sqrt{1+b^2\sigma^2}\)	Standard deviation of the combined latent quantity	Scales the numerator before applying \(\Phi\)

Checks, alternatives, and limitations

Several simple checks can catch errors. First, the result must be between \(0\) and \(1\). Second, setting \(b=0\) should return the outcome-independent dropout probability \(\Phi(a)\). Third, when the outcome variance is zero, the marginal and conditional-at-the-mean probabilities should coincide. Finally, a negative slope must lower the conditional probit index as the outcome increases; reversing the sign or coding response as dropout can reverse the interpretation.

A numerical integration check is useful when implementing the formula. For example, one can evaluate the defining integral over a broad range of outcome values using the normal density, then compare it with the closed form. A discrepancy can indicate that the outcome variance was entered where a standard deviation was required, that the probit indicator was reversed, or that the coefficients and outcome are on inconsistent scales. The formula uses \(\sigma^2\) in \(b^2\sigma^2\), not \(\sigma\).

The closed form relies on both a normal conditional outcome distribution and a probit conditional dropout equation. A logistic dropout equation does not generally have this same closed-form normal average; direct numerical integration or simulation can be used instead. Likewise, a nonnormal outcome distribution generally requires integration under that distribution. A result conditional on a particular history is not automatically the overall dropout probability across a population: averaging over different histories may require a further average over their distribution.

This calculation complements, rather than replaces, the conditional selection equation. The conditional slope describes how the dropout probability changes with the current outcome on the probit-index scale. The marginal probability also depends on the outcome mean and variance. Two histories with the same conditional slope and intercept can therefore have different marginal dropout probabilities if their outcome distributions differ.

Common pitfall

Common mistake: Substitute the mean outcome into the conditional equation and call \(\Phi(a+b\mu)\) the marginal dropout probability. That is the conditional probability at \(Y=\mu\), not generally the average over \(Y\). The correction is to include the variance adjustment: \(\Phi\left((a+b\mu)/\sqrt{1+b^2\sigma^2}\right)\). The two expressions agree when \(b=0\) or \(\sigma^2=0\), but not generally.

Check Your Understanding

Answer each question, then compare with the solutions below.

For \(Y\mid H\sim N(\mu,\sigma^2)\) and dropout probability \(\Phi(a+bY)\), state the marginal dropout probability given \(H\).
If \(b=0\), what is the marginal dropout probability, and why?
For \(Y\mid H\sim N(2,1)\) and dropout probability \(\Phi(0.2+0.5Y)\), calculate the marginal dropout probability to four significant figures.
Explain why \(\Phi(a+b\mu)\) is generally not the marginal dropout probability.

Solutions. (1) It is \(\Phi\left((a+b\mu)/\sqrt{1+b^2\sigma^2}\right)\). (2) It is \(\Phi(a)\), because the conditional dropout probability no longer depends on the outcome. (3) The numerator is \(0.2+0.5(2)=1.2\), and the denominator is \(\sqrt{1+(0.5)^2(1)}=\sqrt{1.25}\). Thus \(P(D=1\mid H)=\Phi(1.2/\sqrt{1.25})=\Phi(1.0733)\approx0.8585\). (4) It evaluates the conditional probability at one outcome value, \(Y=\mu\); the marginal probability averages over the full outcome distribution and includes its variance.