Math Problem Statement

suppose X is a geometric random variable with P (X = x) = (.5)^(x+1), x ∈ {0, 1, 2, . . .}. (i) Find the generating function GX (t). (ii) If Gn(t) is the generating function for Zn where Z0 = 1 and Zn = X1 + · · · + XZn−1 , show that Gn(t) = Gn−1(GX (t)).

Solution

(i) Finding the generating function GX(t)G_X(t)

The generating function of a discrete random variable XX, denoted GX(t)G_X(t), is defined as:

GX(t)=E[tX]=x=0P(X=x)txG_X(t) = \mathbb{E}[t^X] = \sum_{x=0}^{\infty} P(X = x) t^x

Given P(X=x)=(0.5)x+1P(X = x) = (0.5)^{x+1}, we substitute this into the formula for the generating function:

GX(t)=x=0(0.5)x+1txG_X(t) = \sum_{x=0}^{\infty} (0.5)^{x+1} t^x

Factor out 0.50.5 from the sum:

GX(t)=0.5x=0(t2)xG_X(t) = 0.5 \sum_{x=0}^{\infty} \left( \frac{t}{2} \right)^x

Recognize that the summation is a geometric series of the form:

x=0rx=11rforr<1\sum_{x=0}^{\infty} r^x = \frac{1}{1 - r} \quad \text{for} \quad |r| < 1

Here, r=t2r = \frac{t}{2}, so the generating function becomes:

GX(t)=0.5(11t2)G_X(t) = 0.5 \left( \frac{1}{1 - \frac{t}{2}} \right)

Thus, the generating function for XX is:

GX(t)=0.51t2=12tG_X(t) = \frac{0.5}{1 - \frac{t}{2}} = \frac{1}{2 - t}

(ii) Showing that Gn(t)=Gn1(GX(t))G_n(t) = G_{n-1}(G_X(t))

Given that Gn(t)G_n(t) is the generating function for ZnZ_n, and Zn=X1+X2++XZn1Z_n = X_1 + X_2 + \dots + X_{Z_{n-1}}, we aim to show that:

Gn(t)=Gn1(GX(t))G_n(t) = G_{n-1}(G_X(t))

We proceed by induction on nn:

  1. Base Case: n=1n = 1

    By the given definition, Z0=1Z_0 = 1, and so G0(t)=t1=tG_0(t) = t^1 = t.

  2. Inductive Hypothesis:

    Suppose the result holds for n1n-1, i.e., Gn1(t)=Gn2(GX(t))G_{n-1}(t) = G_{n-2}(G_X(t)).

  3. Inductive Step:

    We need to show that Gn(t)=Gn1(GX(t))G_n(t) = G_{n-1}(G_X(t)).

    By the definition of generating functions, Gn(t)=E[tZn]G_n(t) = \mathbb{E}[t^{Z_n}], and since Zn=X1+X2++XZn1Z_n = X_1 + X_2 + \dots + X_{Z_{n-1}}, we consider:

    [ G_n(t) = \mathbb{E}\left[ t^{Z_n} \right] = \mathbb{E} \left[ t^{X_1 + \dots + X_{Z_{n-1}}} \right] ]

    By the Law of Total Expectation, we condition on Zn1Z_{n-1}:

    [ G_n(t) = \mathbb{E} \left[ \mathbb{E}\left[ t^{X_1 + \dots + X_{Z_{n-1}}} \mid Z_{n-1} \right] \right] ]

    Using the fact that the XiX_i are i.i.d., we get:

    E[tX1++XZn1Zn1=k]=(GX(t))k\mathbb{E} \left[ t^{X_1 + \dots + X_{Z_{n-1}}} \mid Z_{n-1} = k \right] = \left( G_X(t) \right)^k

    Thus, the generating function becomes:

    Gn(t)=E[(GX(t))Zn1]G_n(t) = \mathbb{E} \left[ \left( G_X(t) \right)^{Z_{n-1}} \right]

    But this is exactly Gn1(GX(t))G_{n-1}(G_X(t)), so we have:

    Gn(t)=Gn1(GX(t))G_n(t) = G_{n-1}(G_X(t))

Thus, by induction, the relation Gn(t)=Gn1(GX(t))G_n(t) = G_{n-1}(G_X(t)) holds for all n1n \geq 1.


Would you like more details on any part of this? Here are some related questions:

  1. How do you compute expected values from generating functions?
  2. What are the properties of geometric random variables?
  3. Can you explain how the total expectation theorem works in this context?
  4. What are the conditions for the convergence of a generating function?
  5. How does the generating function approach compare to moment-generating functions?

Tip: The generating function is a powerful tool for solving recurrence relations and finding distributions of sums of independent random variables.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Geometric Distribution
Generating Functions
Recurrence Relations

Formulas

Generating function formula: G_X(t) = Σ P(X = x) * t^x
Geometric series formula: Σ r^x = 1 / (1 - r) for |r| < 1
Recurrence relation: Gn(t) = Gn−1(GX(t))

Theorems

Law of Total Expectation
Geometric Series Summation

Suitable Grade Level

Undergraduate (Math or Probability Theory Course)