Math Problem Statement

I need help with 2) and 3)

I need help with 2) and 3)

Solution

Let's solve the questions step by step.

Question 2

We are tasked with expressing E[f(T)]\mathbb{E}[f(T)] as an integral over Ω\Omega. Here is the step-by-step approach:

  1. Understanding the setup:
    T:(E,E)(F,F)T: (E, \mathcal{E}) \to (F, \mathcal{F}) is a measurable mapping.
    f:(F,F)([0,],B(R))f: (F, \mathcal{F}) \to ([0, \infty], \mathcal{B}(\mathbb{R})) is a measurable function.

    The composition f(T):ΩR+f(T): \Omega \to \mathbb{R}_+ is measurable, allowing us to define expectations.

  2. Expected value over Ω\Omega:
    By definition, the expected value of f(T)f(T) is given by: E[f(T)]=Ωf(T(ω))dP(ω),\mathbb{E}[f(T)] = \int_\Omega f(T(\omega)) \, d\mathbb{P}(\omega), where P\mathbb{P} is the probability measure on Ω\Omega.

  3. Expressing using change of measure:
    TT induces a pushforward measure PY=PY1\mathbb{P}_Y = \mathbb{P} \circ Y^{-1} on (E,E)(E, \mathcal{E}), which is the distribution of TT. Using the change of variables formula: E[f(T)]=Ef(y)dPY(y),\mathbb{E}[f(T)] = \int_E f(y) \, d\mathbb{P}_Y(y), where y=T(ω)y = T(\omega) under the mapping.

    This equation expresses E[f(T)]\mathbb{E}[f(T)] as an integral over EE.


Question 3

We are tasked with proving that: i=1nXiBin(nk,p),\sum_{i=1}^n X_i \sim \text{Bin}(nk, p), where X1,,XnX_1, \dots, X_n are i.i.d. random variables, each XiBin(k,p)X_i \sim \text{Bin}(k, p).

  1. Definition of XiX_i:
    Each XiBin(k,p)X_i \sim \text{Bin}(k, p) means that XiX_i counts the number of successes in kk independent trials, each with success probability pp. The probability mass function is: P(Xi=m)=(km)pm(1p)km,m{0,1,,k}.\mathbb{P}(X_i = m) = \binom{k}{m} p^m (1-p)^{k-m}, \quad m \in \{0, 1, \dots, k\}.

  2. Sum of independent binomial variables:
    The sum S=i=1nXiS = \sum_{i=1}^n X_i represents the total number of successes in nknk trials. Since the trials across all XiX_i are independent, SS follows a Binomial distribution: SBin(nk,p).S \sim \text{Bin}(nk, p).

  3. Justification using properties of Binomial distributions:

    • The total number of trials is n×kn \times k, because each XiX_i contributes kk trials.
    • The probability of success pp remains unchanged since each trial is independent.
    • Therefore, the sum of nn independent Bin(k,p)\text{Bin}(k, p) random variables is Bin(nk,p)\text{Bin}(nk, p).
  4. Hint application (Ω={0,1}nk\Omega = \{0, 1\}^{nk}):
    Using Ω={0,1}nk\Omega = \{0, 1\}^{nk}, we can interpret each trial as an indicator variable over Ω\Omega. The sum i=1nXi\sum_{i=1}^n X_i aggregates the results over all nknk trials, confirming that SBin(nk,p)S \sim \text{Bin}(nk, p).


Do you need further clarifications or more detailed steps? Here are 5 related questions to explore:

  1. How does the pushforward measure PY\mathbb{P}_Y simplify calculations in measure theory?
  2. Can you provide another example of changing variables in integrals with a stochastic mapping?
  3. What are the key properties of a Binomial distribution?
  4. How does the Central Limit Theorem relate to summing independent Binomial variables?
  5. What happens if the success probability pp is different for each random variable XiX_i?

Tip: When dealing with sums of independent random variables, always check if they share the same distribution type to simplify computations.

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Math Problem Analysis

Mathematical Concepts

Measure Theory
Probability Theory
Expected Value
Binomial Distribution

Formulas

E[f(T)] = ∫_Ω f(T(ω)) dP(ω)
E[f(T)] = ∫_E f(y) dP_Y(y)
Sum of Binomial distributions: ∑ X_i ~ Bin(nk, p)

Theorems

Change of Variables Formula in Measure Theory
Properties of Binomial Distribution

Suitable Grade Level

University (Advanced Mathematics or Probability)