Moment

Study guides: Exam P-1, Exam FAM, Exam ASTAM, Exam 7

The n-th Moment of a Random Variable X is the Expected Value of its n-th power, E[X^n] (a raw moment); the n-th central moment E[(X - \mu)^n] measures the same thing about the mean \mu. The first four summarise a distribution's location, spread, asymmetry and tail weight.

\mu_n' = E[X^n] = \int_{-\infty}^{\infty} x^n f(x)\,dx

\mu_n = E\big[(X - \mu)^n\big]

• The mean is \mu = \mu_1'. For a discrete variable the integral becomes \sum_x x^n\,p(x).
• The second central moment is the Variance, \mu_2 = E[X^2] - \mu^2. The third gives the skewness \gamma_1 = \mu_3/\sigma^3, computed from raw moments as \mu_3 = E[X^3] - 3\mu E[X^2] + 2\mu^3. The fourth gives the kurtosis \mu_4/\sigma^4, which is 3 for a normal.
• Closed forms worth memorising: exponential E[X^n] = n!\,\theta^n; Gamma E[X^n] = \theta^n\,\alpha(\alpha+1)\cdots(\alpha+n-1); lognormal E[X^n] = e^{n\mu + n^2\sigma^2/2}. Where the moment generating function M_X(t) = E[e^{tX}] exists — the sum or integral defining it must be finite for t in an interval around 0, e.g. t < 1/\theta for the exponential and gamma — it produces every raw moment as E[X^n] = M_X^{(n)}(0). Loss Models' tables give no MGF for the lognormal or the Pareto: take their moments from the formulas directly.
• Moments need not exist. A heavy-tailed Pareto with shape \alpha has E[X^n] finite only for n < \alpha, so a fitted \alpha \le 2 means an infinite variance.
• Moments of several variables — E[XY] and Covariance — are in Moments for Joint Distributions and Moments for Linear Combinations. Setting sample moments equal to these formulas is the Method of Moments.
• Exam 7. Mack (1994) measures a chain-ladder reserve estimate R by its standard error. When the reserve distribution is rather skewed — especially once the standard error exceeds half of R, where a normal interval's lower limit turns negative — he fits a lognormal with the same mean and variance, \sigma^2 = \ln\!\big(1 + \text{s.e.}(R)^2/R^2\big) and \mu = \ln R - \sigma^2/2, and reads percentiles off it.

Skewness of an Exponential Severity Claim sizes are exponential with mean \theta = 1{,}000. Find the first three raw moments, the variance and the skewness.

Answer From E[X^n] = n!\,\theta^n: E[X] = 1{,}000, E[X^2] = 2{,}000{,}000 and E[X^3] = 6 \times 10^9.

\begin{align*}
\sigma^2 &= 2{,}000{,}000 - 1{,}000^2 \\
         &= 1{,}000{,}000 \\
\mu_3 &= 6 \times 10^9 - 3(1{,}000)(2 \times 10^6) + 2(1{,}000)^3 \\
      &= 2 \times 10^9 \\
\gamma_1 &= \frac{2 \times 10^9}{1{,}000^3} \\
         &= 2
\end{align*}

The standard deviation equals the mean (\text{CV} = 1) and the skewness is 2 whatever \theta is: both are scale-free, so every exponential severity has the same shape.

Reserve Percentile from Two Moments A chain-ladder analysis gives total unpaid claims with mean R = \$10.0 million and standard error \$2.5 million. Fit a lognormal with those two moments, as Mack (1994) suggests for a skewed reserve distribution, and find its 95th percentile. Compare with a normal.

Answer Match the CV, 2.5/10 = 0.25, then the mean. Working in $ millions:

\begin{align*}
\sigma^2 &= \ln(1 + 0.25^2) \\
         &= \ln 1.0625 \\
         &= 0.06062 \\
\sigma &= 0.2462 \\
\mu &= \ln 10 - \tfrac{1}{2}(0.06062) \\
    &= 2.2723
\end{align*}

\begin{align*}
x_{0.95} &= e^{\mu + 1.645\sigma} \\
         &= e^{2.2723 + 0.4050} \\
         &= 14.55
\end{align*}

A normal with the same moments gives 10 + 1.645(2.5) = 14.11. The lognormal 95th percentile is $14.55 million, about $0.43 million higher, and its median e^{\mu} = 9.70 sits below the mean. Two distributions sharing a mean and a variance still disagree in the tail — the skewness they don't share is what a risk margin set at a percentile is sensitive to.