这是一份贝叶斯统计推断作业代写的成功案

$$
-\sum_{i} \log \left(1-y_{1} a\right)=s_{1} a+s_{2} a_{2}^{2}+s_{3} a_{3}^{3}+\ldots
$$
where $a$ is any constant such that
$$
|a|<\frac{1}{\max y_{6}}
$$
giving us
$$
g\left(p_{1}^{n_{1}} p_{2}^{n_{2}} \ldots\right)=\sum g_{1}(P, Q) s\left(q_{1}^{x_{1}} q_{2}{ }^{\times *} \ldots\right)
$$
and
$$
s\left(p_{1}^{n_{1}} p_{2}^{n_{2}} \ldots\right)=\sum s_{0}(P, Q) g\left(q_{1}^{x_{1}} q_{2}^{x_{1}} \ldots\right)
$$

STA 144/STAT 451/STAT 506/STA 317 COURSE NOTES ：

$N^{t} e_{j}=0$ for $i<j$
and
$$
N^{j} e_{i}=n(n-1) \ldots(n-j+1)
$$
Hence
$$
V\left(s^{2}\right)=\frac{\mu_{4}-\mu_{2}^{2}}{n}+\frac{2}{n(n-1)}{ }^{\mu_{2}^{2}} .
$$
Using the Pearsonian notation for departure from normality, this can be written as