Multiplicative Inflation of a Severity Distribution

Why Inflation Changes a Severity Distribution

A ground-up loss is the loss amount before applying a deductible, limit, or other coverage modification. When the price level changes, a loss measured in one price base can be represented in another by multiplying it by an inflation factor. This changes not only individual dollar amounts but also the distribution of those amounts. Correctly transforming the distribution is useful when a probability, density, or quantile is required at the new price level.

This tutorial treats inflation as a deterministic multiplicative factor. It does not model uncertainty in future inflation, changes in claim frequency, or changes in the types of losses being measured. Those effects would require additional assumptions beyond a simple severity scale transformation.

Definition: Let $X$ be a nonnegative ground-up loss in a specified price base, with cumulative distribution function $F_X$, survival function $S_X$, and, when it exists, density $f_X$. Let $c>0$ be a deterministic inflation factor. The loss in the new price base is $Y=cX$. If $c$ is obtained from price indices, then $c=I_{\text{new}}/I_{\text{old}}$, where the indices use consistent definitions and bases.

If the price index rises, $c>1$, so each modeled loss amount increases. If the index falls, $0<c<1$, so the amounts decrease. In either case, positivity of $c$ preserves the order of losses: whenever $x_1\le x_2$, $cx_1\le cx_2$. That order-preserving property is central to transforming the distribution and its quantiles.

Distribution, Survival Function, and Density

For any real number $y$, the event that the inflated loss is no greater than $y$ can be rewritten by dividing by the positive factor $c$. This gives the cumulative distribution function directly.

F_Y(y)=P(Y\le y)=P(cX\le y)=P\left(X\le \frac{y}{c}\right) =F_X\left(\frac{y}{c}\right).

The same event transformation gives the survival function, using the convention $S_X(x)=P(X>x)$:

S_Y(y)=P(Y>y)=P\left(X>\frac{y}{c}\right) =S_X\left(\frac{y}{c}\right).

These formulas apply to discrete, continuous, and mixed losses; no density assumption is needed. In particular, an atom at $x_0$ becomes an atom at $cx_0$ with the same probability. Multiplication moves the location of the atom but does not change its probability.

If $X$ has a density, differentiating its transformed cumulative distribution function gives the density of $Y$. The chain rule contributes the factor $1/c$.

f_Y(y)=\frac{1}{c}f_X\left(\frac{y}{c}\right).

This density formula applies where the distributions have densities and the derivative exists. It is not a complete description of a distribution with atoms: for a mixed distribution, retain the transformed continuous density and move each atom from $x_0$ to $cx_0$, preserving its probability. For a nonnegative loss, the support is also scaled: a possible value $x$ of $X$ corresponds to $cx$ for $Y$.

Key result: Under deterministic multiplicative inflation $Y=cX$, with $c>0$, evaluate the original distribution at the threshold divided by $c$: $F_Y(y)=F_X(y/c)$ and $S_Y(y)=S_X(y/c)$. If a density exists, $f_Y(y)=f_X(y/c)/c$.

Quantiles Under Multiplicative Inflation

For a distribution that may have jumps, define its $p$-quantile for $0<p<1$ by the generalized inverse $Q_X(p)=\inf\{x:F_X(x)\ge p\}$. This definition applies to continuous and discrete losses, including values of $p$ at which the cumulative distribution function reaches a jump.

Q_Y(p)=cQ_X(p), \qquad 0<p<1.

To justify the result, let $q=Q_X(p)$. Since $c>0$, the threshold inequality $y\ge cq$ is equivalent to $y/c\ge q$. Also, $F_Y(y)\ge p$ exactly when $F_X(y/c)\ge p$. Thus the smallest $y$ for which $F_Y(y)\ge p$ is $cq$, as claimed. Positivity matters: it ensures that multiplication preserves the ordering used to define the smallest threshold.

This quantile relation is a direct distributional consequence of inflation. It does not require a particular severity family. For a discrete loss, it scales the quantile’s location while retaining the same probability level; it does not interpolate between mass points.

Worked Distribution Transformations

Worked Example: Updating a Discrete Loss Distribution

A hypothetical ground-up loss $X$, in the old price base, equals $500 with probability 0.20, $1,000 with probability 0.50, and $2,000 with probability 0.30. A deterministic price-index ratio gives an inflation factor $c=1.06$. Find the possible inflated losses, the cumulative probability at $1,060, the survival probability above $1,060, and the 65th percentile.

Method and assumptions. Treat the stated loss amounts and probabilities as the complete distribution of $X$, and apply the same positive factor $1.06$ to every outcome. The probabilities sum to $0.20+0.50+0.30=1.00$.

Transform the support. The three possible outcomes become $1.06(500)=530$, $1.06(1{,}000)=1{,}060$, and $1.06(2{,}000)=2{,}120$. Their probabilities remain 0.20, 0.50, and 0.30, respectively.

Cumulative probability. At $1,060, the inflated loss is at most $1,060 for the first two outcomes. Therefore, $F_Y(1{,}060)=0.20+0.50=0.70$.

Survival probability. The survival function uses a strict exceedance, $S_Y(y)=P(Y>y)$. Only the $2,120 outcome is strictly greater than $1,060, so $S_Y(1{,}060)=0.30$. The results are consistent: $F_Y(1{,}060)+S_Y(1{,}060)=1$.

65th percentile. The cumulative probability is 0.20 at $530 and 0.70 at $1,060. The smallest amount at which the cumulative probability is at least 0.65 is therefore $1,060. Equivalently, the original 65th percentile is $1,000, and $1.06(1{,}000)=1{,}060$.

The common mistake avoided here is treating the 65th percentile as an interpolated amount between support points. With the generalized inverse definition, a discrete quantile is a support point when the cumulative probability first reaches or exceeds the target probability.

Worked Example: Exponential Loss and the Threshold Direction

Suppose a hypothetical ground-up loss $X$, in the old price base, has an exponential distribution with rate $\lambda=0.002$ per dollar. Its cumulative distribution function is $F_X(x)=1-e^{-0.002x}$ for $x\ge0$. Inflation is deterministic with factor $c=1.08$. Find $F_Y(1{,}080)$, $S_Y(1{,}080)$, and the density $f_Y(1{,}080)$. Also find the 90th percentile of $Y$.

Method and assumptions. The loss is nonnegative and continuous, and the rate is expressed per old-price-base dollar. Transform thresholds by dividing by $1.08$; transform the density with the additional factor $1/1.08$.

Find the original-price threshold. A new-base threshold of $1,080 corresponds to $1{,}080/1.08=1{,}000$ old-base dollars.

Cumulative probability. $F_Y(1{,}080)=F_X(1{,}000)=1-e^{-0.002(1{,}000)} =1-e^{-2}=0.8646647\ldots$, which is $0.8647$ to four decimal places.

Survival probability. $S_Y(1{,}080)=e^{-0.002(1{,}000)}=e^{-2}=0.1353353\ldots$, which is $0.1353$ to four decimal places. As a check, the cumulative and survival probabilities sum to 1 before rounding.

Density. The original density at $1,000 is $f_X(1{,}000)=0.002e^{-2}=0.0002706706\ldots$ per old-base dollar. Therefore, $f_Y(1{,}080)=f_X(1{,}000)/1.08 =0.002e^{-2}/1.08=0.00025062\ldots$ per new-base dollar. The units change with the price base.

90th percentile. For an exponential loss, $Q_X(p)=-\ln(1-p)/\lambda$. Thus $Q_X(0.90)=-\ln(0.10)/0.002=1{,}151.2925\ldots$ old-base dollars. Scaling gives $Q_Y(0.90)=1.08(1{,}151.2925\ldots)=1{,}243.3960\ldots$, or $1,243.40 to the nearest cent.

The checks are that the threshold was converted to the old price base before evaluating $F_X$, and that the new-base 90th percentile is exactly 1.08 times the old-base percentile.

Worked Example: Transforming a Uniform Density and Quantile

A hypothetical loss $X$ is uniform on the interval from $400 to $1,400 in the old price base. Inflation is deterministic with factor $c=1.05$. Find the support, cumulative distribution function and density of $Y$, and calculate $F_Y(945)$, $f_Y(945)$, and the 80th percentile.

Method and assumptions. For a uniform distribution on $[400,1{,}400]$, the original density is $1/(1{,}400-400)=1/1{,}000$ per dollar within that interval, and zero outside it.

Support and distribution. Scaling both endpoints gives the new support $[1.05(400),1.05(1{,}400)]=[420,1{,}470]$. Within this interval, $F_Y(y)=F_X(y/1.05)=(y/1.05-400)/1{,}000=(y-420)/1{,}050$. Below $420 the cumulative distribution function is 0; above $1,470 it is 1.

Density. The transformed density is $f_Y(y)=f_X(y/1.05)/1.05=1/1{,}050$ per dollar for $420<y<1{,}470$, and zero outside the support. This also agrees with the reciprocal of the new interval width, $1/(1{,}470-420)=1/1{,}050$.

Evaluate at $945. Since $945 lies inside the support, $F_Y(945)=(945-420)/1{,}050=525/1{,}050=0.50$, and $f_Y(945)=1/1{,}050=0.00095238\ldots$ per dollar.

80th percentile. The original 80th percentile is $400+0.80(1{,}000)=1{,}200$. Hence $Q_Y(0.80)=1.05(1{,}200)=1{,}260$. The same result follows from the transformed distribution: $420+0.80(1{,}050)=1{,}260$.

The support check is $420\le945\le1{,}470$, so the interior formulas used for the cumulative distribution and density apply.

Common Trap: Multiplying the Threshold Instead of Dividing

The transformation is $F_Y(y)=F_X(y/c)$, not $F_X(cy)$. A threshold stated in the new price base must be converted back to the old price base before the original distribution is evaluated. Multiplication by $c$ instead transforms an old-base loss amount into a new-base amount.

For the exponential example, the incorrect substitution at a new-base threshold of $1,080 is $F_X(1.08\times1{,}080)=F_X(1{,}166.40)=1-e^{-2.3328}\approx0.9030$ to four decimal places. This is not $F_Y(1{,}080)$: it uses the wrong old-base threshold. The correct calculation is $F_Y(1{,}080)=F_X(1{,}080/1.08)=F_X(1{,}000)=1-e^{-2}\approx0.8647$.

A reliable check is to ask what old-base loss amount would inflate to the specified new-base threshold. Solving $1.08x=1{,}080$ gives $x=1{,}000$, confirming the division.

When the Transformation Applies

A single scale factor is appropriate when the same deterministic price-level adjustment applies to every loss in the modeled distribution. It preserves the probability assigned to each outcome and changes the dollar scale. If inflation varies randomly across claims, depends on claim size, or is combined with changes in the claim population, then $Y=cX$ with a fixed $c$ is not an adequate model without further assumptions.

For calculations, identify the price base of the random variable and of the threshold before substituting. Use $y/c$ to evaluate an original cumulative distribution or survival function; use $f_X(y/c)/c$ for a density; and multiply an original quantile by $c$ to express it in the new base. This distinction is especially important because a density has units of inverse dollars and must change when the dollar scale changes.

Key takeaway: Deterministic multiplicative inflation rescales loss amounts, not their probabilities. Divide a new-base threshold by the inflation factor to evaluate the original distribution; multiply an original quantile or support point by the factor to express it in the new base.

Check Your Understanding

For each problem, treat the stated inflation factor as deterministic and apply it to every ground-up loss.

A hypothetical loss $X$ equals $300 with probability 0.25, $900 with probability 0.50, and $1,500 with probability 0.25. The inflation factor is 1.10. Find the inflated support, $F_Y(990)$, and $S_Y(990)$. (Answer: support $330, $990, and $1,650; $F_Y(990)=0.7500$, $S_Y(990)=0.2500$.)
A hypothetical $X$ is uniform on $[200,800]$, and $Y=1.25X$. Find the support of $Y$, $F_Y(625)$, and $f_Y(625)$. (Answer: support $[250,1{,}000]$; $F_Y(625)=0.5000$; $f_Y(625)=1/750\approx0.001333$ per dollar.)
A hypothetical ground-up loss is exponential with rate 0.004 per old-base dollar, and $Y=0.95X$. Find $F_Y(475)$, $S_Y(475)$, and $f_Y(475)$. (Answer: the old-base threshold is $500; $F_Y(475)=1-e^{-2}=0.8647$, $S_Y(475)=e^{-2}=0.1353$, and $f_Y(475)=(0.004/0.95)e^{-2}\approx0.0005698$ per new-base dollar.)
A discrete loss $X$ equals $1,000, $2,000, or $4,000 with probabilities 0.20, 0.50, or 0.30, respectively. Under inflation factor 1.07, find the 70th percentile of $Y$, using the generalized inverse definition. (Answer: $2,140, since the cumulative probability first reaches 0.70 at that amount.)
A hypothetical loss has an old-base 80th percentile of $1,600. The price index rises from 240 to 258. Find the inflation factor and the new-base 80th percentile. (Answer: $c=258/240=1.075$; new-base percentile $1.075(1{,}600)=\$1{,}720$.)

What You'll Learn