Asymptotic Variance of Gamma Moment Estimates

Why the Joint Variability Matters

A gamma severity model is often summarized by its shape and scale parameters. When both are estimated by the method of moments, each estimate is calculated from the same two sample moments. Their estimation errors are therefore related. A variance calculation that reports only separate standard errors misses part of the uncertainty information, and can give an incorrect result when the estimates are used together.

This tutorial derives the asymptotic covariance matrix for the two estimates. The result applies to a sample of independent, identically distributed, fully observed gamma losses. It is an approximation for large samples, not an exact finite-sample covariance formula.

Definition: Use the gamma shape-scale parameterization $X\sim\operatorname{Gamma}(\alpha,\theta)$, with shape $\alpha>0$, scale $\theta>0$, and density $f_X(x)=x^{\alpha-1}e^{-x/\theta}/(\Gamma(\alpha)\theta^\alpha)$ for $x>0$. Its mean and variance are $E[X]=\alpha\theta$ and $\operatorname{Var}(X)=\alpha\theta^2$.

For a sample $X_1,\ldots,X_n$, define the first and second raw sample moments by $\bar X=n^{-1}\sum_{j=1}^n X_j$ and $M_2=n^{-1}\sum_{j=1}^n X_j^2$. The method-of-moments estimates equate these to the corresponding population raw moments. In particular, $M_2-\bar X^2$ is the sample variance using denominator $n$, not the unbiased sample variance using denominator $n-1$.

Result: The gamma method-of-moments estimates based on the first two raw moments are $\hat\alpha=\bar X^2/(M_2-\bar X^2)$ and $\hat\theta=(M_2-\bar X^2)/\bar X$, provided $\bar X>0$ and $M_2-\bar X^2>0$.

Deriving the Asymptotic Covariance Matrix

First consider the sample-moment vector $(\bar X,M_2)$. For an independent and identically distributed sample with finite fourth moment, its asymptotic covariance matrix is $1/n$ times the covariance matrix of $(X,X^2)$. A gamma distribution has finite moments of every positive integer order, so this condition holds.

The gamma raw moments satisfy $E[X^r]=\theta^r\Gamma(\alpha+r)/\Gamma(\alpha)$. Consequently, $E[X^2]=\alpha(\alpha+1)\theta^2$, $E[X^3]=\alpha(\alpha+1)(\alpha+2)\theta^3$, and $E[X^4]=\alpha(\alpha+1)(\alpha+2)(\alpha+3)\theta^4$. These yield the following covariance matrix for one observation:

\Omega= \begin{pmatrix} \operatorname{Var}(X) & \operatorname{Cov}(X,X^2)\\ \operatorname{Cov}(X,X^2) & \operatorname{Var}(X^2) \end{pmatrix} = \begin{pmatrix} \alpha\theta^2 & 2\alpha(\alpha+1)\theta^3\\ 2\alpha(\alpha+1)\theta^3 & 2\alpha(\alpha+1)(2\alpha+3)\theta^4 \end{pmatrix}.

For example, the off-diagonal entry is $E[X^3]-E[X]E[X^2]=2\alpha(\alpha+1)\theta^3$. The lower-right entry is $E[X^4]-(E[X^2])^2=2\alpha(\alpha+1)(2\alpha+3)\theta^4$. Thus the asymptotic covariance matrix of $(\bar X,M_2)$ is $\Omega/n$.

The estimates are functions of those two sample moments. Write $g_1(m_1,m_2)=m_1^2/(m_2-m_1^2)$ and $g_2(m_1,m_2)=(m_2-m_1^2)/m_1$. At the population moments $m_1=\alpha\theta$ and $m_2=\alpha(\alpha+1)\theta^2$, the Jacobian matrix of these functions is:

J= \begin{pmatrix} \frac{2(\alpha+1)}{\theta} & -\frac{1}{\theta^2}\\[4pt] -\frac{2\alpha+1}{\alpha} & \frac{1}{\alpha\theta} \end{pmatrix}.

The multivariate delta method gives the asymptotic covariance matrix as $J\Omega J^{\mathsf T}/n$. The Jacobian is important: it accounts for how variability in each sample moment affects each parameter estimate, including the shared effect that creates covariance between the estimates.

Find the sample-moment covariance.
Use the gamma raw moments through order four to form $\Omega$, the covariance matrix of $(X,X^2)$.

Differentiate the moment estimates.
Evaluate the Jacobian of $g_1$ and $g_2$ at the population first and second raw moments.

Transform the covariance.
Multiply $J\Omega J^{\mathsf T}$ and divide by $n$ for the large-sample covariance of the estimates.

For clarity, the shape variance entry simplifies to $2\alpha(\alpha+1)$: its three contributions from the matrix multiplication are $4\alpha(\alpha+1)^2$, $-8\alpha(\alpha+1)^2$, and $2\alpha(\alpha+1)(2\alpha+3)$. For the off-diagonal entry, the four contributions simplify to $-2(\alpha+1)(2\alpha+1)\theta$, $4(\alpha+1)^2\theta$, $2(\alpha+1)(2\alpha+1)\theta$, and $-2(\alpha+1)(2\alpha+3)\theta$, whose sum is $-2(\alpha+1)\theta$. The remaining entry is $\theta^2(2\alpha+3)/\alpha$.

Theorem: For independent, identically distributed observations from a gamma distribution with fixed $\alpha>0$ and $\theta>0$, as $n$ grows, $\sqrt n\big((\hat\alpha,\hat\theta)^{\mathsf T}-(\alpha,\theta)^{\mathsf T}\big)$ is asymptotically normal with mean vector zero and covariance matrix $$ \begin{pmatrix} 2\alpha(\alpha+1) & -2(\alpha+1)\theta\\ -2(\alpha+1)\theta & \frac{\theta^2(2\alpha+3)}{\alpha} \end{pmatrix}. $$ Equivalently, the large-sample covariance matrix of $(\hat\alpha,\hat\theta)$ is approximately $$ \operatorname{Cov} \begin{pmatrix} \hat\alpha\\ \hat\theta \end{pmatrix} \approx \frac{1}{n} \begin{pmatrix} 2\alpha(\alpha+1) & -2(\alpha+1)\theta\\ -2(\alpha+1)\theta & \frac{\theta^2(2\alpha+3)}{\alpha} \end{pmatrix}. $$

In practice, the parameters in this approximation are unknown. Substitute $\hat\alpha$ and $\hat\theta$ into the matrix to estimate the covariance. The square roots of its diagonal entries are the estimated asymptotic standard errors. The off-diagonal entry estimates the covariance; it is not itself a standard error. The estimated correlation is that entry divided by the product of the two estimated standard errors.

Worked Applications

Worked Example: Theoretical Precision at a Specified Gamma Model

A hypothetical portfolio of independent, fully observed losses is modeled by a gamma distribution with shape $\alpha=2$ and scale $\theta=3$ dollars. For a sample of $n=100$ losses, calculate the asymptotic covariance matrix of the two method-of-moments estimates and their estimated standard errors if these parameter values are used for planning.

Method. Substitute the specified parameters and sample size into the joint asymptotic covariance formula.

For the shape variance, the numerator is $2\alpha(\alpha+1)=2(2)(3)=12$, so the variance is $12/100=0.12$.

For the covariance, the numerator is $-2(\alpha+1)\theta=-2(3)(3)=-18$, so the covariance is $-18/100=-0.18$.

For the scale variance, the numerator is $\theta^2(2\alpha+3)/\alpha=3^2(7)/2=31.5$, so the variance is $31.5/100=0.315$.

The resulting matrix, with scale measured in dollars, is $$ \begin{pmatrix} 0.12 & -0.18\\ -0.18 & 0.315 \end{pmatrix}. $$ The standard errors are $\sqrt{0.12}=0.3464$ for shape and $\sqrt{0.315}=0.5612$ dollars for scale, rounded to four decimal places. The determinant is $0.12(0.315)-(-0.18)^2=0.0054>0$, a useful positive-definiteness check.

Common mistake avoided: using $2\alpha(\alpha+1)/n$ as the variance of both estimates. The scale variance has its own parameter-dependent expression and the scale units are dollars.

Worked Example: Plugging in Observed Sample Moments

A hypothetical claim study has $n=50$ independent, fully observed losses. The recorded totals are $\sum X_j=300$ dollars and $\sum X_j^2=2{,}400$ dollars squared. Estimate the gamma parameters by moments and calculate the plug-in asymptotic covariance matrix, standard errors, and correlation.

Method. First calculate the two raw sample moments, then estimate the parameters, and finally substitute those estimates in the joint covariance formula.

The first raw moment is $\bar X=300/50=6$ dollars. The second is $M_2=2{,}400/50=48$ dollars squared. The denominator-$n$ sample variance is $M_2-\bar X^2=48-6^2=12$ dollars squared.

Therefore, $\hat\alpha=6^2/12=3$, and $\hat\theta=12/6=2$ dollars. Substitution gives shape variance $2(3)(4)/50=24/50=0.48$, covariance $-2(4)(2)/50=-16/50=-0.32$ dollars, and scale variance $2^2(2(3)+3)/(3\cdot 50)=4(9)/150=0.24$ dollars squared.

The estimated covariance matrix is $$ \begin{pmatrix} 0.48 & -0.32\\ -0.32 & 0.24 \end{pmatrix}. $$ The standard errors are $\sqrt{0.48}=0.6928$ for shape and $\sqrt{0.24}=0.4899$ dollars for scale. The estimated correlation is $-0.32/(\sqrt{0.48}\sqrt{0.24})=-0.9428$, rounded to four decimals. The matrix determinant is $0.48(0.24)-0.32^2=0.0128>0$.

Common mistake avoided: replacing $M_2-\bar X^2=12$ by an unbiased sample variance of $12(50/49)$. The moment equations use the denominator-$n$ variance; using the other convention changes the estimates.

Worked Example: Checking the Effect of Loss Units

For another hypothetical fully observed gamma sample, suppose the fitted parameters are $\hat\alpha=1.5$, $\hat\theta=4$ thousands of dollars, and $n=120$. Calculate the plug-in covariance matrix and the standard errors, keeping the scale in thousands of dollars.

Method. Evaluate each entry separately, retaining the stated scale unit throughout.

The shape variance is $2(1.5)(2.5)/120=7.5/120=0.0625$. The covariance is $-2(2.5)(4)/120=-20/120=-0.1667$, in thousands of dollars. The scale variance is $4^2(2(1.5)+3)/(1.5\cdot120)=16(6)/180=96/180=0.5333$, in thousands of dollars squared.

Thus, to four decimal places, the plug-in matrix is $$ \begin{pmatrix} 0.0625 & -0.1667\\ -0.1667 & 0.5333 \end{pmatrix}. $$ The estimated standard errors are $\sqrt{0.0625}=0.2500$ for shape and $\sqrt{0.5333}=0.7303$ thousands of dollars for scale. Using unrounded entries, the correlation is approximately $-0.9129$. As a check, the determinant from the unrounded entries is $0.0625(0.533333\ldots)-(1/6)^2=0.005556\ldots>0$.

Common mistake avoided: changing the scale from thousands of dollars to dollars in only some entries. If scale were expressed in dollars, it would be 4,000, and both the scale variance and the shape-scale covariance would change units and numerical size accordingly.

Interpretation and Exam Traps

The negative covariance is a consequence of fitting both parameters to the same moments. A sample that makes the fitted mean relatively high, while its dispersion remains comparatively modest, can produce a higher fitted shape and a lower fitted scale. The covariance quantifies this joint movement; it does not mean either estimate is biased downward.

The formula is specific to uncensored, untruncated, independent observations from the stated gamma model and to estimates based on the first two raw sample moments. If observations are dependent, incomplete, or sampled under a truncation mechanism, the moment covariance and possibly the estimating equations must be changed. The approximation is also intended for sufficiently large $n$; it need not describe the exact distribution of estimates in a small sample.

Dropping the covariance. The off-diagonal term is generally nonzero. It must be retained in calculations involving both estimates.
Using the wrong sample variance convention. The raw-moment estimate uses $M_2-\bar X^2$, with denominator $n$.
Confusing scale and rate. Here $\theta$ is the scale in the density; it is not the reciprocal rate.
Ignoring units. The shape variance is dimensionless, the shape-scale covariance has scale units, and the scale variance has squared scale units.
Treating an asymptotic result as exact. The matrix estimates large-sample variability, not the exact finite-sample covariance.

Check Your Understanding

Assume independent, fully observed gamma losses and use the first two raw sample moments unless a question asks only for interpretation.

For $\alpha=2$, $\theta=5$ dollars, and $n=200$, calculate the asymptotic covariance matrix of the two moment estimates. (Answer: $\begin{pmatrix}0.0600&-0.1500\\-0.1500&0.4375\end{pmatrix}$, with the off-diagonal entry in dollars and the scale variance in dollars squared.)
A sample has $n=100$, $\bar X=4$ dollars, and $M_2=24$ dollars squared. Find the moment estimates and the plug-in asymptotic covariance matrix. (Answer: $\hat\alpha=2$, $\hat\theta=2$ dollars; matrix $\begin{pmatrix}0.1200&-0.1200\\-0.1200&0.1400\end{pmatrix}$.)
For fitted values $\hat\alpha=4$, $\hat\theta=1$ thousand dollars, and $n=80$, calculate the two standard errors and covariance. (Answer: shape standard error $0.7071$; scale standard error $0.1854$ thousand dollars; covariance $-0.1250$ thousand dollars.)
Why is it generally incorrect to set the shape-scale covariance to zero when using these estimates jointly? (Answer: both estimates depend on the same two sample moments; the delta-method covariance is $-2(\alpha+1)\theta/n$, which is nonzero for positive shape and scale.)
For $\hat\alpha=1$, $\hat\theta=3$ dollars, and $n=90$, calculate the plug-in correlation between the estimates. (Answer: approximately $-0.8944$.)

What You'll Learn