Zero-Inflated Continuous Distributions

Why Zero-Inflated Continuous Models Arise

Some measurements can be exactly zero for one reason and positive for another. A patient may incur no cost during a period, while patients with any cost have a positive, continuously varying amount. A measurement model for this situation needs to represent both the probability of an exact zero and the distribution of positive values.

A zero-inflated continuous distribution does this with two ingredients: an atom at zero and a continuous law supported on positive values. The central practical distinction is that an interval containing zero may include a point probability, whereas an interval strictly above zero receives probability only from the continuous component.

Definition: Let $0\leq\pi\leq1$, and let $Y$ have a continuous distribution on $(0,\infty)$, with CDF $G$ and density $g$. A zero-inflated continuous random variable $X$ is constructed by setting $X=0$ with probability $\pi$, and setting $X=Y$ with probability $1-\pi$. Equivalently, use an indicator $B$, independent of $Y$, with $P(B=0)=\pi$ and $P(B=1)=1-\pi$, and set $X=BY$. The value $\pi$ is the zero-inflation probability.

Here $Y>0$ with probability one, so the only atom of $X$ is at zero. We use $G(x)=0$ for $x\leq0$, as follows from the positive support of $Y$. If $0<\pi<1$, the conditional distribution of $X$ given $X>0$ is exactly the distribution of $Y$. The endpoint cases $\pi=0$ and $\pi=1$ are also allowed: they give a purely continuous positive law and a point mass at zero, respectively.

The CDF and the Two Components

The CDF records both parts of the model. Below zero, $X$ cannot fall. At zero, the CDF jumps by $\pi$. Above zero, it increases according to the continuous component, scaled by its selection probability $1-\pi$.

Theorem: For the zero-inflated model above, $$ F(x)= \begin{cases} 0, & x<0,\\ \pi+(1-\pi)G(x), & x\geq 0. \end{cases} $$ In particular, $P(X=0)=\pi$, and for $x>0$ the continuous density contribution is $(1-\pi)g(x)$. This contribution is not a density for the entire law, because it does not account for the atom.

Proof. If $x<0$, neither the zero value nor any positive value of $Y$ is at most $x$; hence $P(X\leq x)=0$. If $x\geq0$, the event $X\leq x$ occurs whenever the zero component is selected. If the positive component is selected, it occurs precisely when $Y\leq x$. These alternatives are disjoint, so $$ P(X\leq x)=P(B=0)+P(B=1)P(Y\leq x) =\pi+(1-\pi)G(x). $$ Since $G(0)=0$, this gives $F(0)=\pi$, while $F(0-)=0$. Thus the jump at zero, and therefore the probability at zero, is $\pi$. For $x>0$, differentiating the CDF where $G$ has density $g$ gives the continuous density contribution $(1-\pi)g(x)$. $\square$

The density contribution integrates to $1-\pi$, not to one: $$ \int_0^\infty (1-\pi)g(x)\,dx=1-\pi. $$ The remaining probability, $\pi$, is concentrated at the point zero. This is why a single ordinary density cannot describe the full law when $\pi>0$.

Calculating Interval Probabilities

For intervals above zero, only the continuous component contributes. If an interval extends from below zero through zero, it also contains the atom. Keeping track of whether zero is included is essential, particularly for half-open intervals.

Theorem: For real numbers $a<b$, the interval probability is $$ P(a<X\leq b)= \begin{cases} 0, & b<0,\\ \pi, & a<0,\ b=0,\\ \pi+(1-\pi)G(b), & a<0,\ b>0,\\ (1-\pi)\{G(b)-G(a)\}, & 0\leq a<b. \end{cases} $$

Proof. If $b<0$, the interval contains no possible value of $X$. If $a<0$ and $b=0$, it contains zero but no positive values, so its probability is $\pi$. If $a<0<b$, it contains zero and the positive values satisfying $Y\leq b$. These disjoint contributions have probabilities $\pi$ and $(1-\pi)G(b)$. Finally, if $0\leq a<b$, the interval excludes zero, and the positive component contributes when $a<Y\leq b$. Its probability is $$ (1-\pi)P(a<Y\leq b) =(1-\pi)\{G(b)-G(a)\}, $$ using the CDF interval formula. The formulas also cover $a=0$, since $G(0)=0$ and the interval $(0,b]$ excludes the atom. $\square$

Two useful special cases follow directly. For $b>0$, $P(X\leq b)=\pi+(1-\pi)G(b)$, but $P(0<X\leq b)=(1-\pi)G(b)$. For $0\leq a<b$, the probability of $a<X\leq b$ is $(1-\pi)$ times the corresponding positive-component interval probability. In both cases, the factor $1-\pi$ matters: the continuous law describes values conditional on entering the positive component, not the unconditional frequency of those values.

Worked Examples

Worked Example: A Zero-Inflated Exponential Distribution

Suppose $P(X=0)=0.30$, and conditional on $X>0$, $X$ has an exponential distribution with rate $2$. Its positive-component CDF is $G(x)=1-e^{-2x}$ for $x>0$. Find $P(X\leq1)$, $P(0<X\leq1)$, and $P(0.25<X\leq1)$.

For the first probability, the interval includes the atom at zero: $$ \begin{aligned} P(X\leq1) &=0.30+0.70G(1)\\ &=0.30+0.70(1-e^{-2})\\ &\approx 0.905265302\\ &\approx 0.905265. \end{aligned} $$

The interval $0<X\leq1$ excludes zero, so it receives only the positive-component contribution: $$ \begin{aligned} P(0<X\leq1) &=0.70G(1)\\ &=0.70(1-e^{-2})\\ &\approx 0.605265302\\ &\approx 0.605265. \end{aligned} $$

For the interval bounded away from zero, $$ \begin{aligned} P(0.25<X\leq1) &=0.70\{G(1)-G(0.25)\}\\ &=0.70\{(1-e^{-2})-(1-e^{-0.5})\}\\ &=0.70(e^{-0.5}-e^{-2})\\ &\approx 0.70(0.606530660-0.135335283)\\ &\approx 0.329836764\\ &\approx 0.329837. \end{aligned} $$ The probability for $X\leq1$ exceeds the probability for $0<X\leq1$ by exactly $0.30$, the atom at zero.

Worked Example: A Zero-Inflated Uniform Distribution

Let $P(X=0)=0.25$, with the positive component uniform on $(0,4)$. Thus $G(x)=x/4$ for $0\leq x\leq4$. Find $P(-1<X\leq2.5)$, $P(0.5<X\leq3)$, and $P(X=0\text{ or }1<X\leq3)$.

The first interval contains zero, so add its atom to the positive values through $2.5$: $$ \begin{aligned} P(-1<X\leq2.5) &=0.25+0.75G(2.5)\\ &=0.25+0.75\left(\frac{2.5}{4}\right)\\ &=0.25+0.46875\\ &=0.71875. \end{aligned} $$

The second interval is strictly positive: $$ \begin{aligned} P(0.5<X\leq3) &=0.75\{G(3)-G(0.5)\}\\ &=0.75\left(\frac{3}{4}-\frac{0.5}{4}\right)\\ &=0.75(0.75-0.125)\\ &=0.46875. \end{aligned} $$

The last event is a union of disjoint parts: the atom at zero, and the positive interval $1<X\leq3$. Therefore, $$ \begin{aligned} P(X=0\text{ or }1<X\leq3) &=0.25+0.75\{G(3)-G(1)\}\\ &=0.25+0.75\left(\frac{3}{4}-\frac{1}{4}\right)\\ &=0.25+0.375\\ &=0.625. \end{aligned} $$

Worked Example: A Zero-Inflated Gamma Distribution

Suppose $P(X=0)=0.40$, and conditional on being positive, $X$ has density $g(y)=ye^{-y}$ for $y>0$, a gamma density with shape $2$ and scale $1$. Find $P(0\leq X\leq2)$ and $P(0.5<X\leq2)$.

Integrating the positive density gives its CDF: $$ G(x)=\int_0^x ye^{-y}\,dy=1-(1+x)e^{-x},\qquad x>0. $$ Indeed, an antiderivative is $-(1+y)e^{-y}$, whose derivative is $ye^{-y}$. At $x=2$, $G(2)=1-3e^{-2}\approx0.593994151$. The interval $0\leq X\leq2$ includes the atom, so $$ \begin{aligned} P(0\leq X\leq2) &=0.40+0.60G(2)\\ &=0.40+0.60(1-3e^{-2})\\ &\approx0.756396491. \end{aligned} $$

At $x=0.5$, $G(0.5)=1-1.5e^{-0.5}\approx0.090204010$. Since the second interval excludes zero, $$ \begin{aligned} P(0.5<X\leq2) &=0.60\{G(2)-G(0.5)\}\\ &=0.60\{1.5e^{-0.5}-3e^{-2}\}\\ &\approx0.60(0.503790140)\\ &\approx0.302274084\\ &\approx0.302274. \end{aligned} $$ The factor $0.60$ is the probability of selecting the positive component; the gamma interval probability alone is conditional on that selection.

Common Mistakes and Why the Distinction Matters

A frequent error is to treat the positive-component CDF as the unconditional CDF. For example, using $G(1)$ alone for $P(X\leq1)$ ignores both the probability of entering the positive component and the atom at zero. The correction is to ask two separate questions: does the event include zero, and what part of the positive-component range does it contain? Add $\pi$ if zero is included; multiply a positive-component probability by $1-\pi$.

Confusing density height with point probability. The value $g(0)$, if a density is defined at zero, is not the probability $P(X=0)$. Here the point probability is the CDF jump $\pi$.
Forgetting whether an interval includes zero. The event $0<X\leq b$ excludes the atom; $0\leq X\leq b$ includes it.
Failing to scale the positive component. A probability computed under $G$ is conditional on selecting the positive component. Its unconditional contribution is multiplied by $1-\pi$.
Calling the continuous contribution a full density. Its integral is $1-\pi$; the remaining probability is concentrated at zero.

The zero-inflated formulation is useful whenever exact zeros and positive measurements arise through distinct parts of a data-generating process. It allows the probability of a zero and the distribution of positive magnitudes to be described separately, while still giving a single CDF for all event calculations. This is also why interval endpoints deserve attention: moving an endpoint from $0<X$ to $0\leq X$ changes the probability by $\pi$, even though the positive-density calculation is unchanged.

Key takeaway: A zero-inflated continuous law has a point mass $\pi$ at zero and a positive continuous contribution with total mass $1-\pi$. For positive intervals, scale the positive-component probability by $1-\pi$; add $\pi$ exactly when the event includes zero.

Check Your Understanding

Use the zero probability and positive-component CDF separately in each calculation.

A model has $P(X=0)=0.20$, and its positive component is exponential with rate $1$. Write $F(x)$ for $x<0$ and $x\geq0$.
For the model in Question 1, express $P(0<X\leq2)$ and $P(X\leq2)$ in terms of $e^{-2}$.
A zero-inflated positive uniform distribution has zero probability $0.10$ and positive component uniform on $(0,5)$. Find $P(1<X\leq4)$.
Explain why the integral of the continuous density contribution is not one when the zero probability is positive.
For an interval $-2<X\leq b$, where $b>0$, identify the zero-atom contribution and the positive-component contribution.