Tobit Effects on Observed Outcomes

Why the Observed Outcome Is Not the Latent Outcome

In a left-censored outcome, values below a known threshold are recorded at the threshold rather than at their latent values. For example, a measurement below a reporting limit may be recorded as that limit. A Tobit model describes a latent outcome that can fall below the limit, as well as the observed outcome produced by the recording rule. These are different quantities: a covariate shifts the latent outcome directly, but the observed data also reflect how often values are censored.

This tutorial derives two predictions from the left-censored Gaussian model: the probability that an outcome is uncensored, and the unconditional mean of the observed outcome conditional on the covariates. “Unconditional” here means that the mean averages over both censored and uncensored observations, rather than conditioning on being uncensored.

Model and Definitions

Let $x$ denote a vector of covariates, let $c$ be a known censoring threshold, and let $Y^*$ be the latent outcome. The left-censored Tobit model assumes

Y^* = x'\beta + \varepsilon,\qquad \varepsilon\mid x \sim N(0,\sigma^2),\qquad \sigma>0.

Write $\mu(x)=x'\beta$ for the conditional mean of the latent outcome. The observed outcome is $Y=\max(c,Y^*)$. Thus values at or below $c$ are observed as $c$, and values above $c$ are observed as themselves. Under the continuous Gaussian model, the probability of $Y^*=c$ is zero, so the probability of being uncensored can equivalently be written as $P(Y^*>c\mid x)$ or $P(Y>c\mid x)$.

Definition: Define $U=1$ when an observation is uncensored and $U=0$ when it is censored. The standardized distance of the latent mean above the censoring threshold is $$ a(x)=\frac{\mu(x)-c}{\sigma}. $$ Let $\phi$ and $\Phi$ denote the standard normal density and cumulative distribution function, respectively. The probability of being uncensored is $P(U=1\mid x)$; the unconditional observed mean is $E(Y\mid x)$, averaging over both values of $U$.

The Gaussian, constant-variance assumption is important for the formulas below. In particular, $\sigma$ does not vary with $x$, and the censoring threshold is fixed. These are model assumptions, not consequences of the censoring rule itself.

Probability of Being Uncensored

Standardizing the latent outcome gives $(Y^*-\mu(x))/\sigma\sim N(0,1)$. Therefore,

\begin{aligned} P(U=1\mid x) &=P(Y^*>c\mid x)\\ &=P\left(\frac{Y^*-\mu(x)}{\sigma}> \frac{c-\mu(x)}{\sigma}\,\middle|\,x\right)\\ &=1-\Phi\left(\frac{c-\mu(x)}{\sigma}\right) =\Phi\left(\frac{\mu(x)-c}{\sigma}\right) =\Phi(a(x)). \end{aligned}

The probability depends on how far the latent mean lies above the threshold, measured in standard deviations. If $\mu(x)=c$, then $a(x)=0$ and half of the latent Gaussian distribution lies above the threshold. If the latent mean is far below $c$, the uncensored probability is small; if it is far above $c$, it is large.

Deriving the Unconditional Observed Mean

The observed outcome equals $c$ for latent values at or below $c$, and equals the latent value otherwise. Its conditional expectation can therefore be written as a contribution from the censored outcomes plus a contribution from the uncensored outcomes:

E(Y\mid x) =c\,P(Y^*\leq c\mid x) +\int_c^\infty y f_{Y^*\mid x}(y)\,dy.

The first probability is $1-\Phi(a(x))$. For the integral, set $z=(y-\mu(x))/\sigma$. Its lower limit becomes $(c-\mu(x))/\sigma=-a(x)$, and $y=\mu(x)+\sigma z$. Hence

\begin{aligned} \int_c^\infty y f_{Y^*\mid x}(y)\,dy &=\int_{-a(x)}^\infty \{\mu(x)+\sigma z\}\phi(z)\,dz\\ &=\mu(x)\Phi(a(x)) +\sigma\int_{-a(x)}^\infty z\phi(z)\,dz\\ &=\mu(x)\Phi(a(x))+\sigma\phi(a(x)). \end{aligned}

For the last equality, $\phi'(z)=-z\phi(z)$, so $\int_b^\infty z\phi(z)\,dz=\phi(b)$; also, the normal density is symmetric, giving $\phi(-a)=\phi(a)$. Combining the two parts yields the observed mean.

Result: Under the left-censored Gaussian model, $$ E(Y\mid x) =c\{1-\Phi(a(x))\}+\mu(x)\Phi(a(x))+\sigma\phi(a(x)) =c+\{\mu(x)-c\}\Phi(a(x))+\sigma\phi(a(x)). $$

The expression $E(Y\mid x)-c=\sigma\{a(x)\Phi(a(x))+\phi(a(x))\}$ is nonnegative. The term $a(x)\Phi(a(x))$ can be negative when the latent mean is below the threshold, so the two terms in braces should not be claimed to be individually nonnegative. Their sum is nonnegative; this also follows directly because $Y\geq c$ for every outcome.

Worked Example: A Latent Mean Below the Threshold

Suppose the censoring threshold is $c=0$, the latent conditional mean is $\mu(x)=-1$ outcome units, and the latent standard deviation is $\sigma=1.5$ outcome units. Calculate the probability of being uncensored and the unconditional observed mean.

First, $a(x)=(-1-0)/1.5=-0.6667$. Standard normal values are $\Phi(-0.6667)\approx0.2525$ and $\phi(-0.6667)\approx0.3194$. Thus $P(U=1\mid x)=0.2525$: about 25.25% of outcomes are predicted to exceed the threshold.

Using the observed-mean formula, $$ E(Y\mid x) =(-1)\Phi(-2/3)+(1.5)\phi(-2/3) \approx0.226679 \approx0.2267. $$

The mean observed outcome is about 0.2267 outcome units, even though the latent mean is $-1$. This is because latent outcomes at or below zero are recorded as zero. In the equivalent expression above the threshold, the first contribution $(\mu(x)-c)\Phi(a(x))=-0.2525$ is negative, but the sum of that contribution and $\sigma\phi(a(x))$ is positive. The observed mean is above the threshold, as it must be.

How a Covariate Changes the Predictions

A covariate changes $\mu(x)=x'\beta$, and that changes both the probability of being uncensored and the mean of the recorded outcome. To make the comparison concrete, hold all other covariates fixed and compare two specified values of one covariate. The model-based change in a prediction is the prediction at the second value minus the prediction at the first; it is not generally equal to the corresponding latent regression coefficient.

Worked Example: Comparing Two Covariate Values

Suppose $c=0$, $\sigma=1$, and the latent mean is $\mu(x)=-0.5+1.2x$, where $x$ is measured in units. Compare $x=0$ with $x=1$, holding all other factors fixed.

At $x=0$, $\mu=-0.5$ and $a=-0.5$. Using $\Phi(-0.5)=0.3085$ and $\phi(-0.5)=0.3521$, $$ P(U=1\mid x=0)=0.3085 $$ and $$ E(Y\mid x=0) =(-0.5)(0.3085)+(1)(0.3521) =-0.1543+0.3521 =0.1978. $$

At $x=1$, $\mu=-0.5+1.2=0.7$ and $a=0.7$. With $\Phi(0.7)=0.7580$ and $\phi(0.7)=0.3123$, $$ P(U=1\mid x=1)=0.7580 $$ and $$ E(Y\mid x=1) =(0.7)(0.7580)+(1)(0.3123) =0.5306+0.3123 =0.8429. $$

Increasing $x$ by one unit raises the model-predicted uncensored probability by $0.7580-0.3085=0.4495$, or 44.95 percentage points. It raises the unconditional observed mean by $0.8429-0.1978=0.6451$ outcome units. The latent mean itself rises by 1.2 outcome units, the coefficient on $x$, but the observed mean rises by a different amount because censoring affects the recorded outcomes.

A comparison between covariate values is useful when the change of interest is a particular meaningful step, such as one additional year or a change from one category to another. Predictions should be calculated at both values using the same threshold and standard deviation. The probability difference is in percentage points, while the mean difference is in the units of the outcome.

Worked Example: A Threshold Above the Latent Mean

Suppose the threshold is $c=2$ outcome units and $\sigma=0.8$ outcome units. A covariate change shifts the latent mean from $1.5$ to $2.5$. Calculate the uncensored probabilities and observed means at both values.

For $\mu=1.5$, $a=(1.5-2)/0.8=-0.625$. Using $\Phi(-0.625)=0.2660$ and $\phi(-0.625)=0.3282$, the uncensored probability is $0.2660$. The observed mean is $$ E(Y\mid x) =2+(-0.5)\Phi(-0.625)+(0.8)\phi(-0.625) \approx2-0.1329928+0.2625288 \approx2.1295. $$

For $\mu=2.5$, $a=(2.5-2)/0.8=0.625$. The uncensored probability is $\Phi(0.625)=0.7340$, and $$ E(Y\mid x) =2+(0.5)\Phi(0.625)+(0.8)\phi(0.625) \approx2+0.3670072+0.2625288 \approx2.6295. $$

The covariate change raises the uncensored probability by $0.7340-0.2660=0.4680$, or 46.80 percentage points, and raises the observed mean by $2.6295-2.1295=0.5000$ outcome units. In the first mean calculation, the term involving $\mu-c$ is negative. The full contribution above the threshold is still positive: $-0.1329928+0.2625288\approx0.1295360\approx0.1295$. Both predicted observed means exceed $c=2$, consistent with the rule $Y\geq c$.

Interpretation and a Common Pitfall

The uncensored probability and the unconditional observed mean answer distinct questions. $P(U=1\mid x)$ gives the model-predicted fraction of observations above the censoring threshold at the specified covariates. $E(Y\mid x)$ gives the expected recorded outcome in those same conditions, including observations recorded at $c$. Neither is the conditional mean among only the uncensored observations. That latter quantity conditions on $U=1$ and answers a different question.

A common mistake is to interpret $\beta_j$, the coefficient on a covariate, as the change in the observed mean for a one-unit change in that covariate. It is the change in the latent mean $\mu(x)$, holding other covariates fixed. Because censoring changes the share of observations recorded at the threshold, the change in $E(Y\mid x)$ need not equal $\beta_j$. The calculations above show the distinction: a latent-mean increase of 1.2 units corresponded to an observed-mean increase of 0.6451 units in one comparison. For an empirical application, these predictions also depend on the Tobit model being an appropriate description of the latent conditional distribution and the observation rule.

Key takeaway: For a left-censored Gaussian outcome, the probability of being uncensored is $\Phi(a(x))$, while the unconditional observed mean is $c\{1-\Phi(a(x))\}+\mu(x)\Phi(a(x))+\sigma\phi(a(x))$. Use the first to describe censoring frequency and the second to describe the average recorded outcome.

Check Your Understanding

Use the model definitions and results above to answer these questions.

If $\mu(x)=c$, what is the probability of being uncensored? Explain using $a(x)$.
For $c=0$, $\mu(x)=0$, and $\sigma=1$, calculate $E(Y\mid x)$ using the observed-mean formula.
Why can $(\mu(x)-c)\Phi(a(x))$ be negative even though $E(Y\mid x)\geq c$?
In a covariate comparison, what does the difference between two uncensored probabilities measure, and what units does the difference between observed means have?
Why is a Tobit coefficient generally not the change in the unconditional observed mean for a one-unit increase in a covariate?