Gamma MLE With Left-Truncated Losses

Why Truncation Changes the Gamma Likelihood

Suppose a hypothetical study records a loss only when it exceeds a known reporting threshold. A loss below that threshold does not appear in the data at all. The observed losses are therefore not an ordinary sample from the full loss distribution: each recorded value comes from the distribution conditional on exceeding the threshold. Omitting that conditioning changes the likelihood and can distort both estimated parameters.

This tutorial considers exact observations from a gamma severity model, with a common known threshold. The observations are independent and identically distributed conditional on exceeding that threshold. They are not censored at the threshold: for each recorded loss, its exact value is known.

Definition: Use the gamma shape-scale parameterization $X\sim\operatorname{Gamma}(\alpha,\theta)$, where $\alpha>0$ is the shape and $\theta>0$ is the scale. Its density for $x>0$ is $$ f_X(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\theta^\alpha\Gamma(\alpha)}, $$ where $\Gamma(\alpha)=\int_0^\infty t^{\alpha-1}e^{-t}\,dt$. For a known left-truncation threshold $c\geq 0$, define the upper incomplete gamma function by $\Gamma(\alpha,z)=\int_z^\infty t^{\alpha-1}e^{-t}\,dt$. The probability of being observed is $P(X>c)=\Gamma(\alpha,c/\theta)/\Gamma(\alpha)$.

The scale parameter has the same units as $X$; the shape parameter is dimensionless. For $c=0$, every positive loss is observable and $\Gamma(\alpha,0)=\Gamma(\alpha)$, so the conditional density reduces to the ordinary gamma density. For $c>0$, the normalizing probability depends on both parameters and must be retained.

Conditional Density and Likelihood

Let $x_1,\ldots,x_n$ be exact observed losses, all greater than the same known threshold $c$. Under the assumptions above, each observation has the conditional density $f_X(x\mid X>c)=f_X(x)/P(X>c)$ for $x>c$. Substituting the gamma density and survival probability gives the truncated density.

f_X(x\mid X>c) =\frac{x^{\alpha-1}e^{-x/\theta}} {\theta^\alpha\Gamma(\alpha,c/\theta)},\qquad x>c.

The ordinary gamma normalizer $\Gamma(\alpha)$ cancels, but the upper incomplete gamma function remains. Conditional independence then gives the likelihood for the observed values, conditional on the fact that they were observed above $c$.

L(\alpha,\theta) =\prod_{j=1}^n \frac{x_j^{\alpha-1}e^{-x_j/\theta}} {\theta^\alpha\Gamma(\alpha,c/\theta)}.

This is the appropriate likelihood when the analysis conditions on the observed sample size and the records are exact losses selected by the common threshold. If the sampling design also models how many losses were eligible or observed, that count may supply additional information; it is not part of this conditional likelihood.

Worked Example: A Two-Loss Likelihood With a Known Threshold

A hypothetical portfolio records exact losses $x_1=3$ and $x_2=5$, and records a loss only if it exceeds $c=2$. Evaluate the conditional likelihood at the specified model $X\sim\operatorname{Gamma}(2,1)$. The observations are independent, and the threshold is common and fixed.

Step 1: Find the normalizer. Here $\alpha=2$, $\theta=1$, and $c/\theta=2$. Since $\Gamma(2,z)=e^{-z}(z+1)$, $$ \Gamma(2,2)=3e^{-2}. $$ The conditional density is therefore $f_X(x\mid X>2)=xe^{2-x}/3$ for $x>2$.

Step 2: Evaluate each contribution. At $x_1=3$, the contribution is $3e^{2-3}/3=e^{-1}$. At $x_2=5$, it is $5e^{2-5}/3=5e^{-3}/3$.

Step 3: Multiply. $$ L(2,1)=e^{-1}\frac{5e^{-3}}{3} =\frac{5}{3}e^{-4} \approx 0.030526. $$ The likelihood is a joint density value, not a probability assigned to the two exact points. The calculation is checked by using the two conditional density contributions separately before multiplying. The common mistake avoided here is to use the untruncated gamma density for either observation.

Score Equations for Shape and Scale

The log-likelihood is more convenient than the likelihood for differentiation and numerical optimization. Write $z=c/\theta$. Taking logarithms of the product gives

\ell(\alpha,\theta) =(\alpha-1)\sum_{j=1}^n\log x_j -\frac{1}{\theta}\sum_{j=1}^n x_j -n\alpha\log\theta -n\log\Gamma(\alpha,z).

The shape score involves a derivative of the upper incomplete gamma function with respect to its first argument. Define $\partial_\alpha\Gamma(\alpha,z)=\int_z^\infty t^{\alpha-1}e^{-t}\log(t)\,dt$. The scale score also accounts for the fact that $z=c/\theta$ changes with $\theta$.

Score equations: An interior maximum likelihood estimate $(\hat\alpha,\hat\theta)$ must solve both score equations below, with $z=c/\theta$: $$ \frac{\partial\ell}{\partial\alpha} =\sum_{j=1}^n\log x_j-n\log\theta -n\frac{\partial_\alpha\Gamma(\alpha,z)}{\Gamma(\alpha,z)}=0, $$ $$ \frac{\partial\ell}{\partial\theta} =\frac{\sum_{j=1}^n x_j}{\theta^2} -\frac{n\alpha}{\theta} -\frac{nc}{\theta^2} \frac{z^{\alpha-1}e^{-z}}{\Gamma(\alpha,z)}=0. $$

These equations generally have no closed-form joint solution. In particular, replacing $\Gamma(\alpha,c/\theta)$ by $\Gamma(\alpha)$ would remove terms that depend on the threshold and would solve a different estimation problem.

Why the Incomplete-Gamma Normalizer Appears

The normalizer follows directly from conditioning. For any $x>c$, the probability density among observed losses is the original density at $x$, divided by the probability that a loss is observed. The latter probability is the integral of the original density from $c$ to infinity.

Integrate the gamma density over observable values.
$P(X>c)=\int_c^\infty f_X(x)\,dx$. With $t=x/\theta$, this becomes $\Gamma(\alpha,c/\theta)/\Gamma(\alpha)$.

Divide the original density by that probability.
$f_X(x\mid X>c)=f_X(x)/P(X>c)$ for $x>c$. The complete gamma factors cancel, leaving $\theta^\alpha\Gamma(\alpha,c/\theta)$ in the denominator.

Multiply the conditional contributions.
Conditional independence gives one such density contribution for each exact observed loss. The threshold appears in every contribution when it is common to all records.

This argument also checks that the conditional density integrates to one. Substitution into the conditional density gives $$ \int_c^\infty f_X(x\mid X>c)\,dx =\frac{\int_c^\infty x^{\alpha-1}e^{-x/\theta}\,dx} {\theta^\alpha\Gamma(\alpha,c/\theta)} =1, $$ where the substitution $t=x/\theta$ makes the numerator $\theta^\alpha\Gamma(\alpha,c/\theta)$. Thus the incomplete-gamma term is not an optional adjustment; it is required for a valid conditional density.

Numerical Estimation in Practice

A practical approach is to maximize the log-likelihood numerically over $\alpha>0$ and $\theta>0$. One can solve the two score equations or use a numerical optimizer on $\ell$. Because the parameters must be positive, a convenient implementation optimizes over $\eta=\log\alpha$ and $\xi=\log\theta$, then sets $\alpha=e^\eta$ and $\theta=e^\xi$. The incomplete gamma function and its shape derivative are evaluated numerically.

A reliable calculation should use the log-likelihood rather than multiplying many small density values. It should also check that the reported estimates are positive, that the score equations are close to zero (for an interior solution), and that the log-likelihood is not larger at other plausible starting values. Numerical methods can converge to a local solution or encounter a boundary; a solver's output alone does not establish that a global maximum has been found.

Worked Example: Estimating the Scale in an Exponential Special Case

A hypothetical study observes exact losses $3,5,8$, each exceeding the common threshold $c=2$. Assume the gamma shape is fixed at $\alpha=1$, so $X$ is exponential with scale $\theta$. Estimate $\theta$ by maximum likelihood using the conditional likelihood.

Step 1: Write the conditional density. For $\alpha=1$, $\Gamma(1,z)=e^{-z}$, so $$ f_X(x\mid X>2)=\frac{e^{-x/\theta}}{\theta e^{-2/\theta}} =\frac{1}{\theta}e^{-(x-2)/\theta},\qquad x>2. $$ The threshold shifts the exponential density to the excess $x-2$.

Step 2: Form the log-likelihood. The three excesses are $3-2=1$, $5-2=3$, and $8-2=6$, with sum $1+3+6=10$. Therefore $$ \ell(\theta)=-3\log\theta-\frac{10}{\theta}. $$

Step 3: Differentiate and solve. The score is $-3/\theta+10/\theta^2$. Setting it to zero gives $-3\theta+10=0$, so $\hat\theta=10/3\approx3.333$, in the same units as the losses. The second derivative is $3/\theta^2-20/\theta^3$, which at $10/3$ is negative, confirming a local maximum.

Step 4: Check the fitted likelihood. The likelihood at this estimate is $$ \left(\frac{3}{10}\right)^3e^{-3} =0.027e^{-3} \approx0.001344. $$ The result is checked against the total excess of 10 across 3 observations. The common mistake avoided is estimating the scale from the sum of the raw losses, which would ignore that every recorded loss had to exceed 2.

Worked Example: Evaluating a General-Shape Likelihood Contribution

A hypothetical sample has exact losses $3$ and $4$, each observed only when it exceeds $c=1$. Evaluate their joint conditional likelihood at $X\sim\operatorname{Gamma}(3,2)$, where the parameters are shape $\alpha=3$ and scale $\theta=2$.

Step 1: Compute the threshold argument. Here $z=c/\theta=1/2$. For integer shape 3, $$ \Gamma(3,z)=2e^{-z}\left(1+z+\frac{z^2}{2}\right), \qquad \Gamma(3,1/2)=\frac{13}{4}e^{-1/2}. $$ Thus the denominator shared by each conditional density is $\theta^\alpha\Gamma(\alpha,z)=8(13e^{-1/2}/4)=26e^{-1/2}$.

Step 2: Evaluate the density at each exact loss. At $x=3$, the density is $9e^{-3/2}/(26e^{-1/2})=(9/26)e^{-1}$. At $x=4$, it is $16e^{-2}/(26e^{-1/2})=(8/13)e^{-3/2}$.

Step 3: Multiply and check. $$ L(3,2)=\frac{9}{26}e^{-1}\cdot\frac{8}{13}e^{-3/2} =\frac{36}{169}e^{-5/2} \approx0.017486. $$ The product is based on two conditional densities, each with the same incomplete-gamma normalizer. The common mistake avoided is to use $\Gamma(3)$ in place of $\Gamma(3,1/2)$.

Common Traps and Interpretation

Ignoring truncation. The ordinary gamma density applies only when the sample is not selected by $X>c$, or when $c=0$. For a positive threshold, use the conditional density.
Using a censoring contribution. A truncated loss is absent unless it exceeds the threshold; its value is not reported as “at least $c$.” An exact observed loss contributes a density, not a tail probability.
Using the wrong incomplete-gamma argument. The threshold enters as $c/\theta$, not $c$, because the integration variable is scaled by $\theta$.
Forgetting the scale factor. The conditional denominator is $\theta^\alpha\Gamma(\alpha,c/\theta)$. The incomplete gamma function alone is not the full normalizer in the $x$-scale.
Assuming a closed form for both estimates. Even though the exponential special case has a simple scale estimate when shape is fixed, the general shape-scale equations involve incomplete-gamma functions and normally require numerical solution.

Key takeaway: For independent exact gamma losses selected by a common known threshold $c$, each likelihood contribution is $x^{\alpha-1}e^{-x/\theta}/\{\theta^\alpha\Gamma(\alpha,c/\theta)\}$. The threshold-dependent incomplete-gamma normalizer must be included when estimating either parameter. This treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, assume exact independent observations and a common fixed threshold unless stated otherwise.

A loss $x=3$ is observed only when $X>1$, with $X\sim\operatorname{Gamma}(2,1)$. Find its conditional density contribution. (Answer: $\frac{3}{2}e^{-2}\approx0.2030$.)
A loss $x=7$ is observed only when $X>3$, with $X\sim\operatorname{Gamma}(1,4)$. Find its conditional density contribution. (Answer: $\frac14e^{-1}\approx0.0920$.)
With fixed shape $\alpha=1$, exact losses $4$ and $7$ are recorded above $c=2$. Find the maximum likelihood estimate of the scale. (Answer: $\hat\theta=((4-2)+(7-2))/2=3.5$.)
For $X\sim\operatorname{Gamma}(2,3)$, find the conditional density at $x=6$ given $X>3$. (Answer: $\frac13e^{-1}\approx0.1226$.)
For exact observations $x_1,\ldots,x_n>c$ from $\operatorname{Gamma}(\alpha,\theta)$, state the conditional likelihood and identify the argument of its upper incomplete-gamma normalizer. (Answer: $L=\prod_{j=1}^n x_j^{\alpha-1}e^{-x_j/\theta}/\{\theta^\alpha\Gamma(\alpha,c/\theta)\}$; the argument is $c/\theta$.)

What You'll Learn