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.