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Each individual will have produced j new offspring with probability P_j, $j鈮�0$, 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.

The solution is,

$渭=\underset{j=0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�j{P}_j$

${蟽}^2=\underset{j=0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�(j鈭�渭{)}^2{P}_j$

where $P_j$is probability of producing new offspring.

$X_0=1$

Let,

$X_n=\underset{i=1}{\overset{X_{n鈭�1}}{鈭�}}鈥�Z_i$

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:

$E\left[X_n\right]=E\left[E\left[X_n鈭�X_{n鈭�1}\right]\right.$

$=E\left[E\left[\underset{i=1}{\overset{X_{n鈭�1}}{鈭�}}鈥�Z_i鈭�X_{n鈭�1}\right]\right]$

$=E\left[渭X_{n鈭�1}\right]$

By applying expectation in above:

$E\left(X_n\right)=渭E\left(X_{n鈭�1}\right)$

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

Each individual will have produced j new offspring with probability P_j, $j鈮�0$, 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.

The calculation are

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

$E\left[X_1\right]=渭$

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

$={渭}^2$

Similarly,

$E\left[X_n\right]=渭E\left[X_{n鈭�1}\right]$

$={渭}^n$

Or

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

$={渭}^n脳1$

$={渭}^n$

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

Each individual will have produced j new offspring with probability P_j, $j鈮�0$, 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.

Using law of the total variance, so:

$\mathrm{Var}鈦�\left(X_n\right)=E\left(\mathrm{Var}鈦�\left(X_n鈭�X_{n鈭�1}\right)+\mathrm{Var}鈦�\left(E\left(X_n鈭�X_{n鈭�1}\right)\right)\right.$

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

$\mathrm{Var}鈦�\left(X_n鈭�X_{n鈭�1}\right)=X_{n鈭�1}{蟽}^2$

The conditional variance formula yields,

$\mathrm{Var}鈦�\left(X_n\right)=E\left[X_{n鈭�1}{蟽}^2\right]+\mathrm{Var}鈦�\left(X_{n鈭�1}渭\right)$

$={蟽}^2{渭}^{n鈭�1}+{渭}^2\mathrm{Var}鈦�\left(X_{n鈭�1}\right)$

Therefore it has been shown that$\mathrm{Var}鈦�\left(X_n\right)={蟽}^2{渭}^{n鈭�1}+{渭}^2\mathrm{Var}鈦�\left(X_{n鈭�1}\right)$

Each individual will have produced j new offspring with probability P_j, $j鈮�0$, 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.

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

$\mathrm{Var}鈦�\left(X_n\right)={渭}^2\mathrm{Var}鈦�\left(X_{n鈭�1}\right)+{蟽}^2{渭}^{n鈭�1}$

$={渭}^2\left({渭}^2\mathrm{Var}鈦�\left(X_{n鈭�2}\right)+{蟽}^2{渭}^{n鈭�1}\right)+{蟽}^2{渭}^{n鈭�1}$

$鈥�鈥�鈥�={渭}^2\left({渭}^2\left({渭}^2\mathrm{Var}鈦�\left(X_{n鈭�3}\right)+{蟽}^2{渭}^{n鈭�1}\right)+{蟽}^2{渭}^{n鈭�1}\right)+{蟽}^2{渭}^{n鈭�1}$

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

$\mathrm{Var}鈦�\left(X_n\right)=n{蟽}^2$

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

$\mathrm{Var}鈦�\left(X_n\right)={蟽}^2{渭}^{n鈭�1}+{渭}^2\mathrm{Var}鈦�\left(X_{n鈭�1}\right)$

$={蟽}^2{渭}^{n鈭�1}+{渭}^2{蟽}^2{渭}^{n鈭�2}\left(\frac{{渭}^{n鈭�1}鈭�1}{渭鈭�1}\right)$

$={蟽}^2{渭}^{n鈭�1}\left(1+渭\left(\frac{{渭}^{n鈭�1}鈭�1}{渭鈭�1}\right)\right)$

Now, simplifying above expression:

$\mathrm{Var}鈦�\left(X_n\right)={蟽}^2{渭}^{n鈭�1}\left(1+\frac{{渭}^n鈭�渭}{渭鈭�1}\right)$

$={蟽}^2{渭}^{n鈭�1}\left(\frac{渭鈭�1+{渭}^n鈭�渭}{渭鈭�1}\right)$

$\mathrm{Var}鈦�\left(X_n\right)={蟽}^2{渭}^{n鈭�1}\left(\frac{{渭}^n鈭�1}{渭鈭�1}\right)$

Hence,

$\mathrm{Var}鈦�\left(X_n\right)={蟽}^2{渭}^{n鈭�1}\left(\frac{{渭}^n鈭�1}{渭鈭�1}\right)$

Thus,

$\mathrm{Var}鈦�\left(X_n\right)=\left\{\begin{array}{l}{蟽}^2{渭}^{n鈭�1}\left(\frac{{渭}^n鈭�1}{渭鈭�1}\right)鈥呪赌呪赌呪赌�,渭鈮�1\\ n{蟽}^2鈥呪赌呪赌呪赌�,渭=1\end{array}\right.$

Therefore it has been shown that$\mathrm{Var}鈦�\left(X_n\right)=\left\{\begin{array}{l}{蟽}^2{渭}^{n鈭�1}\left(\frac{{渭}^n鈭�1}{渭鈭�1}\right),渭鈮�1\\ n{蟽}^2,渭=1\end{array}\right.$

Each individual will have produced j new offspring with probability P_j, $j鈮�0$, 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.

Now we need to use the law of total probability:

$P\left(\text{population eventually dies out}\right)=\underset{j=0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�P\left(\text{population eventually dies out}鈭�X_1=j\right)P\left(X_1\right)$

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 $蟺$, Hence:

$P\left(\text{population eventually dies out}鈭�X_1=k\right)={蟺}^j$

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

$P\left(\text{population eventually dies out}\right)=\underset{j=0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�{蟺}^j{P}_j$

Therefore it has been shown that The argument is true and it is$P\left(\text{population eventually dies out}\right)=\underset{j=0}{\overset{\mathrm{鈭�}}{鈭�}}鈥�{蟺}^j{P}_j$