这是一份nottingham诺丁汉大学MATH3029作业代写的成功案例

(a) If $R(p)$ is the reflected form of $Q(p)$ show that the probability-weighted moments have the following relationships $\omega_{R: r, o}=-\omega_{Q: 0, r}$ and $\omega_{Q: 0, s}=-\omega_{Q:-, 0}$.
(b) If a distribution has the form $Q(p)=\lambda+\eta T(p)+\phi H(p)$, where $T(p)$ and $H(p)$ are quantile functions, show that
$$
\omega_{Q: r, o}=\lambda(r+1)+\eta \omega_{T: r, a}+\phi \omega_{H: r, a},
$$
and
$$
\omega_{Q: 0, s}=\lambda(s+1)+\eta \omega_{T: 0, s}+\phi \omega_{H: 0, s}
$$
(c) Using the results of (a) and (b), find the probabilityweighted moments, $\omega_{r, o}$ and $\omega_{0, s}$ for the distribution.
$$
Q(p)=\lambda+(\eta / 2)\left[(1+\delta) p^{\alpha}-(1-\delta)(1-p)^{\alpha}\right] .-1 \leq \delta \leq 1 .
$$

MATH3029 COURSE NOTES ：

If $X$ has an exponential distribution, then the quantile function of the largest observation, returning to the usual notation, is
$$
Q_{(n)}(p)=-\ln \left(1-p^{1 / n}\right)
$$
This distribution will have a median value of $-\ln \left(1-1 / 2^{1 / n}\right)$.
The previous argument also works directly for the smallest observation, using the fact that the probability of $n$ independent observations being greater than or equal to $x$ is $(1-p)^{n}$. Thus
$$
1-p_{(1)}=(1-p)^{n} \text { and so } p=1-\left(1-p_{(1)}\right)^{1 / n}
$$
This leads to a quantile function for $X_{(1)}$ of $Q\left(1-\left(1-p_{(1)}\right)^{1 / n}\right)$. For the exponential example we have on simplifying,
$$
Q_{(1)}(p)=-[\ln (1-p)] / n
$$
This is still an exponential distribution but with a reduced scale factor.