Scaling And Shifting Severity Distributions

Why Scaling and Shifting Matter

A loss amount may be expressed in different units, adjusted by a fixed allowance, or converted by a proportional factor. For instance, a hypothetical reinsurer might model its payment as a fixed amount plus a stated proportion of a covered loss. Such changes can be represented by $Y=aX+b$, where $X$ is the original loss, $a$ is a scale factor, and $b$ is a constant shift. Understanding how this changes the distribution lets an actuary calculate probabilities, moments, and quantiles without rebuilding the model from scratch.

This tutorial focuses on $a>0$, the usual case when larger original losses still produce larger transformed amounts. The formulas do not require $X$ to be continuous: the CDF, mean, variance, and quantile results also apply to discrete or mixed distributions, when the relevant moments exist. The density formula, by contrast, requires a continuous distribution with a density.

Definition: Let $X$ be a random loss and let $a>0$ and $b$ be constants. The positively scaled and shifted loss is $Y=aX+b$. The constant $a$ changes the scale, while $b$ adds the same amount to every possible value. The transformed values of the support of $X$ are obtained by applying $x\mapsto ax+b$.

Distribution, Density, and Moments

Write $F_X(x)=P(X\le x)$ for the CDF of $X$. Since $a>0$, dividing an inequality by $a$ preserves its direction. Therefore, the event that $Y$ is at most $y$ is exactly the event that $X$ is at most $(y-b)/a$.

F_Y(y)=P(Y\le y) =P(aX+b\le y) =P\left(X\le \frac{y-b}{a}\right) =F_X\left(\frac{y-b}{a}\right).

This CDF formula is valid for continuous, discrete, and mixed $X$. It also shows how the support changes: if $X$ can take values in an interval $[L,U]$, then $Y$ can take values in $[aL+b,aU+b]$. For an unbounded or more complicated support, apply the same map to each possible value or support point.

If $X$ is continuous with density $f_X$, differentiating the CDF gives the density of $Y$. The derivative of $(y-b)/a$ with respect to $y$ is $1/a$, so the density includes that factor.

f_Y(y)=\frac{1}{a}f_X\left(\frac{y-b}{a}\right).

The density formula is not a formula for a discrete distribution’s probability mass function. For a discrete $X$, each point $x$ with probability $P(X=x)$ becomes the point $ax+b$ with that same probability. The CDF formula handles both cases.

Theorem: If $Y=aX+b$, where $a>0$, and $E[X]$ and $\operatorname{Var}(X)$ exist and are finite, then $$ E[Y]=aE[X]+b,\qquad \operatorname{Var}(Y)=a^2\operatorname{Var}(X). $$

The mean changes with both the scale and the shift, but the variance changes only with the scale. Adding $b$ moves every outcome by the same amount and therefore does not change how far outcomes lie from their mean. Multiplying by $a$ multiplies each deviation from the mean by $a$; squaring those deviations in the definition of variance produces the factor $a^2$.

Why the Moment Formulas Hold

The expectation rule follows from linearity of expectation. For the variance, first find the transformed mean and then subtract it from $Y$. This gives a direct justification rather than relying on a memorized rule.

Transform the mean.
Linearity gives $E[Y]=E[aX+b]=aE[X]+b$.

Center the transformed variable.
Subtract its mean: $Y-E[Y]=aX+b-(aE[X]+b)=a(X-E[X])$.

Square and take expectations.
By the definition of variance, $\operatorname{Var}(Y)=E[(Y-E[Y])^2]=E[a^2(X-E[X])^2]=a^2\operatorname{Var}(X)$.

The same argument shows why the shift disappears from the variance. It cancels when the mean is subtracted. The square on $a$ is essential: scaling losses by $a$ scales standard deviation by $a$, but variance by $a^2$.

Quantiles Under a Positive Transformation

For a distribution that may have jumps, define the $p$-quantile, for $0<p<1$, as $Q_X(p)=\inf\{x:F_X(x)\ge p\}$. This left-inverse definition is useful for discrete as well as continuous losses. Because $x\mapsto ax+b$ is increasing when $a>0$, it preserves the ordering of all possible loss values, and the corresponding quantile transforms in the same way.

Q_Y(p)=aQ_X(p)+b,\qquad 0<p<1.

For a continuous distribution with a strictly increasing CDF, $Q_X(p)$ is the unique value satisfying $F_X(Q_X(p))=p$. For a discrete distribution, the quantile may be a point at which the CDF jumps over $p$; use the left-inverse definition rather than trying to solve an equality that may not hold.

Worked Example: Transforming an Exponential Loss

A hypothetical claim severity $X$, measured in dollars, has an exponential distribution with rate $\lambda=1/4{,}000$ per dollar. Thus $E[X]=4{,}000$ and $\operatorname{Var}(X)=4{,}000^2=16{,}000{,}000$. A payment measure is defined by $Y=1.5X+200$. Find its support, CDF, density, mean, variance, and 90th percentile.

Method and support. The exponential loss has support $x\ge0$. Since $a=1.5>0$, the transformed support is $y\ge200$. For $y<200$, $F_Y(y)=0$. For $y\ge200$, the inverse argument is $(y-200)/1.5$, so

F_Y(y)=1-\exp\left(-\frac{y-200}{6{,}000}\right),\qquad y\ge200.

Density. The original density is $f_X(x)=\frac{1}{4{,}000}e^{-x/4{,}000}$ for $x\ge0$. The scale factor in the density transformation is $1/1.5$. Hence

f_Y(y)=\frac{1}{6{,}000}\exp\left(-\frac{y-200}{6{,}000}\right),\qquad y\ge200.

Mean and variance. Apply the moment formulas:

E[Y]=1.5(4{,}000)+200=6{,}200\text{ dollars}, $$ $$ \operatorname{Var}(Y)=1.5^2(16{,}000{,}000)=2.25(16{,}000{,}000)=36{,}000{,}000\text{ dollars}^2.

90th percentile. For an exponential loss, $F_X(x)=1-e^{-x/4{,}000}$. Set this equal to $0.90$: $e^{-x/4{,}000}=0.10$, so $Q_X(0.90)=-4{,}000\ln(0.10)=9{,}210.340$ dollars to three decimal places. Therefore,

Q_Y(0.90)=1.5(9{,}210.340371\ldots)+200 =14{,}015.510557\ldots\text{ dollars} \approx \$14{,}015.51.

As a check, inserting the unrounded quantile into the CDF gives $F_Y(Q_Y(0.90))=1-e^{-\ln(10)}=0.90$. The method avoids confusing the scale factor for the density with the factor for the mean: the density has $1/a$, while the mean has $a$.

Worked Example: A Discrete Severity and Its Quantile

A hypothetical portfolio has a discrete loss $X$ with the following distribution: $P(X=1{,}000)=0.20$, $P(X=3{,}000)=0.50$, and $P(X=5{,}000)=0.30$. Define $Y=0.8X+400$. Find the possible values and probabilities of $Y$, its mean and variance, and its 75th percentile.

Transform the support points. Applying $y=0.8x+400$ gives $1{,}000\mapsto1{,}200$, $3{,}000\mapsto2{,}800$, and $5{,}000\mapsto4{,}400$. Because this is a one-to-one increasing transformation, the probabilities stay attached to their corresponding outcomes.

$y$	$P(Y=y)$	$F_Y(y)$
1,200	0.20	0.20
2,800	0.50	0.70
4,400	0.30	1.00

Mean. First, $E[X]=0.20(1{,}000)+0.50(3{,}000)+0.30(5{,}000)=200+1{,}500+1{,}500=3{,}200$. Thus $E[Y]=0.8(3{,}200)+400=2{,}960$ dollars.

Variance. Calculate $E[X^2]=0.20(1{,}000{,}000)+0.50(9{,}000{,}000)+0.30(25{,}000{,}000)=200{,}000+4{,}500{,}000+7{,}500{,}000=12{,}200{,}000$. Since $E[X]^2=3{,}200^2=10{,}240{,}000$, $\operatorname{Var}(X)=12{,}200{,}000-10{,}240{,}000=1{,}960{,}000$. Therefore,

\operatorname{Var}(Y)=0.8^2(1{,}960{,}000) =0.64(1{,}960{,}000) =1{,}254{,}400\text{ dollars}^2.

75th percentile. The CDF is 0.70 at 2,800 and reaches 1.00 at 4,400. The smallest value with CDF at least 0.75 is therefore $Q_Y(0.75)=4{,}400$ dollars. A check is $F_Y(4{,}400)=1.00\ge0.75$, while $F_Y(y)<0.75$ for every $y<4{,}400$. The jump matters: the 75th percentile is not found by interpolating between support points.

Worked Example: Transforming a Uniform Loss

A hypothetical loss $X$ is uniform on $[0,12{,}000]$ dollars. A reporting measure is $Y=0.5X+1{,}000$. Find the support, density, mean, variance, probability $P(4{,}000<Y\le6{,}000)$, and 75th percentile.

Support and density. The transformed endpoints are $0.5(0)+1{,}000=1{,}000$ and $0.5(12{,}000)+1{,}000=7{,}000$. The original density is $1/12{,}000$ on its support. The transformed density is therefore $(1/0.5)(1/12{,}000)=1/6{,}000$ on $[1{,}000,7{,}000]$, and zero outside it.

Mean and variance. A uniform distribution on $[0,12{,}000]$ has mean $6{,}000$ and variance $(12{,}000-0)^2/12=144{,}000{,}000/12=12{,}000{,}000$. Consequently,

E[Y]=0.5(6{,}000)+1{,}000=4{,}000\text{ dollars}, $$ $$ \operatorname{Var}(Y)=0.5^2(12{,}000{,}000)=3{,}000{,}000\text{ dollars}^2.

Probability. Invert both boundaries: $4{,}000<0.5X+1{,}000\le6{,}000$ is equivalent to $6{,}000<X\le10{,}000$. The interval has length $4{,}000$ within the original uniform range of length $12{,}000$, so

P(4{,}000<Y\le6{,}000) =\frac{10{,}000-6{,}000}{12{,}000} =\frac{1}{3} \approx0.3333.

75th percentile. For a uniform distribution on $[0,12{,}000]$, $Q_X(0.75)=0+0.75(12{,}000)=9{,}000$. Thus $Q_Y(0.75)=0.5(9{,}000)+1{,}000=5{,}500$ dollars. It lies in the transformed support, as required.

Common Mistakes and Boundary Cases

A frequent error is to use the transformed threshold in the wrong direction. For example, in the uniform example, $Y\le6{,}000$ requires $X\le(6{,}000-1{,}000)/0.5=10{,}000$, not $X\le0.5(6{,}000)+1{,}000$. The inverse step must undo the shift first and then divide by the scale.

Another common error is to apply the variance rule as $\operatorname{Var}(Y)=a\operatorname{Var}(X)$. This is wrong because variance is based on squared deviations. For example, with $a=0.5$ and $\operatorname{Var}(X)=12{,}000{,}000$, that incorrect calculation gives $6{,}000{,}000$; the correct calculation is $0.5^2(12{,}000{,}000)=3{,}000{,}000$.

The density transformation also has a direction that can be easy to reverse. For positive $a$, the density is $f_X((y-b)/a)/a$, not $a f_X((y-b)/a)$. The factor $1/a$ ensures the transformed density integrates to one over the stretched support. If $a=0$, $Y=b$ with probability one: it is degenerate and does not have an ordinary continuous density. If $a<0$, the order reverses, so the positive-scale CDF and quantile formulas above cannot be used unchanged.

Key takeaway: For $Y=aX+b$ with $a>0$, invert the transformation in the CDF, multiply the mean by $a$ and then add $b$, multiply the variance by $a^2$, and transform quantiles by $aQ_X(p)+b$. Use the density formula only when $X$ has a density.

These rules are useful when a model is reported in a new monetary unit, when a fixed amount is added to a modeled loss, or when a proportional factor is applied. They keep the distributional calculations consistent and help reveal unit errors: means and quantiles retain the units of loss, while variances have squared units.

Check Your Understanding

For each problem, apply the positive affine transformation rules and check the support or CDF where relevant.

A discrete loss $X$ equals $2,000 with probability 0.25 and $6,000 with probability 0.75. Let $Y=2X-500$. Find $E[Y]$ and $\operatorname{Var}(Y)$. (Answer: $E[Y]=\$9{,}500$; $\operatorname{Var}(Y)=\$12{,}000{,}000\text{ dollars}^2$.)
Let $X$ be uniform on $[0,8{,}000]$ dollars and $Y=1.25X+500$. Find the support of $Y$ and $P(Y\le6{,}500)$. (Answer: support $[ \$500,\ \$10{,}500]$; probability $0.6000$.)
Let $X$ be exponential with mean $2,500, and let $Y=0.4X+300$. Find the 80th percentile of $Y$, to the nearest cent. (Answer: $\$1{,}909.44$.)
A continuous loss $X$ has density $f_X(x)=2x/100$ for $0\le x\le10$, and zero otherwise. Let $Y=3X+2$. Find the support and density of $Y$. (Answer: support $[2,32]$; $f_Y(y)=2(y-2)/900$ for $2\le y\le32$, and zero otherwise.)
A loss $X$ has mean $4,000 and variance $900{,}000$ dollars squared. For $Y=0.6X+250$, find the mean, variance, and standard deviation of $Y$. (Answer: mean $\$2{,}650$; variance $\$324{,}000\text{ dollars}^2$; standard deviation $\$569.21$, to the nearest cent.)

What You'll Learn