Gamma Shapes And Density Profiles

Why Gamma Density Profiles Matter

A gamma severity model can describe losses with very different density profiles. Depending on its shape parameter, its density may become unbounded near zero, begin at a finite positive value, or rise to a peak at a positive loss amount before declining. Recognizing which profile applies helps interpret a fitted severity model and prevents errors when identifying its mode.

The profile depends on both the shape and scale parameters. The shape determines the qualitative behavior at the boundary and whether there is an interior mode; the scale stretches the loss axis. The gamma family has a right-skewed loss profile for every positive shape, although the amount of skew changes with shape.

Definition: Let $X$ be a gamma severity with shape $\alpha>0$ and scale $\theta>0$, written $X\sim\operatorname{Gamma}(\alpha,\theta)$. Its density for $x>0$ is $$ f_X(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha}, $$ where $\Gamma(\alpha)=\int_0^\infty t^{\alpha-1}e^{-t}\,dt$. The support is $x>0$, and the density at zero is understood through its right-hand limit. Here $\theta$ is a scale, not a rate: the equivalent rate parameter is $1/\theta$.

For this parameterization, the gamma moments are $E[X]=\alpha\theta$ and $\operatorname{Var}(X)=\alpha\theta^2$. Its coefficient of skewness is $2/\sqrt{\alpha}$. These results show that changing $\theta$ rescales the mean and standard deviation together, while changing $\alpha$ affects both the profile and relative skewness.

Boundary Behavior and the Mode

The behavior near zero is governed by $x^{\alpha-1}$, since $e^{-x/\theta}$ tends to 1 as $x$ approaches zero. The mode can be found by differentiating the log density: this avoids carrying the normalizing constant through the derivative.

\frac{d}{dx}\log f_X(x)=\frac{\alpha-1}{x}-\frac{1}{\theta}.

If $0<\alpha<1$, both terms make this derivative negative for $x>0$. The density therefore decreases throughout its support and tends to infinity as $x$ approaches zero. This infinite boundary value does not mean there is a point mass at zero: the density is still integrable near zero. In fact, the integral of the leading boundary term is finite because $\int_0^\varepsilon x^{\alpha-1}\,dx=\varepsilon^\alpha/\alpha$.

If $\alpha=1$, the density is exponential, $f_X(x)=e^{-x/\theta}/\theta$. It has finite right-hand limit $1/\theta$ at zero and decreases thereafter. If $\alpha>1$, the density tends to zero at the boundary. Setting the log derivative to zero gives a positive mode; the derivative is positive below that point and negative above it.

Theorem: For $X\sim\operatorname{Gamma}(\alpha,\theta)$, the density profile is as follows:

If $0<\alpha<1$, the density decreases on $x>0$ and diverges as $x\downarrow0$; its mode is at the boundary in the limiting sense.
If $\alpha=1$, the density decreases from the finite boundary value $1/\theta$; its mode is at zero.
If $\alpha>1$, the density is zero at the boundary, increases to the interior mode $x=(\alpha-1)\theta$, and then decreases.

Why the Interior Mode Formula Holds

For $\alpha>1$, the density is positive and differentiable for $x>0$, so its stationary points can be found from the derivative of its logarithm. Solving the log-derivative equation gives one candidate, and the sign of that derivative establishes that it is a maximum rather than a minimum.

Differentiate the log density.
The terms constant in $x$ disappear, leaving $(\alpha-1)/x-1/\theta$.

Find the stationary point.
Set the derivative equal to zero: $(\alpha-1)/x=1/\theta$, so $x=(\alpha-1)\theta$.

Check the derivative's sign.
For $0<x<(\alpha-1)\theta$, the derivative is positive; for $x>(\alpha-1)\theta$, it is negative. Thus the density rises before this point and falls after it.

\operatorname{mode}(X)= \begin{cases} 0 & 0<\alpha\leq 1,\\ (\alpha-1)\theta & \alpha>1. \end{cases}

For $0<\alpha<1$, the mode notation refers to the boundary location of the unbounded density, not to a finite density value at zero. For $\alpha=1$, the density has a finite maximum at zero. For $\alpha>1$, the mode is strictly positive. These distinctions matter when a question asks for both the mode and the density's boundary behavior.

Worked Profiles

Worked Example: A Gamma Density That Diverges at Zero

A hypothetical small-equipment insurer models claim severity by $X\sim\operatorname{Gamma}(0.5,2)$, with losses measured in thousands of dollars. Identify the density profile, calculate its mean, variance, and skewness, and evaluate its density at $x=0.5$ and $x=2$. State the parameterization assumption and the method used.

Model and method. The first parameter is shape $\alpha=0.5$ and the second is scale $\theta=2$. Use the gamma density and moment formulas, then examine the boundary limit and log derivative. The common mistake avoided here is assuming that a density unbounded at zero must have a probability mass at zero.

Boundary and mode. Since $\alpha-1=-0.5$, the density contains $x^{-0.5}$ near zero, so $f_X(x)$ tends to infinity as $x\downarrow0$. Also, $$ \frac{d}{dx}\log f_X(x)=\frac{-0.5}{x}-\frac12<0 \qquad (x>0). $$ Thus the density decreases over its support, with its mode at the boundary in the limiting sense.

Moments and skewness. The mean is $0.5(2)=1$ thousand dollars, and the variance is $0.5(2^2)=2$ squared thousand dollars. The coefficient of skewness is $2/\sqrt{0.5}=2.8284$ to four decimals, confirming substantial right skew.

Density values. Since $\Gamma(0.5)=\sqrt{\pi}$, the normalizing denominator is $\sqrt{\pi}\sqrt{2}=\sqrt{2\pi}$. $$ f_X(0.5)=\frac{(0.5)^{-0.5}e^{-0.25}}{\sqrt{2\pi}} =\frac{1.4142(0.7788)}{2.5066} \approx 0.4394, $$ $$ f_X(2)=\frac{2^{-0.5}e^{-1}}{\sqrt{2\pi}} =\frac{0.7071(0.3679)}{2.5066} \approx 0.1038. $$ The density at $0.5$ is larger than at $2$, consistent with the decreasing profile. Values are rounded to four decimals.

Worked Example: The Boundary Profile at Shape One

A hypothetical travel insurer uses an exponential severity model, which is the gamma model $X\sim\operatorname{Gamma}(1,3)$, with losses in hundreds of dollars. Find its density at zero and at $x=3$, its mode, mean, variance, and coefficient of skewness. Explain why the mode is not an interior peak.

Model and method. The shape is $\alpha=1$ and the scale is $\theta=3$. Substitute into the gamma density and moment formulas. The key trap is treating the formula $(\alpha-1)\theta$ as an interior mode even when $\alpha=1$; the positive-mode formula applies only when $\alpha>1$.

Because $\Gamma(1)=1$, the density is $$ f_X(x)=\frac{e^{-x/3}}{3}, \qquad x>0. $$ Its right-hand limit at zero is $1/3$, and $f_X(3)=e^{-1}/3\approx0.1226$. The derivative of its log density is $-1/3$, which is negative for every $x>0$. Therefore the density decreases from its finite boundary maximum; there is no interior peak.

The mean is $1(3)=3$ hundreds of dollars, the variance is $1(3^2)=9$ squared hundreds of dollars, and the coefficient of skewness is $2/\sqrt{1}=2$. The density value at $x=3$ is below $1/3$, as required by the decreasing profile.

Worked Example: Locating an Interior Gamma Mode

A hypothetical property insurer models a claim amount in thousands of dollars by $X\sim\operatorname{Gamma}(4,2)$. Find the boundary limit, mode, density at the mode, mean, variance, and coefficient of skewness. Check that the calculated density profile agrees with the result.

Model and method. Use shape $\alpha=4$ and scale $\theta=2$. Apply the log-derivative mode calculation, then substitute the mode into the density. The mistake avoided is using $\alpha\theta$, the mean, as if it were the mode.

Since $\alpha>1$, the density tends to zero as $x\downarrow0$, and its mode is $$ (4-1)(2)=6 \text{ thousand dollars}. $$ The derivative of the log density is $3/x-1/2$; it is positive for $0<x<6$ and negative for $x>6$, verifying the maximum.

The density at the mode is $$ f_X(6)=\frac{6^3e^{-6/2}}{\Gamma(4)2^4} =\frac{216e^{-3}}{6(16)} =\frac{216(0.049787)}{96} \approx 0.1120 $$ per thousand dollars, rounded to four decimals. The mean is $4(2)=8$ thousand dollars, the variance is $4(2^2)=16$ squared thousand dollars, and the coefficient of skewness is $2/\sqrt{4}=1$. In particular, the mean is 8 and the mode is 6; they are not interchangeable.

Worked Example: Comparing Profiles at a Fixed Mean

Three hypothetical portfolios are each assigned a gamma severity model with mean 12 thousand dollars. Their shape parameters are $0.5$, $1$, and $4$. For each model, find the scale, boundary behavior, mode, and coefficient of skewness. Which profile has the greatest relative skewness?

Model and method. Assume all three models use the shape-scale convention and have the same mean. From $E[X]=\alpha\theta=12$, solve $\theta=12/\alpha$. The common trap is holding the scale fixed while comparing shapes, which would not keep the mean fixed.

For $\alpha=0.5$, the scale is $12/0.5=24$. The density diverges at zero and decreases from the boundary; its boundary mode is zero in the limiting sense. Its skewness is $2/\sqrt{0.5}=2.8284$.

For $\alpha=1$, the scale is $12/1=12$. The density has finite boundary value $1/12$, decreases from zero, and has mode zero. Its skewness is $2$.

For $\alpha=4$, the scale is $12/4=3$. The density tends to zero at the boundary and has interior mode $(4-1)(3)=9$. Its skewness is $2/\sqrt{4}=1$.

All three means are 12 thousand dollars, but their profiles differ: the shape-$0.5$ model has the greatest relative skewness and a boundary divergence, while the shape-4 model has a positive mode. This comparison confirms that a fixed mean alone does not determine the density profile.

Right Skew and Interpretation

For every $\alpha>0$, the coefficient of skewness $2/\sqrt{\alpha}$ is positive. The gamma density therefore has a longer right side than left side in the sense captured by its positive standardized third moment. For $\alpha<1$, the density is largest near zero and declines, yet its positive skewness still reflects the extended right tail. For $\alpha>1$, the peak is positive, but the density also declines to the right of that peak.

The mode, mean, and scale answer different questions. The mode locates the density's maximum, the mean describes its probability-weighted average, and the scale controls the size of losses on the horizontal axis. For $\alpha>1$, the mean $\alpha\theta$ is greater than the mode $(\alpha-1)\theta$ by exactly $\theta$. For $\alpha\leq1$, the mode lies at the boundary while the mean remains positive.

Common Mistakes and Exam Checks

Applying the interior-mode formula outside its range. $(\alpha-1)\theta$ is an interior mode only for $\alpha>1$. At $\alpha=1$, the density has a finite boundary maximum; below one, it diverges at the boundary.
Calling an infinite density a point mass. A density value describes local probability per unit of loss, not probability at a single value. A continuous gamma variable has no atom at zero; its probability on an interval is found by integrating the density.
Confusing scale and rate. With scale $\theta$, the exponential factor is $e^{-x/\theta}$, and the interior mode is $(\alpha-1)\theta$. If a model instead supplies rate $\lambda=1/\theta$, the mode is $(\alpha-1)/\lambda$.
Confusing mean with mode. The mean $\alpha\theta$ is not the location of the density maximum. For shape above one, the mode is smaller by $\theta$.
Comparing shapes without checking what is held fixed. If the mean must remain constant, the scale must change with shape according to $\theta=E[X]/\alpha$.

Key takeaway: First compare the shape parameter with one. Below one, the gamma density decreases from an infinite boundary limit; at one, it decreases from a finite boundary value; above one, it rises from zero to the interior mode $(\alpha-1)\theta$ and then falls. In every case, the gamma severity has positive skewness.

Check Your Understanding

For each model, identify the profile before calculating. Answers are included so you can check the boundary behavior, mode, and parameter use.

A hypothetical repair-cost severity follows $\operatorname{Gamma}(2,5)$, with costs in hundreds of dollars. Find its mean, boundary behavior, and mode. (Answer: mean $10$ hundreds of dollars; density tends to zero at the boundary; mode $5$ hundreds of dollars.)
A model is $X\sim\operatorname{Gamma}(0.75,4)$. Find its mean, variance, boundary behavior, and coefficient of skewness to four decimals. (Answer: mean $3$, variance $12$, density diverges at zero and decreases, skewness $2.3094$.)
A hypothetical equipment-loss model has shape $3$ and scale $2$. Find its mode and density at the mode. (Answer: mode $4$; $f_X(4)=4^2e^{-2}/(\Gamma(3)2^3)=e^{-2}\approx0.1353$.)
A gamma severity has mean $20$ and variance $200$, with amounts measured in the same monetary units. Determine its shape, scale, mode, and skewness. (Answer: shape $2$, scale $10$, mode $10$, skewness $1.4142$.)
A gamma model has shape $1.2$ and scale $5$. Find the boundary limit, mode, and mean. Explain why its density does not diverge at zero. (Answer: density tends to zero; mode $1$; mean $6$; the exponent $\alpha-1=0.2>0$, so the density's power of $x$ tends to zero at the boundary.)

What You'll Learn