In a branching process, $Z_n$ represents the number of individuals in the $n$-th generation, and $X$ denotes the number of offspring produced by an individual. Given that $E[X] = 1$ and $Var[X] = \sigma^2$, we want to prove that $Var[Z_n] = n\sigma^2$.

Proof:

1. Initial Condition: Start with $Z_0 = 1$. This indicates that there is one individual at generation 0.

2. Recursion Relation: The number of individuals in the next generation can be expressed as: $Z_{n} = \sum_{i=1}^{Z_{n-1}} X_i$ where $X_i$ represents the number of offspring produced by the $i$-th individual in generation $n-1$.

3. Expectation: [ E[Z_n] = E\left[\sum_{i=1}^{Z_{n-1}} X_i\right] = E[Z_{n-1}] \cdot E[X] ] Since $E[X] = 1$, [ E[Z_n] = E[Z_{n-1}] \quad \text{and recursively} \quad E[Z_n] = E[Z_0] = 1. ] Therefore, $E[Z_n] = 1$ for all $n$.

4. Variance: Now we compute the variance: $Var[Z_n] = Var\left(\sum_{i=1}^{Z_{n-1}} X_i\right).$ By the law of total variance, we can express this as: [ Var[Z_n] = E[Var[Z_n | Z_{n-1}]] + Var[E[Z_n | Z_{n-1}]]. ]

5. First Term: The first term can be computed as follows: [ Var[Z_n | Z_{n-1}] = Z_{n-1} \cdot Var[X] = Z_{n-1} \cdot \sigma^2. ] Therefore, [ E[Var[Z_n | Z_{n-1}]] = E[Z_{n-1}] \cdot \sigma^2 = 1 \cdot \sigma^2 = \sigma^2. ]

6. Second Term: The second term involves the expectation: [ Var[E[Z_n | Z_{n-1}]] = Var[Z_{n-1}] = Var[Z_{n-1}]. ]

7. Putting It Together: Combining these, we have: [ Var[Z_n] = \sigma^2 + Var[Z_{n-1}]. ] This gives us a recursive relation: [ Var[Z_n] = Var[Z_{n-1}] + \sigma^2. ]

8. Base Case: Starting with $Var[Z_0] = 0$ (since $Z_0 = 1$), $Var[Z_1] = 0 + \sigma^2 = \sigma^2,$ $Var[Z_2] = \sigma^2 + \sigma^2 = 2\sigma^2,$ $Var[Z_3] = 2\sigma^2 + \sigma^2 = 3\sigma^2,$ Continuing this way, we find that: $Var[Z_n] = n\sigma^2.$

Conclusion:

Thus, we have proved that $Var[Z_n] = n\sigma^2$.

If you have any questions or need further details, feel free to ask! Here are some related questions you might consider:

1. What assumptions are made about the distribution of offspring in this branching process?
2. How does the proof change if $E[X] \neq 1$?
3. Can this result be generalized to other stochastic processes?
4. What is the significance of the variance in a branching process?
5. How does the variance of the offspring distribution affect the overall process?

Tip: Understanding the law of total variance is crucial for analyzing such recursive processes!