The Inverse Transform Theorem

Why the Inverse Transform Method Works

Simulation often starts with a source of uniform random numbers. The inverse transform method converts a uniform value into a value from a specified continuous distribution. In actuarial work, that can provide simulated claim severities for a loss model, provided the distribution meets the conditions needed for the ordinary inverse CDF.

The central idea is that a CDF converts a value on the loss scale into a probability scale. If the CDF is continuous and strictly increasing on the distribution’s support, it has an ordinary inverse: each probability level corresponds to exactly one value. Applying that inverse to a uniform probability therefore produces values whose probabilities follow the target CDF.

Definition: Let $X$ have CDF $F_X(x)=P(X\le x)$. Suppose $F_X$ is continuous and strictly increasing on its support, with its range spanning the probabilities between 0 and 1. Its ordinary inverse $F_X^{-1}(u)$ is the unique value $x$ such that $F_X(x)=u$. If $U\sim\operatorname{Uniform}(0,1)$, the inverse transform method defines $Y=F_X^{-1}(U)$.

The theorem concerns an ideal continuous uniform random variable $U$. A computer’s pseudo-random uniform numbers are finite-precision approximations; using them is a practical simulation implementation, not a change to the probability result. The value $U=0$ or $U=1$ has probability zero under an ideal continuous uniform distribution. For finite-support distributions, these endpoints can be assigned endpoint values if an implementation can produce them.

Under the definition’s continuity and strict-increase conditions, the transformed variable $Y=F_X^{-1}(U)$ has CDF $F_X$. The proof compares the event that the transformed value is no greater than $x$ with an event involving $U$.

Start with the desired CDF.
For any $x$ in the support, $P(Y\le x)=P(F_X^{-1}(U)\le x)$.

Use monotonicity.
Because $F_X$ is strictly increasing, its inverse is increasing. Thus $F_X^{-1}(U)\le x$ holds exactly when $U\le F_X(x)$.

Apply the uniform distribution.
Since $F_X(x)$ lies between 0 and 1, $P(U\le F_X(x))=F_X(x)$.

Conclude equality of distributions.
Combining the steps gives $P(Y\le x)=F_X(x)$, which is the defining CDF of $X$.

P(Y\le x) =P\bigl(F_X^{-1}(U)\le x\bigr) =P\bigl(U\le F_X(x)\bigr) =F_X(x).

For $x$ below the support, the probability on both sides is 0; for $x$ above the support, it is 1. This completes the argument across the real line. The equality in the second step depends on using an ordinary inverse of a strictly increasing CDF. If the CDF has jumps or flat portions, that step cannot be justified with the same ordinary inverse argument. The treatment of discrete and mixed distributions requires a different inverse convention.

Worked Examples

Worked Example: Inverting a Bounded Severity Distribution

Method: Solve the CDF equation for the loss as a function of the uniform probability. A hypothetical loss $X$, measured in dollars, has CDF $F_X(x)=1-(1-x/10)^3$ for $0\le x\le10$, with $F_X(x)=0$ below 0 and $F_X(x)=1$ above 10. Its density on the support is $f_X(x)=\frac{3}{10}(1-x/10)^2$, which is nonnegative and integrates to 1. Generate a value when $U=0.875$. No interest or mortality assumption applies.

Set $u=F_X(x)$ and rearrange:

u=1-(1-x/10)^3 \quad\Longrightarrow\quad 1-u=(1-x/10)^3 \quad\Longrightarrow\quad F_X^{-1}(u)=10\left[1-(1-u)^{1/3}\right].

Substitute $u=0.875$. Then $1-u=0.125$, and $(0.125)^{1/3}=0.5$. Therefore,

F_X^{-1}(0.875)=10(1-0.5)=5.

The generated loss is $5$ dollars. Check by evaluating its CDF at 5: $F_X(5)=1-(1-5/10)^3=1-(0.5)^3=0.875$, matching the input probability. This example avoids the common mistake of treating a probability input as though it were already a loss value.

Worked Example: Generating an Exponential Loss

Method: Invert the CDF and substitute the uniform value. A hypothetical claim severity $X$, measured in thousands of dollars, has an exponential distribution with rate $\lambda=0.2$ per thousand dollars. Thus $F_X(x)=1-e^{-0.2x}$ for $x\ge0$. Use $U=0.8$. No interest or mortality assumption applies.

First solve $u=1-e^{-0.2x}$ for $x$:

e^{-0.2x}=1-u, \qquad -0.2x=\ln(1-u), \qquad F_X^{-1}(u)=\frac{-\ln(1-u)}{0.2}.

For $u=0.8$, $1-u=0.2$, so

F_X^{-1}(0.8) =\frac{-\ln(0.2)}{0.2} =\frac{1.6094379\ldots}{0.2} =8.047189\ldots

Rounded only at the final step to four decimal places, the generated severity is $8.0472$ thousand dollars, or approximately $\$8{,}047.19$. Check the CDF: $e^{-0.2(8.047189\ldots)}=e^{\ln(0.2)}=0.2$, so $F_X(x)=1-0.2=0.8$. The common mistake avoided here is using $-\ln(U)/\lambda$, which corresponds to a different uniform input under this CDF convention.

Worked Example: A Quadratic CDF and a Tail Probability

Method: Invert the CDF, then translate a loss threshold into a uniform threshold. A hypothetical severity $X$, measured in thousands of dollars, has CDF $F_X(x)=x^2/16$ for $0\le x\le4$, with CDF 0 below 0 and 1 above 4. It is continuous and strictly increasing on its support; its density is $f_X(x)=x/8$. Use $U=0.36$ to generate a loss, and determine the probability that a generated loss exceeds 3. No interest or mortality assumption applies.

Solving $u=x^2/16$ on the nonnegative support gives $x=4\sqrt{u}$. Therefore,

F_X^{-1}(0.36)=4\sqrt{0.36}=4(0.6)=2.4.

The simulated loss is $2.4$ thousand dollars. To find the probability of exceeding 3, use the monotone transformation: $X>3$ exactly when $U>F_X(3)$. Evaluate $F_X(3)=9/16=0.5625$. Since $U$ is uniform,

$$ P(X>3)=P(U>0.5625)=1-0.5625=0.4375. $$

A direct check gives $1-F_X(3)=1-9/16=7/16=0.4375$. The common mistake avoided is to confuse the particular generated value $2.4$ with the probability of exceeding a separate threshold.

Conditions, Scope, and Exam Traps

The ordinary inverse transform theorem is most straightforward when the target CDF is continuous and strictly increasing over its support. Before inverting, check that the proposed CDF is valid and that the inverse is well-defined for the uniform inputs. For a bounded support, confirm that the CDF reaches 0 and 1 at the appropriate endpoints. For an unbounded support, its limiting values must approach 0 and 1.

For example, a CDF that is strictly increasing only on the interior of a support may still have a flat section outside that support. That does not cause difficulty if the inverse is defined on the probability range and the support endpoints are handled consistently. The essential issue for the ordinary inverse is whether each probability strictly between 0 and 1 identifies a unique support value. A discrete CDF has jumps, and a mixed CDF can have jumps as well as continuous portions; in those cases an ordinary inverse is not generally the appropriate direct tool.

A particularly common exam error is reversing the direction of the calculation. If a problem gives $U=u$ and asks for a simulated value, calculate $F_X^{-1}(u)$, not $F_X(u)$. In the first worked example, $F_X(0.875)=1-(0.9125)^3=0.240201171875$, approximately $0.2402$. That is the CDF evaluated at the loss $0.875$; it is not the simulated loss generated from $U=0.875$, which is 5.

A second trap is mishandling the complement in a decreasing exponential term. From $u=1-e^{-\lambda x}$, the correct rearrangement is $e^{-\lambda x}=1-u$, giving $x=-\ln(1-u)/\lambda$. Keeping the complement visible until the logarithm is taken helps prevent the error.

Key takeaway: When $F_X$ is continuous and strictly increasing on its support, the inverse transform $Y=F_X^{-1}(U)$, with $U\sim\operatorname{Uniform}(0,1)$, has CDF $F_X$. Prove it by translating $Y\le x$ into $U\le F_X(x)$, then use the uniform CDF.

Check Your Understanding

For each problem, first check the CDF conditions and then apply the appropriate inversion or probability argument.

A hypothetical loss is uniform on $[2,14]$ dollars. Find the generated loss for $U=0.25$. (Answer: $\$5$)
A hypothetical severity in dollars has $F_X(x)=1-(1-x/6)^2$ for $0\le x\le6$. Find $F_X^{-1}(0.75)$. (Answer: $\$3$)
A hypothetical exponential waiting-time loss, measured in years, has rate $\lambda=0.5$ per year. Find the generated value for $U=0.5$, rounded to four decimals. (Answer: $1.3863$ years)
A hypothetical severity in thousands of dollars has $F_X(x)=(x/5)^3$ for $0\le x\le5$. Find the generated value for $U=0.008$, and find $P(X>4)$. (Answer: $1$ thousand dollars; $0.4880$)
A target distribution is discrete, so its CDF has jumps. Does the ordinary inverse theorem proved here apply directly under its stated conditions? (Answer: No; the continuity and strict-increase conditions fail, so a discrete inverse convention is needed.)

The Inverse Transform Theorem

What You'll Learn

Why the Inverse Transform Method Works