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# 数学代写| 随机过程代考|Martingale convergence theorem

A population starts with one individual at time $n=0: Z_{0}=1$.

After one unit of time (at time $n=1$ ) the sole individual produces $Z_{1}$ identical clones of itself and dies. $Z_{1}$ is an $\mathbb{N}_{0}$-valued random variable.

(a) If $Z_{1}$ happens to be equal to 0 the population is dead and nothing happens at any future time $n \geq 2$.

(b) If $Z_{1}>0$, a unit of time later, each of $Z_{1}$ individuals gives birth to a random number of children and dies. The first one has $Z_{1,1}$ children, the second one $Z_{1,2}$ children, etc. The last, $Z_{1}^{\text {th }}$ one, gives birth to $Z_{1, Z_{1}}$ children. We assume that the distribution of the number of children is the same for each individual in every generation and independent of either the number of individuals in the generation and of the number of children the others have. This distribution, shared by all $Z_{n, i}$ and $Z_{1}$, is called the offspring distribution. The total number of individuals in the second generation is now
$$Z_{2}=\sum_{k=1}^{Z_{1}} Z_{1, k}$$
(c) The third, fourth, etc. generations are produced in the same way. If it ever happens that $Z_{n}=0$, for some $n$, then $Z_{m}=0$ for all $m \geq n$ – the population is extinct. Otherwise,
$$Z_{n+1}=\sum_{k=1}^{Z_{n}} Z_{n, k}$$

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## 数学代写| 随机过程代考|martingale with

Theorem. If $\left{S_{n}\right}$ is a martingale with $\mathbb{E}\left(S_{n}^{2}\right)<M<\infty$ for some $M$ and all $n$, then there exists a random variable $S$ such that $S_{n}$ converges to $S$ almost surely and in mean square.
This result has a more general version which, amongst other things,
(i) deals with submartingales,
(ii) imposes weaker moment conditions,
(iii) explores convergence in mean also,
but the proof of this is more difficult. On the other hand, the proof of (1) is within our grasp, and is only slightly more difficult than the proof of the strong law for independent sequences, Theorem (7.5.1); it mimics the traditional proof of the strong law and begins with a generalization of Chebyshov’s inequality. We return to the theory of martingales in much greater generality

## 数学代写| 随机过程代考|Markov chains

Example. Markov chains. Suppose that the chain $X_{0}, X_{1}, \ldots$ of Example (7.7.7) is irreducible and persistent, and let $\psi$ be a bounded function mapping $S$ into $\mathbb{R}$ which satisfies equation (7.7.8). Then the sequence $\left{S_{n}\right}$, given by $S_{n}=\psi\left(X_{n}\right)$, is a martingale and satisfies the condition $\mathbb{E}\left(S_{n}^{2}\right) \leq M$ for some $M$, by the boundedness of $\psi$. For any state $i$, the event $\left{X_{n}=i\right}$ occurs for infinitely many values of $n$ with probability 1 . However, $\left{S_{n}=\psi(i)\right} \supseteq$ $\left{X_{n}=i\right}$ and so
$$S_{n} \stackrel{\text { a.s. }}{\longrightarrow} \psi(i) \text { for all } i,$$
which is clearly impossible unless $\psi(i)$ is the same for all $i$. We have shown that any bounded solution of equation (7.7.8) is constant.

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