Heteroskedastic Tobit Models

Why Allow the Variance to Change?

In a Tobit model, the observed outcome is censored even though the underlying outcome is latent. The probability of falling below the censoring point depends on the latent mean and the latent error variance. If the variance differs across covariate values, a model that forces it to be constant can misrepresent both the frequency of censoring and the distribution of observed positive outcomes. A heteroskedastic Tobit model makes that variance dependence explicit.

The focus here is the left-censored Gaussian model with censoring point zero. The latent outcome is continuous, so an observation at zero represents censoring rather than a latent outcome exactly equal to zero. Conditional on covariates, observations are assumed independent. The mean and variance specifications must both be stated: a variance equation is not a substitute for a mean equation.

Definition: Let $y_i^*$ be the latent outcome, $x_i$ a vector of mean covariates, and $z_i$ a vector of variance covariates for observation $i$. A heteroskedastic Gaussian Tobit model is $$ y_i^*=x_i'\beta+\varepsilon_i,\qquad \varepsilon_i\mid x_i,z_i\sim N(0,\sigma_i^2),\qquad \log(\sigma_i^2)=z_i'\gamma, $$ with observed outcome $y_i=\max(0,y_i^*)$. Thus $\sigma_i=\exp(z_i'\gamma/2)>0$. The vectors $x_i$ and $z_i$ may overlap, but need not.

The coefficient vector $\beta$ determines the conditional latent mean, $x_i'\beta$. The vector $\gamma$ determines how the conditional latent variance changes with $z_i$. In particular, a one-unit increase in a variance covariate, holding the other entries of $z_i$ fixed, changes the log variance by its corresponding coefficient. A coefficient in $\gamma$ is not directly a change in the variance measured in outcome-squared units.

Likelihood Contributions

Write $\mu_i=x_i'\beta$, and define $d_i=1$ when $y_i>0$ and $d_i=0$ when $y_i=0$. For an uncensored observation, the observed value equals the latent value, so its contribution is the Gaussian density evaluated at $y_i$. For a censored observation, the data reveal only that $y_i^*\leq 0$, so its contribution is the probability of that event.

Likelihood: Under the model above, observation $i$ contributes $$ L_i(\beta,\gamma)= \left[\frac{1}{\sigma_i}\phi\!\left(\frac{y_i-\mu_i}{\sigma_i}\right)\right]^{d_i} \left[\Phi\!\left(\frac{-\mu_i}{\sigma_i}\right)\right]^{1-d_i}, $$ where $\phi$ and $\Phi$ are the standard normal density and distribution functions. For conditionally independent observations, the log likelihood is $$ \ell(\beta,\gamma)= \sum_{i=1}^n\left\{ d_i\left[-\log(\sigma_i)+\log\phi\!\left(\frac{y_i-\mu_i}{\sigma_i}\right)\right] +(1-d_i)\log\Phi\!\left(\frac{-\mu_i}{\sigma_i}\right) \right\}. $$

The likelihood for a censored observation is a probability, not a density at zero. Its probability is $\Phi(-\mu_i/\sigma_i)$, since standardizing $y_i^*\leq 0$ gives $\varepsilon_i/\sigma_i\leq-\mu_i/\sigma_i$. For an uncensored observation, the density includes the factor $1/\sigma_i$; omitting it would give an incorrect likelihood and would fail to account for the changing scale.

Worked Example: Computing Censored and Uncensored Contributions

Suppose one hypothetical observation has $\mu_i=1.2$ and $\sigma_i=2$, while another has $\mu_i=1.2$, $\sigma_i=2$, and observed outcome $y_i=3$. Compute the contribution for each observation.

For the first observation, $y_i=0$, so $d_i=0$. Its contribution is $$ L_i=\Phi\!\left(\frac{-1.2}{2}\right)=\Phi(-0.6)\approx 0.2743. $$ This is the model probability that the latent outcome is at or below zero. Its log-likelihood contribution is $\log(0.2743)\approx -1.293$.

For the second observation, $y_i=3>0$, so $d_i=1$. The standardized residual is $(3-1.2)/2=0.9$, and its density contribution is $$ L_i=\frac{1}{2}\phi(0.9)\approx \frac{0.2661}{2}=0.1330. $$ Its log-likelihood contribution is approximately $\log(0.1330)=-2.017$. The two observations use different parts of the likelihood: a probability for the censored outcome and a density for the uncensored outcome.

Maximum likelihood estimation chooses $\beta$ and $\gamma$ jointly to maximize the sum of these contributions, typically by numerical optimization. Joint estimation matters because both $\mu_i$ and $\sigma_i$ enter the standardized quantity $\mu_i/\sigma_i$ that determines the censoring probability. In general, there is no separate closed-form regression calculation for the mean followed by a variance calculation that reproduces this joint likelihood estimate.

How Variance Enters Marginal Effects

With censoring at zero, the probability of a positive observed outcome is $P(y_i>0\mid x_i,z_i)=\Phi(\mu_i/\sigma_i)$. The expected observed outcome includes both the zero outcomes and the positive outcomes. Let $a_i=\mu_i/\sigma_i$. Integrating the latent Gaussian outcome over its positive range gives

E[y_i\mid x_i,z_i]=\mu_i\Phi(a_i)+\sigma_i\phi(a_i).

This is not simply the latent mean $\mu_i$: censored observations contribute zero to the observed outcome. If a covariate $w$ appears in both the mean and variance equations, its effect on the expected observed outcome has two components. Write $\partial\mu_i/\partial w=\beta_w$ and $\partial\log(\sigma_i^2)/\partial w=\gamma_w$, so that $\partial\sigma_i/\partial w=\sigma_i\gamma_w/2$. Differentiating the expectation yields

\frac{\partial E[y_i\mid x_i,z_i]}{\partial w} =\Phi(a_i)\beta_w+\phi(a_i)\frac{\partial\sigma_i}{\partial w} =\Phi(a_i)\beta_w+\frac{\sigma_i\gamma_w}{2}\phi(a_i).

For the first equality, differentiate $\mu_i\Phi(a_i)+\sigma_i\phi(a_i)$ and use $\partial a_i/\partial w=(\partial\mu_i/\partial w)/\sigma_i-(a_i/\sigma_i)(\partial\sigma_i/\partial w)$. The terms involving $a_i\phi(a_i)$ cancel, leaving $\Phi(a_i)\partial\mu_i/\partial w+\phi(a_i)\partial\sigma_i/\partial w$. This cancellation explains why the effect has a mean component and a scale component. If $w$ appears only in the mean equation, set $\gamma_w=0$; if it appears only in the variance equation, set $\beta_w=0$.

Worked Example: A Covariate Changes Both Mean and Variance

Consider a hypothetical observation at which $\mu_i=1$, $\sigma_i=1$, $\beta_w=0.4$, and $\gamma_w=0.6$. Thus $a_i=1$. Find the derivative of the expected observed outcome with respect to $w$.

Using $\Phi(1)\approx0.8413$ and $\phi(1)\approx0.2420$, $$ \frac{\partial E[y_i\mid x_i,z_i]}{\partial w} =\Phi(1)(0.4)+\frac{1(0.6)}{2}\phi(1) \approx (0.8413)(0.4)+(0.3)(0.2420). $$

Using unrounded normal values, the two components are approximately $0.3365$ and $0.07259$, respectively, giving a total derivative of approximately $0.4091$ outcome units per unit of $w$. The mean component alone would be about $0.3365$. Because $w$ also raises the log variance, omitting the scale component would not report the model-implied effect on the expected observed outcome.

Constant-Variance and Heteroskedastic Specifications

The constant-variance Tobit is nested within this specification when $z_i'\gamma$ is constant across observations. For example, an intercept-only variance equation permits one common positive variance, while additional entries in $z_i$ allow the variance to vary systematically. The key contrast is not merely whether standard errors change: the variance specification changes the likelihood used to estimate the mean parameters and the model-implied censoring probabilities.

Feature	Constant-variance Tobit	Heteroskedastic Tobit
Latent variance	One common $\sigma^2$	$\sigma_i^2=\exp(z_i'\gamma)$
Censoring probability	$\Phi(-\mu_i/\sigma)$	$\Phi(-\mu_i/\sigma_i)$
Likelihood scale for positive outcomes	Common density scale	Observation-specific density scale
Marginal effects	Depend on the mean specification and common scale	Can include a scale effect when covariates affect variance

Worked Example: Different Censoring Rates at the Same Latent Mean

Suppose two hypothetical covariate groups have the same latent mean, $\mu_i=1.5$, but different latent standard deviations: $0.75$ for group A and $1.5$ for group B. Calculate their probabilities of a zero and their expected observed outcomes.

For group A, $a_i=1.5/0.75=2$. Its probability of being censored is $\Phi(-2)\approx0.02275$. Its expected observed outcome is $$ 1.5\Phi(2)+0.75\phi(2) \approx 1.5(0.97725)+0.75(0.05399) \approx 1.506. $$

For group B, $a_i=1.5/1.5=1$. Its probability of being censored is $\Phi(-1)\approx0.1587$, and its expected observed outcome is $$ 1.5\Phi(1)+1.5\phi(1) \approx 1.5(0.8413)+1.5(0.2420) \approx 1.625. $$

Despite having the same latent mean, the groups have different censoring probabilities and expected observed outcomes because their variances differ. A constant-variance model cannot represent both group-specific distributions at this same latent mean. In an estimation problem, it may compensate by changing fitted mean coefficients, so those coefficients need not recover the intended latent-mean relationships.

What Variance Misspecification Does

If the true variance depends on covariates but a constant-variance Tobit is fitted, its likelihood generally does not describe the data-generating distribution. In ordinary uncensored linear regression, a correctly specified conditional mean can sometimes remain consistently estimated despite heteroskedasticity. Tobit likelihood estimation is different: the probability of censoring depends on $\mu_i/\sigma_i$, and the density of positive observations also depends on $\sigma_i$. A wrongly imposed common scale therefore affects the likelihood information used to estimate $\beta$, not just the precision assigned to an otherwise unchanged estimate. The fitted coefficients can be distorted, although the direction and size of distortion depend on the covariates and censoring pattern; there is no universal correction factor.

Marginal effects are also affected. A constant-variance model assigns no covariate-driven scale component to the effect on the expected observed outcome. If the true variance changes with a covariate, its effects on censoring and on the observed-outcome distribution may therefore be misrepresented. Conversely, adding a variance equation without a substantive reason can make estimates less stable, especially when the data provide little information about variation among censored observations. The variance covariates and functional form should be chosen and assessed as part of the model, not treated as an automatic remedy.

Common Mistakes

Using a density contribution for a censored observation. The observed zero says $y_i^*\leq0$, so the likelihood contribution is $\Phi(-\mu_i/\sigma_i)$, not a Gaussian density evaluated at zero.
Forgetting the density scale. An uncensored observation contributes $\phi((y_i-\mu_i)/\sigma_i)/\sigma_i$. The $1/\sigma_i$ factor is essential.
Confusing log variance with log standard deviation. Under $\log(\sigma_i^2)=z_i'\gamma$, the standard deviation is $\exp(z_i'\gamma/2)$, not $\exp(z_i'\gamma)$.
Reporting only the latent-mean coefficient as an observed-outcome effect. When a covariate enters the variance equation, the derivative of $E[y_i\mid x_i,z_i]$ also contains the scale component.
Assuming variance misspecification only changes standard errors. In censored likelihood estimation, variance affects censoring probabilities and uncensored densities, so it can also affect fitted mean parameters and marginal effects.

Key takeaway: A heteroskedastic Tobit model specifies the latent variance as a positive function of covariates and includes that variance in both censored probabilities and uncensored densities. When covariates affect the variance, their effects on observed outcomes can differ from their effects on the latent mean alone.

Check Your Understanding

Try these questions using the model and notation above.

For an observation censored at zero, explain why its likelihood contribution is a cumulative probability rather than a density.
If $\log(\sigma_i^2)=z_i'\gamma$, write $\sigma_i$ in terms of $z_i'\gamma$.
For an uncensored observation, identify the factor in its likelihood contribution that would be missing if one used only the standard normal density of the standardized residual.
Suppose a covariate appears in both the mean and variance equations. Name the two components of its derivative on the expected observed outcome.
Why can a constant-variance Tobit produce distorted mean coefficients when the true variance changes across covariates?

Feature	Constant-variance Tobit	Heteroskedastic Tobit
Latent variance	One common \(\sigma^2\)	\(\sigma_i^2=\exp(z_i'\gamma)\)
Censoring probability	\(\Phi(-\mu_i/\sigma)\)	\(\Phi(-\mu_i/\sigma_i)\)
Likelihood scale for positive outcomes	Common density scale	Observation-specific density scale
Marginal effects	Depend on the mean specification and common scale	Can include a scale effect when covariates affect variance

What You'll Learn