# 统计代写|随机过程代写stochastic process代考|MTH7090

## 统计代写|随机过程代写stochastic process代考|Distribution of Total Number of Progeny

Let $X_n$ denote the size of the $n$th generation $n=0,1,2, \ldots$ and $X_0=1$.
Then the random variable $Y_n=\sum_{k=0}^n X_k=1+X_1+\ldots+X_n$
denotes the total number of progeny, i.e. the total number of descendents upto and including the $n$th generation and also including the ancestors.
Theorem 5.4 The G.F. $R_n(s)$ of $Y_n$ satisfies the recursion relation
$$R_n(s)=s \varnothing\left(R_{n-1}(s)\right),$$
$\varnothing$ being the G.F. of the offspring distribution.
Proof Let $Z_n=X_1+X_2+\ldots+X_n$ and $G_n(s)$ be its G.F.
Then $R_n(s)=s G_n(s)\left(Y_n=1+Z_n\right.$ and G.F. of 1 is $\left.s\right)$.
We have $G_n(s)=\sum_{k=0}^{\infty} P\left(Z_n=k\right) s^k$.
$P\left(Z_n=k\right)=\sum_{i=0}^{\infty} P$ [total number of descendents in the succeeding $(n-1)$
generations following the first generation is $k-i$ given $\left.X_1=i\right] P\left[X_1=i\right]$.
If the process starts with one ancestor then the probability of having $r$ descendents in succeeding $m$ generations is the coefficient of $s^r$ in $G_m(s)$ and if it starts with $i$ ancestors then the probability of having $r$ descendents in the succeeding $m$ generations will be the coefficient of $s^r$ in $\left[G_m(s)\right]^i$.
Thus
\begin{aligned} P\left(Z_n\right.&=k)=\sum_{i=0}^{\infty} \text { coefficient of } s^{k-i} \text { in }\left{G_{n-1}(s)\right}^i p_i \ &\left.=\sum_{i=0}^{\infty} p_i \text { (coefficient of } s^k \text { in }\left{s G_{n-1}(s)\right}^i\right) \ &=\text { coefficient of } s^k \text { in } \sum_{i=0}^{\infty} p_i\left{s G_{n-1}(s)\right}^i \ &=\text { coefficient of } s^k \text { in } \varnothing\left(s G_{n-1}(s)\right) \end{aligned}

## 统计代写|随机过程代写stochastic process代考|Continuous Parameter Branching Process

Let $X(t)$ denote the size of the population at time $t$ so that $X(t)$ takes only nonnegative integral values. We consider only homogeneous branching process, that is, processes for which $p_{i j}\left(t_1, t_2\right)=p_{i j}\left(t_2-t_1\right)$. Let us consider at present, processes with a single type of object. Since in branching processes every object evolves independently of the others we may assume that at the initial instant only one object exists. With the passage of time $t$ it either disappears or turns into $k$ individuals that are all of the same type and constitute the first generation.
Every object of the first generation “lives” independently of the other, and the same probability laws obtain for it as for the initial object “the ancestor”. At some instant of time the object either disappears or turns into second generation objects and so forth. The entire process is described by a single integer valued stochastic process $X(t)$ that is equal to the number of objects existing at the instant $t$. By assumption $X(0)=1$ a.s.

The set of possible states of the process is the sequence of natural numbers 0 , $1,2,3, \ldots$ Here 0 be an absorbing state. If $X(t)=0$ then $X(t)=0$ for all $t>t_0$. If $X(t)=0$ a.s. for some $t$, then the process is said to be degenerate process. In other case $X(t)$ vanishes after a finite interval of time with probability $<1$. This probability is called the probability of degeneration of the process. It is possible that $X(t)$ increases without bound in time. In case of nuclear reaction this situation can be interpreted as explosive. Thus we can ask the questions: What is the probability of degeneration and what is the $\lim _{t \rightarrow \infty} X(t)$ ?

We state the formal definition of a continuous parameter Markov Branching process as follows.

Definition 5.1 A Stochastic process ${X(t), t \geq 0}$ with values on the set of nonnegative integers is called a continuous parameter Markov branching process if
(i) it is a stationary Markov chain
(ii) its transition probabilities $p_{i j}(t)=P(X(t)=j \mid X(0)=i$ ) satisfy the relation $p_{i j}(t)=p_{1 k}^{(t)}(t)$, the $i$ fold convolution of the probabilities.

# 随机过程代考

## 统计代写|随机过程代写stochastic process代考|Distribution of Total Number of Progeny

$$R_n(s)=s \varnothing\left(R_{n-1}(s)\right)$$
Ø作为后代分布的 $\mathrm{GF}$ 。

$P\left(Z_n=k\right)=\sum_{i=0}^{\infty} P[$ 后代的总数 $(n-1)$

## 统计代写|随机过程代写stochastic process代考|Continuous Parameter Branching Process

(i) 它是一个平稳的马尔可夫链
(ii) 它的转移概率p一世j(吨)=P(X(吨)=j∣X(0)=一世) 满足关系p一世j(吨)=p1k(吨)(吨)， 这一世概率的折叠卷积。

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