Conditional mean of Weibull to the power of N How can I (is it possible to) derive the conditional expectation of a variable $w$ that follows a 2 parameter Weibull Distribution $W(\lambda,k)$ with $\lambda$ scale parameter and $k$ shape parameter?
We know that:
$E[w^n]=\lambda^n \Gamma(1+\frac{n}{k})$
What is (is there) an expression for?
$E[w^n  |  \underline{w}<w<\bar{w}]$
 A: As always, the effect of the scale parameter $\lambda\gt 0$ is merely to change the units of measurement of the variable.  Consequently the answer must be a multiple of $\lambda^n$ and that multiple is found by assuming $\lambda=1.$
By definition, the survival function of a Weibull variable $w$ with unit scale and shape parameter $k \gt 0$ is given by
$$\Pr(w \gt x) = \exp\left(-x^k\right).$$
Therefore, assuming $n\gt 0,$
$$\Pr(w^n \gt x) = \Pr(w \gt x^{1/n}) = \exp\left(-\left(x^{1/n}\right)^k\right) = \exp\left(-x^{k/n}\right)$$
shows $w^n$ follows a Weibull distribution with shape parameter $k/n$ (and scale parameter $1$).
These considerations have reduced the question to finding the expectation of a Weibull variable $X,$ of unit scale and shape parameter $k/n,$ that has been truncated at the values $l=\underline{w}\,\lambda^{-n}$ and $u=\bar{w}\,\lambda^{-n}.$  This expectation will equal
$$E[X] = C^{-1}\int_l^u x\,\mathrm{d}\left(1-\exp\left(-x^{k/n}\right)\right) = \frac{k}{nC}\int_l^u x^{k/n}\exp\left(-x^{k/n}\right)\,\mathrm{d}x\tag{*}$$
where the normalizing constant is
$$C = \Pr(l \le X \le u) = \exp\left(-l^{k/n}\right) - \exp\left(-u^{k/n}\right).$$
Multiply this expression for $E[X]$ by $\lambda ^n$ to account for the scale factor in $w.$ 
To evaluate $(*),$ change the variable to $y=x^{k/n},$ giving
$$\eqalign{
E[w^n\mid \underline{w}\le w \le \bar{w}] &= \lambda ^n\frac{k}{nC}\int_{l^{k/n}}^{u^{k/n}} y\,e^{-y}\,\mathrm{d}\left(y^{n/k}\right)\\
&= \lambda ^n\frac{1}{C}\int_{l^{k/n}}^{u^{k/n}} y^{n/k}e^{-y}\,\mathrm{d}y \\
&= \lambda ^n\frac{\gamma\left(n/k+1, u^{k/n}\right) - \gamma\left(n/k+1, l^{k/n}\right)}{\exp\left(-l^{k/n}\right) - \exp\left(-u^{k/n}\right)}\\
&= \lambda ^n\frac{\gamma\left(n/k+1, \bar{w}^{k/n}\lambda^{-k}\right) - \gamma\left(n/k+1, \underline{w}^{k/n}\lambda^{-k}\right)}{\exp\left(-\underline{w}^{k/n}\lambda^{-k}\right) - \exp\left(-\bar{w}^{k/n}\lambda^{-k}\right)}
}$$
where $\gamma(s,t) = \int_0^t x^{s-1}e^{-x}\,\mathrm{d}x$ is the incomplete Gamma function.
