https://file.aiquickdraw.com/mathbot/file/1728690830607wyhogyus.jpg

This excerpt from a textbook focuses on **conditional probability density functions (p.d.f.)**, particularly in a sequence of **random variables**. Here's a breakdown of the content:

### Conditional p.d.f. of \( Z \) given \( W = (X_1, X_2) = (x_1, x_2) \):
- The problem involves five random variables \( X_1, \dots, X_5 \), with the assumption that these variables are **conditionally independent** and **identically distributed (i.i.d.)** given a random variable \( Z = z \).
- The **conditional p.d.f.** of the sequence \( (X_3, X_4, X_5) \) given \( (X_1 = x_1, X_2 = x_2) \) and \( Z = z \) is computed using **Bayes' Theorem**.

### Step-by-Step Process:
1. **Finding the conditional p.d.f. of** \( (X_3, X_4, X_5) \) given \( Z = z \) and \( W = (x_1, x_2) \):
   \[
   g_1(y|z, w) = z^3 e^{-z(x_3 + x_4 + x_5)}, \quad \text{for } x_3, x_4, x_5 > 0.
   \]

2. **Using the previous p.d.f.** of \( Z \) given \( W = w \) from Eq. (3.7.13):
   \[
   f_2(z|w) = \frac{1}{2}(2 + x_1 + x_2)^3 z^2 e^{-z(2 + x_1 + x_2)}.
   \]

3. **Using the p.d.f.** of the last three observations \( X_3, X_4, X_5 \) given \( X_1, X_2 \), as calculated in a previous example (3.7.14):
   \[
   f_1(y|w) = \frac{60(2 + x_1 + x_2)^3}{(2 + x_1 + \dots + x_5)^6}.
   \]

4. **Final step - Applying Bayes’ Theorem** to combine the p.d.f.s:
   \[
   g_2(z|y, w) = \frac{z^3 e^{-z(x_3 + x_4 + x_5)} \frac{1}{2}(2 + x_1 + x_2)^3 z^2 e^{-z(2 + x_1 + x_2)}}{\frac{60(2 + x_1 + x_2)^3}{(2 + x_1 + \dots + x_5)^6}}.
   \]
   Simplifying gives the final result:
   \[
   g_2(z|y, w) = \frac{1}{120}(2 + x_1 + \dots + x_5)^6 z^5 e^{-z(2 + x_1 + \dots + x_5)}, \quad \text{for } z \geq 0.
   \]

This section illustrates how Bayes' Theorem is used to combine different conditional distributions in a sequence of random variables.

### Do you want more details or have any questions about this?

### Here are five related questions to explore further:
1. What are the key assumptions behind using Bayes' Theorem in this context?
2. How is the i.i.d. assumption used in simplifying these probability density functions?
3. Can you explain the role of the exponential function in these probability density functions?
4. What are some applications of conditional p.d.f.s in statistical inference or machine learning?
5. How would you compute the marginal p.d.f. of \( Z \) in this case?

### Tip: 
When simplifying expressions that involve exponentials and products of terms, focus on common factors to reduce complexity effectively.

Conditional probability density functions (p.d.f.) of Z given (X1, X2) and the use of Bayes' Theorem in the sequence of random variables.

This page explores conditional probability density functions (p.d.f.) for Z given (X1, X2) in a sequence of random variables using Bayes' Theorem. It includes key concepts such as conditional probability, exponential distribution, and the use of i.i.d. assumptions. Detailed step-by-step calculations are provided.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation