# Bounded Pareto Distribution Moment Generating Function

I am interested in the Bounded Pareto Distribution but I can't find any reference that supplies me the moment generating function. This does however seem like something that is known.

More general : Is there some (online/book/...) reference where I can find a list of all (or at least many) distributions and their most important properties (like moments, PDF, CDF,...) I have been mainly using wikipedia but it has now let me down.

## EDIT:

From the provided comment, I have found that a general formula for the moment generating formula for the MGF might not be avaible. I do however still wonder how to obtain it numerically. When I use the moments (for which there does exist a general formula) and the taylor series expansion to obtain the MGF, I find that (when the upper bound is somewhat large) the moments quickly explode and (at least numerically) the Taylor series diverges.

• One canonical reference for probability distributions is Johnson-Kotz-Balakrishnan's discrete/continuous (both multivariate and univariate) books. – StubbornAtom Apr 9 at 9:39
• Don't think the mgf exists; it doesn't exist in the usual sense even for the standard pareto distribution. – StubbornAtom Apr 9 at 10:43
• For a bounded Pareto it must exists since if $L \leq X \leq H$ then we have for any $t: \mathbb{E}[e^{tX}] \leq \max\{e^{-tL}, e^{tH}\}$ – Darkwizie Apr 9 at 10:45
• You can see it form a incomplete gamma integral when the argument of the mgf is negative. – BGM Apr 9 at 11:54
• @Darkwizie - yes, and since $0 \le L$ that simplifies to $\mathbb{E}[e^{tX}] \leq e^{tH}$ – Henry Apr 9 at 16:59

I'm assuming that if you had a general formula for the moments that you wouldn't need the MGF.

Using Mathematica the general formula for the $k$-th moment is found in the following manner:

f[x_] := \[Alpha] L^\[Alpha] x^(-\[Alpha] - 1) /(1 - (L/H)^\[Alpha])
muk = Integrate[x^k f[x], {x, L, H}, Assumptions -> {\[Alpha] > 0, L > 0, H > L, k >= 0}]


$$\mu_k=\frac{\alpha \left(H^{\alpha } L^k-H^k L^{\alpha }\right)}{(\alpha -k) \left(H^{\alpha }-L^{\alpha }\right)}$$

This formula clearly has a problem when $k=\alpha$. So when $k=\alpha$ we can solve for that moment directly:

mualpha = Integrate[x^\[Alpha] f[x], {x, L, H}, Assumptions -> {\[Alpha] > 0, L > 0, H > L}]


$$\frac{\alpha L^{\alpha } \log \left(\frac{H}{L}\right)}{1-\left(\frac{L}{H}\right)^{\alpha }}$$

This is also what one obtains when taking the limit of the general formula as $k\rightarrow\alpha$.

* 2nd Update*

There is a relatively closed-form solution for the MGF but one needs to be a bit more careful when extracting the moments. First, the closed-form solution:

$$\text{mgf}=\sum _{k=0}^{\infty } \frac{t^k \left(\alpha \left(H^{\alpha } L^k-H^k L^{\alpha }\right)\right)}{k! \left((\alpha -k) \left(H^{\alpha }-L^{\alpha }\right)\right)}$$

$$=-\frac{\alpha \left(L^{\alpha } (-H t)^{\alpha } \Gamma (-\alpha ,0,-H t)-H^{\alpha } (-L t)^{\alpha } \Gamma (-\alpha ,0,-L t)\right)}{L^{\alpha }-H^{\alpha }}$$

where $\Gamma(a,z_0,z_1)$ is the generalized incomplete gamma function (just as suggested in a comment by @BGM).

mgf = Sum[(((H^\[Alpha] L^k - H^k L^\[Alpha]) \[Alpha])/((H^\[Alpha] -
L^\[Alpha]) (-k + \[Alpha]))) t^k/k!, {k, 0, \[Infinity]}]


To obtain the $k$-th moment one would usually differentiate $k$ times and then set $t=0$ but that won't work with this MGF. If $\alpha$ is not a positive integer, then you'll need to take the $k$-th derivative followed by taking the limit of that derivative as $t\rightarrow 0$. Using Mathematica commands that translates to the following when $k=3$:

Limit[D[mgf, {t, 3}], t -> 0]


$$\frac{\alpha \left(L^3 H^{\alpha }-H^3 L^{\alpha }\right)}{(\alpha -3) \left(H^{\alpha }-L^{\alpha }\right)}$$

If $\alpha$ is a positive integer (say $\alpha=5$, then two limits need to be taken:

$$\lim_{\alpha \to 5} \, \lim_{t\to 0} \, \frac{\partial ^3\text{mgf}}{\partial t^3}=\frac{5 \left(H^5 L^3-H^3 L^5\right)}{2 \left(H^5-L^5\right)}$$

with the associated Mathematica commands:

Limit[Limit[D[mgf, {t, 3}], t -> 0], \[Alpha] -> 5]

• I really like/appreciate your answer (therefore I have upvoted it) but I do in fact take interest in having a way to compute the MGF – Darkwizie Apr 9 at 19:22
• However, I accept your answer :)! – Darkwizie Apr 9 at 19:22
• @Darkwizie. General form of the MGF added. – JimB Apr 10 at 4:15
• Wonderful, thank you! – Darkwizie Apr 10 at 21:05