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The image discusses conditional probability density functions (p.d.f.) of random variables and the use of Bayes' theorem to compute a conditional p.d.f. for a set of random variables in sequence. Let me break it down step by step:

### Problem Context:
- The setup is to compute the conditional p.d.f. of \( Z \), given \( X_1, X_2, X_3, X_4, X_5 \), where each of these random variables are conditionally independent and identically distributed (i.i.d.) given \( Z = z \).
- The main goal is to find the conditional p.d.f. of \( Z \) given the entire sequence \( (X_1, X_2, X_3, X_4, X_5) \), i.e., \( p(Z | X_1, X_2, X_3, X_4, X_5) \).

### Step-by-Step Breakdown:

1. **First Conditional p.d.f.:**
   - The text starts by calculating the conditional p.d.f. \( g_1(y | z, w) \), where \( w = (X_1, X_2) = (x_1, x_2) \), for the remaining variables \( X_3, X_4, X_5 \).
   - This is given by the product of the conditional densities of \( X_3, X_4, X_5 \), all conditional on \( Z = z \). Since \( X_3, X_4, X_5 \) are conditionally i.i.d., we get the expression:
     \[
     g_1(y | z, w) = z^3 e^{-(x_3 + x_4 + x_5)} \quad \text{for} \quad x_3, x_4, x_5 > 0
     \]

2. **Second Conditional p.d.f.:**
   - The conditional p.d.f. of \( Z \) given \( W = w \) is denoted as \( f_2(z | w) \). From equation 3.7.13, this is provided as:
     \[
     f_2(z | w) = \frac{1}{2} (2 + x_1 + x_2)^2 z^2 e^{-z(2 + x_1 + x_2)}
     \]

3. **Third Conditional p.d.f.:**
   - The conditional p.d.f. of the last three observations \( X_3, X_4, X_5 \) given the first two \( X_1, X_2 \) is taken from Example 3.7.14 and given by:
     \[
     f_1(y | w) = \frac{60 (2 + x_1 + x_2)^3}{(2 + x_1 + \dots + x_5)^6}
     \]

4. **Applying Bayes’ Theorem:**
   - To combine the information from all these conditional distributions, Bayes' theorem is used. The theorem combines the conditional p.d.f.s into:
     \[
     g_2(z | y, w) = \frac{z^3 e^{-(x_3 + x_4 + x_5)} \frac{1}{2} (2 + x_1 + x_2)^2 z^2 e^{-z(2 + x_1 + x_2)}}{\frac{60 (2 + x_1 + x_2)^3}{(2 + x_1 + \dots + x_5)^6}}
     \]
     Simplifying this, you get:
     \[
     g_2(z | y, w) = \frac{1}{120} (2 + x_1 + \dots + x_5)^5 z^5 e^{-z(2 + x_1 + \dots + x_5)}
     \]

This results in the final conditional p.d.f. \( g_2(z | y, w) \).

### Key Takeaways:
- The formula relies heavily on the independence assumption of the random variables given \( Z \), as well as the structure of Bayes’ theorem to combine the conditional distributions.
- The exponential terms in the distributions reflect the typical behavior of exponential families, and the powers of \( z \) correspond to the dimensionality of the conditioning variables.

Would you like more details on how the terms were derived or how Bayes' theorem applies here?

### Related Questions:
1. What are the key properties of i.i.d. random variables in conditional distributions?
2. How does Bayes' theorem work in the context of continuous random variables?
3. What is the role of the exponential distribution in this context?
4. Why is the conditional independence assumption important in this problem?
5. How can we generalize this approach to larger sequences of random variables?

### Tip:
When using Bayes' theorem for continuous distributions, always ensure that the normalizing constant accounts for the total probability over the support of the distribution.

The user is studying the use of Bayes' theorem and conditional probability density functions (p.d.f.) for random variables, specifically the calculation of p.d.f. of Z given multiple random variables X1, X2, X3, X4, and X5.

Explore how to apply Bayes' theorem to compute the conditional probability density function (p.d.f.) of a random variable Z, given five other random variables (X1, X2, X3, X4, and X5). This detailed solution covers conditional probability, independence, and key formulas related to conditional p.d.f.s.

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