Why Transform a Severity Variable?
A severity model describes a nonnegative loss \(X\). In some applications, the quantity of interest is instead a monotone function of that loss: for example, a logarithm used to represent a compressed scale, or a power used to express a different severity scale. To use the transformed variable, we need its distribution, not just the formula connecting it to \(X\).
The central method is to translate an event about the transformed variable back into an event about \(X\). A strictly monotone transformation has an inverse, so the event can be written using that inverse. Differentiating the resulting CDF gives the density when suitable regularity conditions hold. The direction of the inequality changes for a decreasing transformation, a detail that is particularly important in exam calculations.
The Inverse-Transformation Method
First identify the range of \(Y\). On the interior of that range, translate \(Y\le y\) into an event about \(X\). If \(g\) is increasing, the inequality keeps its direction; if \(g\) is decreasing, it reverses. Since \(X\) is continuous, probabilities at a single point are zero, so for either direction the CDF can be expressed using \(F_X\) or its upper tail.
The absolute-continuity condition matters. Differentiability of \(h\) alone is not enough to guarantee that the transformed distribution has a density. A common sufficient condition is that \(g\) is continuously differentiable with a nonzero derivative on the interior of its domain; then its inverse is continuously differentiable there. The density formula is an almost-everywhere statement, and endpoints where the derivative is unbounded or undefined can be handled by the CDF and limits.
When \(g\) is differentiable and \(g'(x)\ne 0\), the inverse-function rule gives \(h'(y)=1/g'(h(y))\). Thus an equivalent form is $$ f_Y(y)=\frac{f_X(h(y))}{|g'(h(y))|}. $$ This form shows why the absolute value is essential: a density cannot be negative, whether \(g\) increases or decreases.
Why the Density Formula Holds
The formula follows from how probabilities over intervals change under the inverse map. Take an interval \((a,b]\) inside the range of \(Y\). For an increasing transformation, \(a<Y\le b\) is equivalent to \(h(a)<X\le h(b)\). Therefore, $$ P(a<Y\le b)=\int_{h(a)}^{h(b)} f_X(x)\,dx =\int_a^b f_X(h(y))h'(y)\,dy. $$
The last equality is the change-of-variables formula for the locally absolutely continuous, increasing inverse \(h\). For a decreasing transformation, the corresponding \(X\)-interval has its endpoints reversed. Its probability is still the integral of \(f_X(h(y))|h'(y)|\) over \((a,b]\). In both cases, probabilities over intervals are obtained by integrating the stated nonnegative function, which establishes the density formula almost everywhere. This argument also explains why the regularity condition cannot be omitted: without the change-of-variables result for \(h\), differentiating a CDF formally need not produce a density for the whole distribution.
Use the support of \(X\) and the behavior of \(g\) to determine the possible values of \(Y\).
Set \(x=h(y)=g^{-1}(y)\) on the interior of the range.
For increasing \(g\), use \(F_X(h(y))\); for decreasing \(g\), use \(1-F_X(h(y))\), since \(X\) is continuous.
Use the absolute value of the inverse derivative, and verify that the CDF has the correct endpoint limits and that the density integrates to 1.
Logarithmic Transformations
A logarithm requires a positive argument. For example, if \(X>0\), the transformation \(Y=\ln X\) is increasing, has range over all real numbers, and has inverse \(h(y)=e^y\). More generally, \(Y=\ln(1+cX)\), where \(c>0\) and \(X\ge 0\), has inverse \(h(y)=(e^y-1)/c\) and range \(y\ge 0\). In both cases, the density follows by multiplying the original density at the inverse point by the derivative of that inverse.
Worked Example: A Logarithmic Transformation of an Exponential Severity
In a hypothetical model, let \(X\) be a loss measured in thousands of dollars with exponential distribution of rate \(\lambda=1\) per thousand dollars. Thus \(f_X(x)=e^{-x}\) for \(x\ge 0\), and \(F_X(x)=1-e^{-x}\). Define \(Y=\ln(1+2X)\). Derive its CDF and density, and evaluate both at \(y=\ln 3\).
Method and support. The transformation is strictly increasing for \(x\ge0\), and \(Y\ge0\). Solving \(y=\ln(1+2x)\) gives the inverse \(h(y)=(e^y-1)/2\), with derivative \(h'(y)=e^y/2\). This inverse is continuously differentiable on the interior of the range, so the density formula applies.
For \(y<0\), \(F_Y(y)=0\). For \(y\ge0\), $$ F_Y(y)=F_X\left(\frac{e^y-1}{2}\right) =1-\exp\left(-\frac{e^y-1}{2}\right). $$
Differentiate on \(y>0\): $$ f_Y(y)=e^{-(e^y-1)/2}\frac{e^y}{2} =\frac{e^y}{2}\exp\left(-\frac{e^y-1}{2}\right). $$ The density is 0 for \(y<0\); its value at the single endpoint \(y=0\) does not affect probabilities.
At \(y=\ln 3\), \(e^y=3\), so the exponent is \(-(3-1)/2=-1\). Consequently, \(F_Y(\ln 3)=1-e^{-1}\approx0.6321\), and \(f_Y(\ln 3)=(3/2)e^{-1}\approx0.5518\). The CDF starts at 0 when \(y=0\) and tends to 1 as \(y\) tends to infinity, as required. This method avoids treating the logarithm itself as the inverse: the derivative used in the density is that of \(h\).
Power Transformations
For \(X>0\) and \(r>0\), \(Y=X^r\) is increasing, with inverse \(h(y)=y^{1/r}\). Its derivative on \(y>0\) is \((1/r)y^{1/r-1}\). If \(X\) has density \(f_X\), then $$ f_Y(y)=f_X\left(y^{1/r}\right)\frac{1}{r}y^{1/r-1},\qquad y>0, $$ provided the inverse satisfies the stated regularity condition. If instead \(r<0\), the power is decreasing on \(X>0\); the same absolute-derivative density formula applies, but the CDF must use the upper tail of \(X\).
Worked Example: Cubing a Uniform Severity
Let \(X\) be a hypothetical severity with uniform distribution on \((0,2)\), so \(f_X(x)=1/2\) on that interval and \(F_X(x)=x/2\) for \(0<x<2\). Define \(Y=X^3\). Find the support, CDF, and density, and check the density.
Method. The transformation is increasing on the support, and it maps \((0,2)\) to \((0,8)\). Its inverse is \(h(y)=y^{1/3}\), with derivative \(h'(y)=(1/3)y^{-2/3}\) for \(y>0\). The inverse is locally absolutely continuous on the interior of the transformed support.
For \(0<y<8\), translate \(Y\le y\) into \(X\le y^{1/3}\). Thus $$ F_Y(y)=F_X(y^{1/3})=\frac{y^{1/3}}{2}. $$ The CDF equals 0 for \(y\le0\) and 1 for \(y\ge8\). Differentiation gives $$ f_Y(y)=\frac{1}{2}\cdot\frac{1}{3}y^{-2/3} =\frac{1}{6y^{2/3}},\qquad 0<y<8. $$
For a check, integrate the density: $$ \int_0^8 \frac{1}{6y^{2/3}}\,dy =\left[\frac{1}{2}y^{1/3}\right]_0^8 =\frac{1}{2}(2)-0=1. $$ Also, \(F_Y(1)=1/2\) and \(f_Y(1)=1/6\). The density becomes unbounded as \(y\) approaches 0 from above, but its integral is finite; an unbounded density at an endpoint does not by itself invalidate the distribution.
Worked Example: A Decreasing Reciprocal-Power Transformation
A hypothetical loss variable \(X\) is uniform on \([1,3]\), with density \(1/2\) on that interval. Set \(Y=X^{-2}\). Find its CDF and density, then calculate \(F_Y(4/25)\) and \(f_Y(4/25)\).
Method and support. The transformation is strictly decreasing for \(x>0\). It maps \([1,3]\) to \([1/9,1]\), and its inverse is \(h(y)=y^{-1/2}\), with derivative \(h'(y)=-(1/2)y^{-3/2}\). The inverse is continuously differentiable in the interior of the range.
For \(1/9<y<1\), \(Y\le y\) is equivalent to \(X\ge y^{-1/2}\), not \(X\le y^{-1/2}\). Since \(X\) is continuous, $$ F_Y(y)=P(X\ge y^{-1/2}) =1-F_X(y^{-1/2}) =1-\frac{y^{-1/2}-1}{2} =\frac{3-y^{-1/2}}{2}. $$
The density is $$ f_Y(y)=f_X(y^{-1/2})|h'(y)| =\frac{1}{2}\cdot\frac{1}{2}y^{-3/2} =\frac{1}{4y^{3/2}},\qquad \frac{1}{9}<y<1. $$ At \(y=4/25\), \(y^{-1/2}=5/2\). Therefore \(F_Y(4/25)=(3-5/2)/2=1/4\), and \(f_Y(4/25)=1/[4(4/25)^{3/2}]=1/[4(8/125)]=125/32=3.90625\).
Check the total probability: $$ \int_{1/9}^{1}\frac{1}{4y^{3/2}}\,dy =\left[-\frac{1}{2\sqrt{y}}\right]_{1/9}^{1} =-\frac12+\frac32=1. $$ The CDF is 0 at the lower endpoint and 1 at the upper endpoint, providing a second check.
Common Mistake: Reversing the Wrong Step
For a decreasing transformation, the inverse relationship reverses the event inequality. In the reciprocal-power example, the incorrect calculation \(F_Y(y)=F_X(y^{-1/2})\) would give \(F_Y(4/25)=F_X(5/2)=3/4\). The correct event is \(X\ge5/2\), so its probability is \(1-F_X(5/2)=1/4\). The incorrect expression even decreases as \(y\) increases, so it cannot be a CDF.
Other checks help catch errors before they spread: verify that the CDF is nondecreasing, that it has limits 0 and 1 at the ends of the transformed support, and that the density is nonnegative and integrates to 1. Do not omit the absolute value in the density formula, and do not apply a single inverse formula across a transformation that is not one-to-one on the support.
Check Your Understanding
For each problem, identify the transformed support and derive the requested distribution. The final answers are included for checking.
- A hypothetical severity \(X\) is exponential with rate \(\lambda=2\), and \(Y=\ln X\). Give \(F_Y(y)\) and \(f_Y(y)\). (Answer: support \((-\infty,\infty)\); \(F_Y(y)=1-e^{-2e^y}\), \(f_Y(y)=2e^y e^{-2e^y}\).)
- Let \(X\) be uniform on \((0,4)\) and \(Y=\sqrt{X}\). Give the support, CDF, and density of \(Y\). (Answer: support \((0,2)\); \(F_Y(y)=y^2/4\) and \(f_Y(y)=y/2\) for \(0<y<2\).)
- Let \(X\) be uniform on \([2,6]\) and \(Y=1/X\). Derive the CDF and density on the transformed support. (Answer: support \([1/6,1/2]\); \(F_Y(y)=(6-1/y)/4\) and \(f_Y(y)=1/(4y^2)\) in the interior.)
- Let \(X\) be exponential with rate \(\lambda=1\), and set \(Y=X^2\). Derive its CDF and density, then check that the density integrates to 1. (Answer: support \([0,\infty)\); \(F_Y(y)=1-e^{-\sqrt y}\), \(f_Y(y)=e^{-\sqrt y}/(2\sqrt y)\) for \(y>0\); the density integral is 1.)
- Let \(X\) be uniform on \((1,e^2)\) and \(Y=\ln X\). Find the support, CDF, and density. (Answer: support \((0,2)\); \(F_Y(y)=(e^y-1)/(e^2-1)\), \(f_Y(y)=e^y/(e^2-1)\) for \(0<y<2\).)