Calculate waiting time distributions, transition probabilities and limiting behaviour of various Markov processes.

这是一份Bath巴斯大学MA20225作业代写的成功案

$B_{a}{ }^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(z_{1}-M_{n}\right)^{2}$
where $M_{a}$ is the sample mean as defined earlies. We may expand the above expression,
$$
S_{n}{ }^{t}=\frac{1}{\pi} \sum_{i=1}^{n} x_{i}{ }^{n}-\frac{2}{n} M_{a} \sum_{i=1}^{n} x_{i}+M_{a}{ }^{n}=\sum_{n} \sum_{i=1}^{n} x_{i}{ }^{2}-M_{a^{n}}
$$
This is a more useful form of $S_{\mathrm{m}}{ }^{2}$ for the ealeulation of its expectntion
$$
B\left(S_{m}{ }^{2}\right)=\frac{1}{x} B\left(\sum_{i=1}^{n} x_{4}^{7}\right)-E\left(M_{n}{ }^{2}\right)
$$

MA20225 COURSE NOTES ：

$$
\sum_{j=1}^{k} P_{2}\left(C_{3} \mid A_{1} C_{2}\right)=0.0
$$
Part (e) is mast ensily done by direct sabestitution, $\sum_{j=1}^{k} P\left(A_{j}^{f}\right]=\sum_{j=1}^{\infty}\left[1-P\left(A_{j}\right)\right]=k-\sum_{j=1}^{k} P\left(A_{j}\right)$
and gince we are told that the $\mathcal{A}{j} / s$ ser nutually exclusive and eollectizely exhaustive, we have $$ \sum{k=1}^{k} P\left(A_{j}^{r}\right)=k-1.0
$$
For part (ل), we ean uee the definition ol cotditional probabtlity and the given properties of listo 1 and 2 to $w$ rite
$$
\sum_{i=1}^{k} \sum_{i=1}^{2} P\left(A_{1}\right) P\left(B_{j} \mid A_{i}\right)=\sum_{i=1}^{k} \sum_{i=1}^{t} P\left(A_{4} B_{j}\right)=\sum_{i=1}^{k} P\left(A_{i}\right)=1.0
$$