这是一份uwa西澳大学STAT3061的成功案例

随机过程及其应用|STAT3061 Random Processes and their Applications代写 UWA代写

For $p \geq 1 / 2$, we have to consider $f_{i}=\mathbb{P}{i}\left(T{i}<\infty\right)$. Writing $$ \left.\mathbb{P}{0}\left(T{0}<\infty\right)=1-q+q \mathbb{P}{1} \text { (hit } 0\right), $$ we see that if $\mathbb{P}{1}($ hit 0$)<1$, the chain is transient. But $$ \mathbb{P}{i}(\text { hit } i-1)=\frac{1-p}{p}, i \geq 1 $$ see Section 1.5. Hence, for $p>1 / 2$ the chain is transient. It remains to check the case $p=1 / 2$. Here, $f{i}=1$, and the chain is recurrent. The invariance equations $$ \pi_{i}=\frac{1}{2} \pi_{i-1}+\frac{1}{2} \pi_{i+1}, i>1,
$$
have the general solution $\pi_{i}=A+B i, i \geq 1$. At $i=1,0$ they have the form
$$
\pi_{1}=q \pi_{0}+\frac{1}{2} \pi_{2}, \pi_{0}=(1-q) \pi_{0}+\frac{1}{2} \pi_{1},
$$
which yields $B=0$ and
$$
\pi_{i} \equiv A, i \geq 1, \pi_{0}=\frac{1}{2 q} A
$$

STAT3061 COURSE NOTES ：

these probabilities do not depend on the value of $i$ because of the homogeneous property of the chain. Conditioning on the first jump and using the strong Markov property we get
$$
a=q+p b a^{2}, b=q+p b a
$$
whence
$$
b=\frac{q}{1-p a}, \text { and } a=q+\frac{p q a^{2}}{1-p a} .
$$
Thus,
$$
p(1+q) a^{2}-(p q+1) a+q=0
$$
and the solutions are
$$
a=1 \text { and } a=\frac{q}{1-q^{2}}
$$
We are interested in the minimal solution
$$
\frac{q}{1-q^{2}}<1 \text { if and only if } q<\frac{\sqrt{5}-1}{2} .
$$
Therefore, the chain is recurrent if and only if $q \geq(\sqrt{5}-1) / 2$ and transient if and only if $q<(\sqrt{5}-1) / 2$.