Generalized Inverses For Discrete And Mixed Laws

Why Generalized Inverses Are Needed

The inverse transform method begins with a uniform random variable and maps it to a desired distribution. For a continuous, strictly increasing cumulative distribution function, this may look like ordinary function inversion. But a discrete distribution has jumps, and a mixed distribution may have both jumps and stretches where its cumulative distribution function is constant. In those cases, an ordinary inverse may not exist as a single-valued function.

The generalized inverse gives a consistent rule for these cases. It selects the smallest outcome whose cumulative probability reaches a specified probability level. This convention matters at jump boundaries: a uniform value exactly equal to a cumulative probability is assigned to the outcome at that jump.

Definition: Let $F_X(x)=P(X\leq x)$ be the cumulative distribution function of $X$. For $0<u<1$, the generalized inverse, also called the quantile function, is $$ F_X^{-1}(u)=\inf\{x:F_X(x)\geq u\}. $$ It is the least threshold at which the cumulative probability is at least $u$. The notation $F_X^{-1}$ here denotes a generalized inverse, not necessarily an ordinary algebraic inverse.

The definition applies to any distribution on the real line. For a discrete distribution, it selects an outcome at a cumulative-probability jump. For a mixed distribution, it may select either an atom or a value in a continuous component. Uniform variates equal to 0 or 1 occur with probability zero under a continuous uniform distribution; in implementation, endpoint handling can be specified separately if a generator can return those values.

Discrete Laws: Probability Intervals Become Outcomes

Suppose $X$ takes ordered values $x_1<x_2<\cdots<x_n$, with probabilities $p_j=P(X=x_j)$. Define the cumulative probabilities $c_j=\sum_{k=1}^{j}p_k$, with $c_0=0$ and $c_n=1$. Then the generalized inverse assigns $x_j$ to every $u$ in the interval $(c_{j-1},c_j]$. The interval has length $p_j$, so a uniform input lands in it with exactly the probability required for $x_j$.

F_X^{-1}(u)=x_j \quad\text{when}\quad c_{j-1}<u\leq c_j.

Worked Example: Inverting a Three-Outcome Loss Distribution

A hypothetical coverage has loss $X$ equal to $\$0$, $\$4$, or $\$10$, with probabilities $0.20$, $0.50$, and $0.30$, respectively. Find the generalized inverse at $u=0.12$, $u=0.20$, $u=0.68$, and $u=0.70$. State the probability interval assigned to each possible loss.

Method. Calculate the cumulative probabilities in increasing order of loss: $$ F_X(0)=0.20,\qquad F_X(4)=0.20+0.50=0.70,\qquad F_X(10)=0.70+0.30=1. $$ Thus the generalized inverse is $0$ for $0<u\leq0.20$, $4$ for $0.20<u\leq0.70$, and $10$ for $0.70<u<1$.

Since $0.12$ and $0.20$ are both in $(0,0.20]$, their outputs are $\$0$. Since $0.68$ and $0.70$ are both in $(0.20,0.70]$, their outputs are $\$4$. In particular, $u=0.70$ maps to $\$4$, not $\$10$, because the definition uses the first outcome at which the cumulative probability is at least $u$.

Check. The input intervals have lengths $0.20$, $0.50$, and $0.30$, matching the three stated probabilities. The common mistake avoided here is using a strict upper boundary, which would incorrectly assign the exact value $u=0.70$ to the next loss.

Mixed Laws: Atoms and Continuous Components

A mixed law can place positive probability at specific points and distribute the remaining probability continuously. The generalized inverse must account for both components. At an atom, the cumulative distribution function jumps; within a continuous component, it may increase smoothly. If there is a gap in the possible outcomes, the cumulative distribution function stays constant across that gap, and the quantile moves from the lower atom to the beginning of the continuous component as $u$ passes the atom’s cumulative probability.

Worked Example: An Atom Followed by a Uniform Loss Component

A hypothetical loss $X$ is $\$2$ with probability $0.35$. With probability $0.65$, it is uniform on $[\$5,\$9]$. Find $F_X^{-1}(0.35)$ and $F_X^{-1}(0.80)$, identify the gap in the support, and check the second answer by substitution.

Method. The cumulative distribution function is zero below $\$2$, jumps to $0.35$ at $\$2$, and remains $0.35$ until the continuous component begins at $\$5$. For $5\leq x\leq9$, the conditional uniform cumulative probability is $(x-5)/4$. Weighting it by the component probability $0.65$ gives $$ F_X(x)=0.35+0.65\frac{x-5}{4},\qquad 5\leq x\leq9. $$ At $u=0.35$, the first point where the cumulative probability reaches $u$ is $x=2$, so $F_X^{-1}(0.35)=\$2$.

For $u=0.80$, the quantile is in the continuous component. Solve $0.35+0.65(x-5)/4=0.80$: $$ \frac{x-5}{4}=\frac{0.80-0.35}{0.65}=\frac{0.45}{0.65}=\frac{9}{13}, \qquad x=5+\frac{36}{13}=\frac{101}{13}\approx \$7.77. $$ The support gap is $(\$2,\$5)$: there is no probability between the atom at $\$2$ and the start of the continuous component.

Check. Substitution gives $$ F_X\left(\frac{101}{13}\right) =0.35+0.65\frac{(101/13)-5}{4} =0.35+0.65\frac{9}{13} =0.35+0.45 =0.80. $$ The answer is rounded to the nearest cent only at the final step. A tempting but incorrect shortcut is to interpolate from $\$5$ to $\$9$ using $u=0.80$ as if the continuous component had total probability 1. That would give $\$8$. In fact, $F_X(8)=0.35+0.65(3/4)=0.8375$, so $\$8$ cannot be the $0.80$ quantile.

Why the Generalized Inverse Produces the Target Law

Let $U$ be uniform on $(0,1)$, and define $Y=F_X^{-1}(U)$. The key fact is that $Y$ has the same cumulative distribution function as $X$. For any real $x$, the event that the generated value is at most $x$ corresponds to the event that the uniform input is at most $F_X(x)$:

\{F_X^{-1}(U)\leq x\}=\{U\leq F_X(x)\}.

To see why, if $U\leq F_X(x)$, then $x$ is one of the values whose cumulative probability is at least $U$. The infimum of those values cannot exceed $x$. Conversely, if the generalized inverse is at most $x$, the cumulative probability has reached the level $U$ by $x$; hence $U\leq F_X(x)$. Therefore $$ P(Y\leq x)=P(U\leq F_X(x))=F_X(x), $$ because a uniform random variable lies in $(0,a]$ with probability $a$ for $0\leq a\leq1$. This proves that $Y$ has the target distribution, including its atoms and continuous portions.

Theorem: If $U\sim\operatorname{Uniform}(0,1)$ and $F_X^{-1}(u)=\inf\{x:F_X(x)\geq u\}$, then $F_X^{-1}(U)$ has cumulative distribution function $F_X$. The result does not require $F_X$ to be continuous or strictly increasing.

The proof also explains why the convention at jump boundaries is important. The set of uniform inputs assigned to an atom has length equal to the size of the jump. Flat portions of the cumulative distribution function do not receive uniform probability of their own; instead, the generalized inverse skips the corresponding range of outcomes.

Worked Example: Inverting a Law with Two Atoms and a Continuous Component

A hypothetical payment $X$ is $\$0$ with probability $0.20$, is uniform on $[\$1,\$3]$ with probability $0.50$, and is $\$5$ with probability $0.30$. Find the generalized inverse at $u=0.10$, $u=0.45$, $u=0.70$, and $u=0.85$. Then verify that the probability assigned to each component is correct.

Method. The cumulative distribution is $0$ below $0$, equals $0.20$ from $0$ up to but not including $1$, and increases continuously on $[1,3]$. On that interval, $$ F_X(x)=0.20+0.50\frac{x-1}{2}. $$ At $x=3$, the cumulative probability is $0.20+0.50=0.70$. It stays at $0.70$ until the atom at $5$, where it jumps to $1$. Thus $F_X^{-1}(u)=0$ for $0<u\leq0.20$, $F_X^{-1}(u)=1+4(u-0.20)$ for $0.20<u\leq0.70$, and $F_X^{-1}(u)=5$ for $0.70<u<1$.

For $u=0.10$, the quantile is $\$0$. For $u=0.45$, it is $$ 1+4(0.45-0.20)=1+4(0.25)=\$2. $$ For $u=0.70$, it is $1+4(0.70-0.20)=\$3$. For $u=0.85$, it is $\$5$. The endpoint $u=0.70$ maps to $\$3$, the last value in the continuous component, because the distribution has reached $0.70$ there; values just above $0.70$ map to $\$5$.

Check. The set of inputs mapping to $0$ has length $0.20$. The continuous component receives inputs in $(0.20,0.70]$, an interval of length $0.50$. The atom at $5$ receives inputs in $(0.70,1)$, of length $0.30$. These lengths match the component probabilities. The common mistake avoided is assigning $u=0.70$ to the final atom instead of using the least-outcome convention.

A Practical Inversion Procedure

For a distribution whose cumulative distribution function is given in pieces, the following sequence makes boundary handling explicit. This is useful whether the inversion is performed by hand or implemented numerically.

List the probability jumps.
Record the locations of atoms and the cumulative probability immediately after each jump. For a discrete law, calculate cumulative sums in increasing outcome order.

Locate the input probability.
Determine which jump interval or continuous segment contains $u$, using lower-open and upper-closed intervals where appropriate.

Solve within that segment.
If the cumulative distribution is continuous and increasing there, solve $F_X(x)=u$. If $u$ falls within a jump’s cumulative-probability interval, choose the atom.

Check the result.
Substitute the proposed quantile into $F_X$ and confirm that it reaches $u$; for a continuous segment, it should equal $u$. Confirm that no smaller outcome already reaches the level.

For a complicated distribution, inversion within a continuous component may require numerical root-finding rather than an algebraic solution. The defining condition remains the same: find the least $x$ for which $F_X(x)\geq u$. The method is not ordinary inversion across a jump or a flat region, since no unique ordinary inverse exists there.

Common Errors and Exam Checks

Using the wrong boundary convention. The generalized inverse uses the condition $F_X(x)\geq u$. At a cumulative-probability boundary, use the smallest outcome that has reached that level.
Ignoring the weight of a mixture component. A conditional uniform component with mixture probability $w$ contributes $w$ times its conditional cumulative probability to the overall distribution function.
Treating a support gap as continuous probability. A gap in possible outcomes makes the cumulative distribution function flat, not linearly increasing.
Failing to check an answer. Substitution into $F_X$ often reveals a misplaced component weight or a boundary error immediately.

These checks are especially useful in simulation work: a seemingly small boundary or mixture-weight error changes the distribution of generated outcomes. For an actuarial model, that can affect simulated claim counts, payment amounts, or aggregate loss results downstream. The generalized inverse provides the link from a uniform input to the specified law, but it only preserves that law when the cumulative probabilities and interval boundaries are handled correctly.

Check Your Understanding

For each problem, use the least-outcome convention and check any continuous-segment quantile by substitution.

A discrete loss takes values $\$1,\$4,\$9$ with probabilities $0.15,0.25,0.60$. Find $F_X^{-1}(0.15)$, $F_X^{-1}(0.40)$, and the interval of uniform inputs that maps to $\$4$. (Answer: $\$1,\$4,\ (0.15,0.40]$)
A loss is $\$3$ with probability $0.40$, and otherwise is uniform on $[\$7,\$11]$. Find $F_X^{-1}(0.40)$ and $F_X^{-1}(0.75)$. (Answer: $\$3,\ \$28/3\approx \$9.33$)
A payment is $\$0$ with probability $0.10$, uniform on $[\$2,\$6]$ with probability $0.60$, and $\$8$ with probability $0.30$. Find $F_X^{-1}(0.55)$, $F_X^{-1}(0.70)$, and $F_X^{-1}(0.7001)$. (Answer: $\$5,\ \$6,\ \$8$)
A mixed loss has an atom at $-\$2$ with probability $0.20$, and is otherwise uniform on $[\$0,\$4]$. Find $F_X^{-1}(0.65)$, then calculate $F_X(1)$. (Answer: $F_X^{-1}(0.65)=\$2.25;\ F_X(1)=0.20+0.80(1/4)=0.4000$)
A loss is $\$5$ with probability $0.25$, and otherwise uniform on $[\$9,\$13]$. Find the support gap and the $0.85$ quantile, then verify the quantile by substitution. (Answer: gap $(\$5,\$9)$; $F_X^{-1}(0.85)=\$12.20$, since $0.25+0.75(12.20-9)/4=0.85$)

What You'll Learn