can anyone help me on this?

A claim size distribution of an insurance portfolio follows a Pareto distribution given by:


I am trying to derive a formula for the $r^{th}$ moment $\alpha_{r}$ of the Pareto distribution in terms of its $(r-1)^{th}$ moment $\alpha_{r-1}$ and from this expression find, $\alpha_2$, $\alpha_3$ and $\alpha_4$ using the known result for the mean $\mu$ of the above Pareto distribution.

I found that the mean of the Pareto distribution is given by:

$$\mu=\int_{\beta}^{\infty} (u-\beta)\frac{\alpha}{\beta}\frac{\beta^{^{(\alpha+1)}}}{u^{(\alpha+1)}} du$$which is:

$$\alpha\beta^{\alpha}\int_{\beta}^{\infty}\frac{1}{u^{\alpha}}-\frac{1}{u^{(\alpha+1)}} du=\frac{\beta}{\alpha-1}$$

I also find that the second moment is given by:

$$\alpha_2=\alpha\beta^{\alpha}\int_{0}^{\infty}\frac{x^{2}}{(\beta+x)^{\alpha+1}} dx$$, where $\beta+x=u$ then,

$$\alpha\beta^{\alpha}\int_{\beta}^{\infty}\frac{u^{2}-2u\beta+\beta^{2}}{(u)^{\alpha+1}} du=\frac{2\beta^{2}}{(\alpha-2)(\alpha-1)}$$

I am not sure if this is correct and I could not find anything regarding $\alpha_3$ and $\alpha_4$ which I believe they are the skewness and kurtosis of this Pareto distribution.

In terms of the $r^{th}$ moment formula I really cannot understand what I should do to derive the formula. I read that there are the $r^{th}$ moment about the origin that for a continues is given by

$$\alpha_r=E(x^{r})=\int_{}^{} x^r f(x) dx$$

And also the $r^{th}$ moment about the mean is given by:


Can anyone give me any tip in how to continue and let me know if what I did is correct and can be somehow of any use for this question?



In general for positive integers $k$, $$\begin{align*} \operatorname{E}[X^k] &= \alpha\beta^\alpha \int_{x=0}^\infty x^k (x+\beta)^{-(\alpha+1)} \, dx \\ &= \alpha\beta^\alpha \int_{x=\beta}^\infty (x-\beta)^k x^{-(\alpha+1)} \, dx \\ &= \alpha\beta^\alpha \int_{x=\beta}^\infty x^{-(\alpha+1)} \sum_{i=0}^k \binom{k}{i} x^{k-i} (-\beta)^i \\ &= \alpha\beta^\alpha \sum_{i=0}^k \binom{k}{i}(-\beta)^i \int_{x=\beta}^\infty x^{k-(i+\alpha+1)} \, dx \\ &= \alpha\beta^\alpha \sum_{i=0}^k \binom{k}{i}(-\beta)^i \left[ \frac{x^{k-(\alpha+i)}}{k-(\alpha+i)} \right]_{x=\beta}^\infty \\ &= \alpha\beta^\alpha \sum_{i=0}^k \binom{k}{i}\frac{-(-\beta)^i \beta^{k-(\alpha+i)}}{k-(\alpha+i)}, \quad k < \alpha \\ &= \alpha\beta^\alpha \beta^{k-\alpha} \sum_{i=0}^k \binom{k}{i} \frac{(-1)^{i+1}}{k-(\alpha+i)} \\ &= \alpha\beta^k \sum_{i=0}^k \binom{k}{i} \frac{(-1)^i}{\alpha+i - k}. \end{align*}$$

So for $k = 1$, we get $$\operatorname{E}[X] = \alpha\beta \left( \frac{1}{\alpha-1} - \frac{1}{\alpha} \right) = \frac{\beta}{\alpha-1},$$ and for $k = 2$, $$\operatorname{E}[X^2] = \alpha\beta^2 \left( \frac{1}{\alpha-2} - \frac{2}{\alpha-1} + \frac{1}{\alpha} \right) = \frac{2\beta^2}{(\alpha-2)(\alpha-1)}.$$ Next, $$\operatorname{E}[X^3] = \alpha\beta^3 \left(\frac{1}{\alpha-3} - \frac{3}{\alpha-2} + \frac{3}{\alpha-1} - \frac{1}{\alpha} \right) = \frac{6\beta^3}{(\alpha-3)(\alpha-2)(\alpha-1)}.$$ Now you should suspect a pattern; namely, $$\operatorname{E}[X^k] = \frac{k! \beta^k}{\prod_{i=1}^k (\alpha-i)} = \beta^k \binom{\alpha-1}{k}^{-1}, \quad k < \alpha.$$ The proof of this last step is not too difficult, so I have left it as an exercise.

  • $\begingroup$ Hi, how the first formula fot he general k is given in terms of its $(r-1)^{th}$ moment $\alpha_{r-1}$? I will continue the exercise for $\alpha_4$. Thank you very much for your help. $\endgroup$ – user290335 Mar 10 '18 at 14:03

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