Q.7.44 - A First Course in Probability

Question

Consider a population consisting of individuals able to produce offspring of the same kind. Suppose that by the end of its lifetime, each individual will have produced j new offspring with probability P_j, $j\ge 0$, independently of the number produced by any other individual. The number of individuals initially present, denoted by X₀, is called the size of the zeroth generation. All offspring of the zeroth generation constitute the first generation, and their number is denoted by X1. In general, let X_n denote the size of the nth generation. Let $\mu =\sum _{j=0}^{\mathrm{\infty }} j{P}_j$ and ${\sigma }^2=\sum _{j=0}^{\mathrm{\infty }} (j-\mu {)}^2{P}_j$ denote, respectively, the mean and the variance of the number of offspring produced by a single individual. Suppose that X₀ = 1— that is, initially there is a single individual in the population 

(a) Show that $E\left[X_n\right]=\mu E\left[X_{n-1}\right]$.

(b) Use part (a) to conclude that $E\left[X_n\right]={\mu }^n$

(c) Show that $\mathrm{Var}\left(X_n\right)={\sigma }^2{\mu }^{n-1}+{\mu }^2\mathrm{Var}\left(X_{n-1}\right)$

(d) Use part (c) to conclude that 

$\mathrm{Var}\left(X_n\right)=\left\{\begin{array}{l}{\sigma }^2{\mu }^{n-1}\left(\frac{{\mu }^n-1}{\mu -1}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }\mu \ne 1\\ n{\sigma }^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }\mu =1\end{array}\right.$

The model just described is known as a branching process, and an important question for a population that evolves along such lines is the probability that the population will eventually die out. Let π denote this probability when the population starts with a single individual. That is, 

$\left.\pi =P\text{ {population eventually dies out }\mid X_0=1\right)$

(e) Argue that π satisfies 

$\pi =\sum _{j=0}^{\mathrm{\infty }} P_j{\pi }^j$

Step-by-Step Solution

Answer

1. It has been showed that $E\left(X_n\right)=\mu E\left(X_{n-1}\right)$.
2. It has been shown and conclude that $E\left(X_n\right)={\mu }^n$.
3. It has been shown that $\mathrm{Var}\left(X_n\right)={\sigma }^2{\mu }^{n-1}+{\mu }^2\mathrm{Var}\left(X_{n-1}\right)$.
4. It has been shown that $\mathrm{Var}\left(X_n\right)=\left\{\begin{array}{l}{\sigma }^2{\mu }^{n-1}\left(\frac{{\mu }^n-1}{\mu -1}\right),\mu \ne 1\\ n{\sigma }^2,\mu =1\end{array}\right.$
5. It has been shown that The argument is true and it is $P\left(\text{ population eventually dies out }\right)=\sum _{j=0}^{\mathrm{\infty }} {\pi }^j{P}_j$.

Step 1: Given information (Part a)

Each individual will have produced j new offspring with probability P_j, $j\ge 0$, independently of the number produced by any other individual. The number of individuals initially present, denoted by X₀, is called the size of the zeroth generation. 

Step 2: Solution (Part a)

The solution is,

$\mu =\sum _{j=0}^{\mathrm{\infty }} j{P}_j$

${\sigma }^2=\sum _{j=0}^{\mathrm{\infty }} (j-\mu {)}^2{P}_j$

where $P_j$ is probability of producing new offspring.

$X_0=1$

Let,

$X_n=\sum _{i=1}^{X_{n-1}} Z_i$

where $Z_i$ is the number of offspring of the $i^{th}$ individual of the ${(n-1)}^{st}$ generation, so:

By conditioning on $X_{n-1}$, it can be evaluated that:

$E\left[X_n\right]=E\left[E\left[X_n\mid X_{n-1}\right]\right.$

$=E\left[E\left[\sum _{i=1}^{X_{n-1}} Z_i\mid X_{n-1}\right]\right]$

$=E\left[\mu X_{n-1}\right]$

By applying expectation in above:

$E\left(X_n\right)=\mu E\left(X_{n-1}\right)$

Step 3: Final answer (Part a)

Therefore, it has been showed that $E\left(X_n\right)=\mu E\left(X_{n-1}\right)$.

Step 4: Given information (Part b)

Each individual will have produced j new offspring with probability P_j, $j\ge 0$, independently of the number produced by any other individual. The number of individuals initially present, denoted by X₀, is called the size of the zeroth generation. 

Step 5: Solution (Part b)

The calculation are 

Suppose that the claim is true for $n-1$. Using the relation from part (a), so:

$E\left[X_1\right]=\mu$

$E\left[X_2\right]=\mu E\left[X_1\right]$

$={\mu }^2$

Similarly,

$E\left[X_n\right]=\mu E\left[X_{n-1}\right]$

$={\mu }^n$

Or

$E\left(X_n\right)={\mu }^{n}E\left(X_0\right)$

$={\mu }^n\times 1$

$={\mu }^n$

Step 6: Final answer (Part b)

It has been shown and conclude that $E\left(X_n\right)={\mu }^n$

Step 7: Given information (Part c)

Each individual will have produced j new offspring with probability P_j, $j\ge 0$, independently of the number produced by any other individual. The number of individuals initially present, denoted by X₀, is called the size of the zeroth generation. 

Step 8:Solution (Part c)

Using law of the total variance, so:

$\mathrm{Var}\left(X_n\right)=E\left(\mathrm{Var}\left(X_n\mid X_{n-1}\right)+\mathrm{Var}\left(E\left(X_n\mid X_{n-1}\right)\right)\right.$

Now, given that $X_{n-1},X_n$ is just the sum of $X_{n-1}$ independent random variables each having the distribution $\left\{P_j,j\ge 0\right\}$, hence,

$\mathrm{Var}\left(X_n\mid X_{n-1}\right)=X_{n-1}{\sigma }^2$

The conditional variance formula yields,

$\mathrm{Var}\left(X_n\right)=E\left[X_{n-1}{\sigma }^2\right]+\mathrm{Var}\left(X_{n-1}\mu \right)$

$={\sigma }^2{\mu }^{n-1}+{\mu }^2\mathrm{Var}\left(X_{n-1}\right)$

Step 9: Final answer (Part c)

Therefore it has been shown that $\mathrm{Var}\left(X_n\right)={\sigma }^2{\mu }^{n-1}+{\mu }^2\mathrm{Var}\left(X_{n-1}\right)$

Step 10: Given information (Part d)

Each individual will have produced j new offspring with probability P_j, $j\ge 0$, independently of the number produced by any other individual. The number of individuals initially present, denoted by X₀, is called the size of the zeroth generation. 

Step 11: Solution (Part d)

Now, from result of part (c), we have:

$\mathrm{Var}\left(X_n\right)={\mu }^2\mathrm{Var}\left(X_{n-1}\right)+{\sigma }^2{\mu }^{n-1}$

$={\mu }^2\left({\mu }^2\mathrm{Var}\left(X_{n-2}\right)+{\sigma }^2{\mu }^{n-1}\right)+{\sigma }^2{\mu }^{n-1}$

$\dots \dots \dots ={\mu }^2\left({\mu }^2\left({\mu }^2\mathrm{Var}\left(X_{n-3}\right)+{\sigma }^2{\mu }^{n-1}\right)+{\sigma }^2{\mu }^{n-1}\right)+{\sigma }^2{\mu }^{n-1}$

Now Expanding terms in above expression and putting $\mu =1$, so:

$\mathrm{Var}\left(X_n\right)=n{\sigma }^2$

Step 12: Solution (Part d)

Not let's assume that $\mu \ne 1$, as $Var\left(X_1\right)={\sigma }^2$ which is true, suppose that the claim is true for $n-1$, using the relation from part (c), we have:

$\mathrm{Var}\left(X_n\right)={\sigma }^2{\mu }^{n-1}+{\mu }^2\mathrm{Var}\left(X_{n-1}\right)$

$={\sigma }^2{\mu }^{n-1}+{\mu }^2{\sigma }^2{\mu }^{n-2}\left(\frac{{\mu }^{n-1}-1}{\mu -1}\right)$

$={\sigma }^2{\mu }^{n-1}\left(1+\mu \left(\frac{{\mu }^{n-1}-1}{\mu -1}\right)\right)$

Now, simplifying above expression:

$\mathrm{Var}\left(X_n\right)={\sigma }^2{\mu }^{n-1}\left(1+\frac{{\mu }^n-\mu }{\mu -1}\right)$

$={\sigma }^2{\mu }^{n-1}\left(\frac{\mu -1+{\mu }^n-\mu }{\mu -1}\right)$

$\mathrm{Var}\left(X_n\right)={\sigma }^2{\mu }^{n-1}\left(\frac{{\mu }^n-1}{\mu -1}\right)$

Hence,

$\mathrm{Var}\left(X_n\right)={\sigma }^2{\mu }^{n-1}\left(\frac{{\mu }^n-1}{\mu -1}\right)$

Thus,

$\mathrm{Var}\left(X_n\right)=\left\{\begin{array}{l}{\sigma }^2{\mu }^{n-1}\left(\frac{{\mu }^n-1}{\mu -1}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em},\mu \ne 1\\ n{\sigma }^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em},\mu =1\end{array}\right.$

Step 13: Final answer (Part d)

Therefore it has been shown that $\mathrm{Var}\left(X_n\right)=\left\{\begin{array}{l}{\sigma }^2{\mu }^{n-1}\left(\frac{{\mu }^n-1}{\mu -1}\right),\mu \ne 1\\ n{\sigma }^2,\mu =1\end{array}\right.$

Step 14: Given information (Part e)

Each individual will have produced j new offspring with probability P_j, $j\ge 0$, independently of the number produced by any other individual. The number of individuals initially present, denoted by X₀, is called the size of the zeroth generation. 

Step 15: Solution (part e)

Now we need to use the law of total probability:

$P\left(\text{ population eventually dies out }\right)=\sum _{j=0}^{\mathrm{\infty }} P\left(\text{ population eventually dies out }\mid X_1=j\right)P\left(X_1\right)$

Notice that if we comprehend that $X_1=j$, believe the branching process as j independent branching processes that begin from the beginning. The probability that each of them fails eventually is $\pi$, Hence:

$P\left(\text{ population eventually dies out }\mid X_1=k\right)={\pi }^j$

On the other side it is given that $P\left(X_1=k\right)=P_j$, therefore:

$P\left(\text{ population eventually dies out }\right)=\sum _{j=0}^{\mathrm{\infty }} {\pi }^j{P}_j$

Step 16: Final answer (Part e)

Therefore it has been shown that The argument is true and it is $P\left(\text{ population eventually dies out }\right)=\sum _{j=0}^{\mathrm{\infty }} {\pi }^j{P}_j$