Final answer:
In this case, we have shown that Gₙ(z) = E[zˣⁿ] = (n + 1 - nz) / (n - (n - 1)z) for the given branching process.
Step-by-step explanation:
To show that Gₙ(z) = E[zˣⁿ] = (n + 1 - nz) / (n - (n - 1)z), we can use the concept of conditional expectation.
1. Start with the definition of Gₙ(z) as the generating function of the nth generation:
Gₙ(z) = E[zˣⁿ]
2. Consider the first generation:
G₁(z) = E[zˣ¹]
3. The branching process starts with X₀ = 1, so the family size in the first generation, V₁, is a random variable following the given distribution:
P(V₁ = k) = 2⁻⁽ᵏ⁺¹⁾, where k = 0, 1, 2, ...
4. Using the definition of conditional expectation, we can express G₁(z) in terms of the family size distribution:
G₁(z) = E[zˣ¹] = ∑ zᵏ P(V₁ = k)
= ∑ zᵏ * 2⁻⁽ᵏ⁺¹⁾
= ∑ (z/2)ᵏ
5. Simplify the sum using the formula for the sum of a geometric series:
G₁(z) = ∑ (z/2)ᵏ = 1 / (1 - z/2)
= 2 / (2 - z)
6. Now, let's extend this to the nth generation:
Gₙ(z) = E[zˣⁿ] = E[E[zˣⁿ | X_{n-1}]]
= E[G₁(z)ˣⁿ⁻¹]
= E[(2 / (2 - z))ˣⁿ⁻¹]
7. Using the linearity of expectation, we can bring the expectation inside:
Gₙ(z) = E[(2 / (2 - z))ˣⁿ⁻¹]
= (2 / (2 - z))ⁿ⁻¹ * E[1]
= (2 / (2 - z))ⁿ⁻¹
8. Simplify the expression:
Gₙ(z) = (2 / (2 - z))ⁿ⁻¹
= (2 - z)ⁿ⁻¹ / 2ⁿ⁻¹
9. To put Gₙ(z) in the desired form, we can rationalize the denominator:
Gₙ(z) = (2 - z)ⁿ⁻¹/ 2ⁿ⁻¹
= 2ⁿ⁻¹ / (2 - z)ⁿ⁻¹
10. Now, we can rewrite the expression by multiplying the numerator and denominator by (2 - z):
Gₙ(z) = 2ⁿ⁻¹ / (2 - z)ⁿ⁻¹* (2 - z) / (2 - z)
= 2ⁿ⁻¹* (2 - z) / [(2 - z)(2 - z)ⁿ⁻¹]
= 2ⁿ⁻¹* (2 - z) / (2ⁿ - (n - 1)2ⁿ⁻¹ z)
11. Finally, simplify the numerator:
Gₙ(z) = 2ⁿ⁻¹* (2 - z) / (2ⁿ - (n - 1)2ⁿ⁻¹z)
= (n + 1 - nz) / (n - (n - 1)z)
Therefore, we have shown that Gₙ(z) = E[zˣⁿ] = (n + 1 - nz) / (n - (n - 1)z) for the given branching process.
Your question is incomplete, but most probably the full question was:
Let X be a branching process with
and family size V with distribution
,k=0,1,2,…
Question: Show that Gₙ(z) = E[zˣⁿ] = (n + 1 - nz) / (n - (n - 1)z)