这是一份kcl伦敦大学学院 6CCM341A作业代写的成功案
Note that for any random variable $y$, defined by $y=a r+b$, we may obtain $f_{v}^{T}(s)$ in terms of $f_{r}^{T}(s)$ from the definition of the $s$ transform as. follows:
$f_{\nu}{ }^{T}(s)=E\left(e^{-w}\right)=E\left(e^{-a-s^{-s} e^{-\infty}}\right)=e^{-\Delta t} \int_{-\infty}^{n} e^{-\operatorname{anv} f_{r}\left(r_{0}\right) d r_{0}}$
We may recognize the integral in the above equation to obtain $f_{y}^{T}(s)=e^{-\Delta b_{p}}{ }^{T}(a s)$
We shall apply this relation to the ease where $y$ is the standardized random variable for $r$,
$$
y=\frac{r-E(r)}{\sigma_{p}}=\frac{r-n E(x)}{\sqrt{n} \sigma_{n}} \quad a=\frac{1}{\sqrt{n} \sigma_{s}} \quad b=-\frac{\sqrt{n} E(x)}{\sigma_{z}}
$$
6CCM341A COURSE NOTES ：
and if $a$ and $b$ are integers with $b>a$, there follows
$\operatorname{Prob}(a \leq k \leq b)=\sum_{k_{0}=a}^{b}\left(\begin{array}{l}n \ k_{0}\end{array}\right) P v_{k}(1-P)^{a-k_{0}}$
Should this quantity be of interest, it would generally require a very unpleasant ealculation. So we might, for large $n$, turn to the eentral limit theorem, noting that
$$
k=x_{1}+x_{2}+\cdots+x_{n}
$$