Higher Moments: Skewness and Kurtosis

Why Examine Moments Beyond the Mean and Variance?

The mean and variance summarize a loss distribution’s average and spread, but two distributions can have the same values for both and still differ in important ways. One may be more asymmetric, for example, or have a different pattern of large deviations from its mean. The standardized third and fourth central moments provide two additional summaries: skewness and kurtosis.

These summaries are useful for comparing loss distributions, but they do not replace the full distribution. In particular, a single moment does not determine all aspects of a loss model or its extreme outcomes. The calculations below assume the moments being used exist and are finite.

Definition: For a loss random variable $X$ with finite mean $E[X]$, its $r$th central moment is $m_r=E[(X-E[X])^r]$. The second central moment is the variance, $m_2=\operatorname{Var}(X)$. If $0<m_2<\infty$ and the third central moment is finite, the standardized skewness is $\gamma_1=m_3/m_2^{3/2}$. If the fourth central moment is finite, the Pearson kurtosis is $\beta_2=m_4/m_2^2$, and the excess kurtosis is $\gamma_2=\beta_2-3$. Skewness and kurtosis are dimensionless.

For a discrete loss taking values $x_j$ with probabilities $p_j$, each central moment is calculated as $m_r=\sum_j p_j(x_j-E[X])^r$. For a continuous loss with density $f_X$, the corresponding calculation is $m_r=\int (x-E[X])^r f_X(x)\,dx$, when the integral exists. A distribution with zero variance has no standardized skewness or kurtosis because the denominators are zero.

Calculating the Standardized Moments

The third central moment retains the sign of deviations: losses above the mean contribute positively, while losses below the mean contribute negatively. Skewness therefore describes the direction and degree of asymmetry as summarized by the third moment. A positive value indicates that positive deviations contribute more to the signed third moment; a negative value indicates the opposite. Zero skewness does not, by itself, prove that a distribution is symmetric.

The fourth central moment raises deviations to an even power, so both positive and negative deviations contribute positively. Pearson kurtosis summarizes the fourth-power-weighted deviations relative to the variance squared. Excess kurtosis simply subtracts 3 from Pearson kurtosis; this convention makes the excess value for a normal distribution equal to zero. The two conventions must not be confused.

\gamma_1=\frac{E[(X-E[X])^3]}{\operatorname{Var}(X)^{3/2}}, \qquad \beta_2=\frac{E[(X-E[X])^4]}{\operatorname{Var}(X)^2}, \qquad \gamma_2=\beta_2-3.

When raw moments $E[X^r]$ are available, central moments can be found by expanding powers around the mean. In particular, writing $M_r=E[X^r]$ and $M_1=E[X]$,

m_2=M_2-M_1^2,\qquad m_3=M_3-3M_1M_2+2M_1^3,

$$ m_4=M_4-4M_1M_3+6M_1^2M_2-3M_1^4. $$

These expressions follow from expanding $(X-M_1)^r$ and taking expectations term by term. For example, $(X-M_1)^3=X^3-3M_1X^2+3M_1^2X-M_1^3$; taking expectations and using $E[X]=M_1$ gives $m_3=M_3-3M_1M_2+2M_1^3$. The fourth-moment expression follows by the same expansion of $(X-M_1)^4$.

Worked Example: Calculating Skewness and Kurtosis From a Loss Table

A hypothetical annual loss $X$, measured in thousands of dollars, is 0 with probability 0.50, 2 with probability 0.25, and 6 with probability 0.25. Calculate the mean, variance, skewness, Pearson kurtosis, and excess kurtosis. Use the discrete distribution directly, and round the standardized values to four decimal places.

Method and mean. Multiply each loss by its probability and add: $E[X]=0(0.50)+2(0.25)+6(0.25)=0+0.50+1.50=2$. The mean loss is 2 thousand dollars.

Variance. The deviations from the mean are $-2$, $0$, and $4$. Thus, $\operatorname{Var}(X)=0.50(-2)^2+0.25(0)^2+0.25(4)^2=2+0+4=6$. The standard deviation is $\sqrt{6}$ thousand dollars.

Third central moment and skewness. Weight the cubed deviations by their probabilities: $m_3=0.50(-2)^3+0.25(0)^3+0.25(4)^3=-4+0+16=12$. Therefore, $\gamma_1=12/6^{3/2}=12/(6\sqrt{6})=2/\sqrt{6}=\sqrt{2/3}\approx0.8165$.

Fourth central moment and kurtosis. Weight the fourth powers: $m_4=0.50(-2)^4+0.25(0)^4+0.25(4)^4=8+0+64=72$. Hence Pearson kurtosis is $\beta_2=72/6^2=72/36=2$, and excess kurtosis is $\gamma_2=2-3=-1$.

Check. The probabilities sum to 1, the variance is positive, and the skewness is positive, consistent with the substantial loss above the mean contributing a positive cubed deviation. The calculation avoids dividing the third moment by the variance rather than by the variance raised to the $3/2$ power.

Why Standardize?

Central moments depend on the units in which losses are measured. If losses are expressed in dollars instead of thousands of dollars, the third and fourth central moments change substantially. Standardizing divides by the appropriate power of the standard deviation, so the units cancel. A fixed shift also leaves central moments unchanged because it does not change deviations from the mean.

More generally, let $Y=a+bX$, where $a$ is a constant and $b>0$. Then $E[Y]=a+bE[X]$, so $Y-E[Y]=b(X-E[X])$. Consequently, $m_r(Y)=b^r m_r(X)$, while $\operatorname{Var}(Y)=b^2\operatorname{Var}(X)$. Substitution into the standardized formulas shows that both skewness and kurtosis are unchanged. If $b<0$, skewness changes sign because the third central moment changes sign; kurtosis is unchanged.

Worked Example: Comparing Distributions With the Same Mean and Variance

Two hypothetical loss models, $A$ and $B$, are measured in the same units. Model $A$ takes values 0, 2, and 4 with probabilities 0.25, 0.50, and 0.25. Model $B$ takes values 0 and 3 with probabilities $1/3$ and $2/3$. Calculate their means, variances, skewnesses, and Pearson kurtoses. Round standardized values to four decimal places.

Model A. Its mean is $E[A]=0(0.25)+2(0.50)+4(0.25)=2$. Deviations from 2 are $-2,0,2$, so its variance is $0.25(4)+0.50(0)+0.25(4)=2$. Its third central moment is $0.25(-8)+0.50(0)+0.25(8)=0$, giving skewness $0/2^{3/2}=0$. Its fourth central moment is $0.25(16)+0.50(0)+0.25(16)=8$, giving Pearson kurtosis $8/2^2=2$.

Model B. Its mean is $E[B]=0(1/3)+3(2/3)=2$. Its deviations from 2 are $-2$ and $1$. Its variance is $(1/3)(4)+(2/3)(1)=4/3+2/3=2$. Its third central moment is $(1/3)(-8)+(2/3)(1)=-8/3+2/3=-2$, so its skewness is $-2/2^{3/2}=-1/\sqrt{2}\approx-0.7071$. Its fourth central moment is $(1/3)(16)+(2/3)(1)=16/3+2/3=6$, so its Pearson kurtosis is $6/4=1.5$.

Comparison and check. Both models have mean 2 and variance 2, but their skewnesses and kurtoses differ. Model A has zero skewness, whereas Model B has negative skewness. Their excess kurtoses are respectively $2-3=-1$ and $1.5-3=-1.5$. The calculation confirms that mean and variance alone do not identify these higher-moment summaries.

Interpreting Kurtosis Carefully

Kurtosis is sometimes described loosely as a measure of “peakedness.” For actuarial work, it is more precise to say that the fourth central moment gives especially strong weight to observations far from the mean, on either side. A higher Pearson kurtosis than another model’s value indicates a higher standardized fourth moment, not a complete description of the distribution’s shape. It does not, on its own, specify a particular exceedance probability or determine a tail quantile.

The normal distribution provides a useful reference point for the two kurtosis conventions: its Pearson kurtosis is 3 and its excess kurtosis is 0. That reference does not imply that every distribution with excess kurtosis zero is normal. Likewise, zero skewness alone does not establish symmetry. These statistics summarize particular moments, not every feature of a loss distribution.

Worked Example: Separating a Fixed Loss Component From Random Variation

A hypothetical annual loss is $X=100{,}000+100{,}000Y$ dollars, where $Y=0$ with probability 0.80 and $Y=1$ with probability 0.20. Calculate the mean, variance, skewness, and Pearson kurtosis of $X$. Round standardized values to four decimal places.

Mean and variance. The loss is $100,000 with probability 0.80 and $200,000 with probability 0.20. Its mean is $E[X]=100{,}000(0.80)+200{,}000(0.20)=80{,}000+40{,}000=\$120{,}000$. The deviations are $-\$20{,}000$ and $\$80{,}000$. Thus, $\operatorname{Var}(X)=0.80(20{,}000)^2+0.20(80{,}000)^2=320{,}000{,}000+1{,}280{,}000{,}000=\$1{,}600{,}000{,}000$. The standard deviation is $\sqrt{1{,}600{,}000{,}000}=\$40{,}000$.

Third central moment and skewness. The third central moment is $m_3=0.80(-20{,}000)^3+0.20(80{,}000)^3=-6{,}400{,}000{,}000{,}000+102{,}400{,}000{,}000{,}000=\$96{,}000{,}000{,}000{,}000$. Therefore, $\gamma_1=m_3/(40{,}000)^3=96{,}000{,}000{,}000{,}000/64{,}000{,}000{,}000{,}000=1.5000$.

Fourth central moment and kurtosis. The fourth central moment is $m_4=0.80(20{,}000)^4+0.20(80{,}000)^4=128{,}000{,}000{,}000{,}000{,}000+8{,}192{,}000{,}000{,}000{,}000{,}000=8{,}320{,}000{,}000{,}000{,}000{,}000$. Since $\operatorname{Var}(X)^2=(1{,}600{,}000{,}000)^2=2{,}560{,}000{,}000{,}000{,}000{,}000$, Pearson kurtosis is $8.32/2.56=3.2500$, and excess kurtosis is $0.2500$.

Check. The fixed $100,000 component shifts the mean but does not change deviations from that mean. The positive scale factor converts the deviations from $Y$ to dollars, but cancels in the standardized moments. Thus the positive skewness and kurtosis agree with those of a Bernoulli variable with success probability 0.20.

Common Exam Trap

A frequent error is to divide $m_3$ by the variance instead of by the variance to the $3/2$ power. In the first worked example, that incorrect calculation would give $12/6=2$. The correct calculation is $12/6^{3/2}=12/(6\sqrt{6})=\sqrt{2/3}\approx0.8165$. Another common error is to report Pearson kurtosis as excess kurtosis, or vice versa: for that example, the Pearson value is 2 and the excess value is $2-3=-1$.

Before interpreting a result, check the requested convention, confirm that the denominator uses the correct power of the variance, and verify that all central moments were calculated around the same mean. If a problem supplies raw moments, use the central-moment conversion formulas rather than treating raw moments as though they were central moments.

Key takeaway: Skewness standardizes the third central moment, and kurtosis standardizes the fourth. They can distinguish loss distributions that share a mean and variance, but each is a limited summary and should be interpreted using its stated convention.

Check Your Understanding

For each problem, calculate the requested quantities before checking the answer.

A hypothetical loss $X$, in thousands of dollars, is 0 with probability 0.75 and 4 with probability 0.25. Find its mean, variance, skewness, and Pearson kurtosis. (Answer: Mean $=1$; variance $=3$; skewness $=1.1547$; Pearson kurtosis $=2.3333$.)
A loss $X$ takes values 1, 2, and 5 with probabilities 0.25, 0.50, and 0.25. Find its mean, variance, third central moment, skewness, and excess kurtosis. (Answer: Mean $=2.5$; variance $=2.25$; third central moment $=3$; skewness $=0.8889$; excess kurtosis $= -0.8148$.)
Loss model $A$ has skewness 0.60 and Pearson kurtosis 4.10. Loss model $B$ has skewness $-0.20$ and excess kurtosis 0.40. Which has negative skewness, and what is model $B$'s Pearson kurtosis? (Answer: Model $B$ has negative skewness; its Pearson kurtosis is $3.40$.)
A loss $X$ has mean $50,000, variance $\$400{,}000{,}000$, third central moment $8{,}000{,}000{,}000{,}000$, and fourth central moment $1{,}600{,}000{,}000{,}000{,}000{,}000$. Calculate its standard deviation, skewness, and Pearson kurtosis. (Answer: Standard deviation $=\$20{,}000$; skewness $=1.0000$; Pearson kurtosis $=10.0000$.)
Let $Y$ have skewness $-0.75$ and Pearson kurtosis 5.2. Define $Z=10+3Y$. State the skewness and excess kurtosis of $Z$, and explain the effect of the transformation. (Answer: Skewness $=-0.75$; excess kurtosis $=2.2$. The positive scale and shift leave both standardized quantities unchanged, while excess kurtosis is Pearson kurtosis minus 3.)

What You'll Learn