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I am using this function (which is type of Bounded Pareto Distribution) with domain of support $x \in[1, H]$:

$$f(x) =\frac{\alpha x^{-(\alpha +1)}}{1-\left(\frac{1}{H}\right)^{\alpha }}$$

and am attempting to find $E[Y]$, where $Y=\min\{X_1,\ldots,X_n\}$ is the minimum of an i.i.d. sample $(X_1,\ldots,X_n)$ of size $n$ with distribution $f$, hence $Y$ has the following density function:

$$g(x) = \frac{\alpha n H^{\alpha } \left(H^{\alpha }-1\right)^{-n} \left(\left(\frac{H}{x}\right)^{\alpha }-1\right)^n}{x H^{\alpha }-x^{\alpha +1}}$$

Now, I am having no luck with the following integral:

$$E[Y]=\int_1^H \frac{\alpha n H^{\alpha } \left(H^{\alpha }-1\right)^{-n} \left(\left(\frac{H}{x}\right)^{\alpha }-1\right)^n}{H^{\alpha }-x^{\alpha }} \, dx$$

Typically, I run into these types of integrals often - Does anyone know how to solve this? Any help (in general) would be appreciated. Also, a way to approximate this would be helpful as well.

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    A couple of the things you have there are constants and can be pulled out of the integral. What remains... I'm not sure how to integrate for general $\alpha,n$, but [this post](http://math.stackexchange.com/questions/1999869/evaluate-int-frac11xndx-for-n-in-mathbb-r) may help you. – 2017-01-01
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    Ok - thanks for the reference - Need to digest the material in your reference a bit – 2017-01-01
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    How about the "Delta Method" - Are you familiar with that? – 2017-01-01
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    Your notation does not explain the relevance (if any) of $n$. Why $f_{1:n}(x) $? What then does $X_{1:n}$ denote? 1st order statistic in a sample of size $n$? Why would it have the same density notation $f_{1:n}(x)$ as the parent? – 2017-01-02
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    You are right - my bad - that was a typo on my part. $f_{1:n}(x)$ definitely has a different density than its parent (it has to) - Like I said, typo - thanks – 2017-01-02
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    What is $X_{1:n}$? This is the usual notation for an ordered sample of size $n$ (thus, a random *vector*) but you seem to be thinking of a real valued random variable. Please explain. – 2017-01-02
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    Sorry about the confusion - I'm trying to compute the minimum of $n$ samples of that random variable. So the $1:n$ means the minimum - $n:n$ would be the maximum – 2017-01-02

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So far, your calculations appear to be correct. We then observe $$\frac{H^\alpha ((H/x)^\alpha-1)^n}{H^\alpha-x^\alpha} = (H/x)^\alpha \cdot \frac{((H/x)^\alpha-1)^n}{(H/x)^\alpha-1} = (H/x)^\alpha ((H/x)^\alpha - 1)^{n-1}.$$ We can now proceed to integrate term-by-term the respective binomial expansion $$\begin{align*} \operatorname{E}[Y] &= \frac{\alpha n}{(H^\alpha - 1)^n} \sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} \int_{x=1}^H \left(\frac{H}{x}\right)^{\alpha(n-k)} \, dx \\ &= \frac{\alpha n}{(H^\alpha-1)^n} \sum_{k=0}^{n-1} (-1)^k \binom{n-1}{k} \begin{cases} \frac{H^{\alpha(n-k)} - H}{\alpha(n-k) - 1}, & \alpha(n-k) \ne 1, \\ H \log H, & \alpha(n-k) = 1. \end{cases} \end{align*} $$ From this point, I relied on Mathematica to complete the summation when $\alpha$ is not the reciprocal of some integer between $1$ and $n$; and was able to express the answer in terms of the beta and incomplete beta functions: $$\operatorname{E}[Y] = \frac{Hn}{(H^\alpha-1)^n} \left((-1)^n B(n, 1 - 1/\alpha) - B(1/H; 1/\alpha-n, n) \right),$$ where $$B(z;a,b) = \int_{t=0}^z t^{a-1} (1-t)^{b-1} \, dt,$$ and $$B(a,b) = B(1;a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$ When $\alpha = 1/m$ for some $m \in \{1, \ldots, n\}$, then the above formula will be undefined, but you can, I think, either attempt a limiting approach or try to achieve some kind of cancellation in the above expression.

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    Thanks for your help! At first glance, this looks correct, but I need some time to digest this... – 2017-01-02
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    I am unclear how you were able to express this in terms of the Beta and Incomplete Beta functions. Could you go into a little more detail? Thanks. – 2017-02-17