Math Problem Statement

Calculate the conditional probability density function (p.d.f.) of Z given W = (x1, x2), with three more observations X3, X4, and X5, using Bayes' theorem in the context of conditioning on random variables in sequence.

Solution

The image you uploaded is a section discussing the conditional probability density function (p.d.f.) of random variables in a sequence using Bayes' theorem. The content involves:

  1. Conditional p.d.f. of ZZ given W=(x1,x2)W = (x_1, x_2), denoted f2(zw)f_2(z|w), which was calculated as: f2(zw)=12(2+x1+x2)3z2ez(2+x1+x2)f_2(z|w) = \frac{1}{2}(2 + x_1 + x_2)^3 z^2 e^{-z(2 + x_1 + x_2)}

  2. Conditional p.d.f. of the last three observations (X3,X4,X5)(X_3, X_4, X_5) given ZZ, W=(x1,x2)W = (x_1, x_2), denoted g1(yz,w)g_1(y|z,w), and calculated as: g1(yz,w)=z3ez(x3+x4+x5)forx3,x4,x5>0g_1(y|z, w) = z^3 e^{-z(x_3 + x_4 + x_5)} \quad \text{for} \quad x_3, x_4, x_5 > 0

  3. Conditional p.d.f. of the last three observations given the first two, f1(yw)f_1(y|w), calculated as: f1(yw)=60(2+x1+x2)3(2+x1++x5)6f_1(y|w) = \frac{60 (2 + x_1 + x_2)^3}{(2 + x_1 + \dots + x_5)^6}

  4. Finally, the combination of these results using Bayes' theorem (Example 3.7.17) to find g2(zy,w)g_2(z|y, w): g2(zy,w)=z3ez(x3+x4+x5)12(2+x1+x2)3z2ez(2+x1+x2)60(2+x1+x2)3(2+x1++x5)6g_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}} Simplified to: g2(zy,w)=1120(2+x1++x5)5z5ez(2+x1++x5)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)} for z>0z > 0.

This discussion focuses on conditioning random variables in sequence and calculating their conditional distributions using Bayes' theorem.

Would you like further clarification on any specific part or a breakdown of the steps?

Here are five questions that might help deepen your understanding:

  1. How is Bayes' theorem applied in the context of random variables here?
  2. What does it mean for the random variables X1,,X5X_1, \dots, X_5 to be conditionally i.i.d. given ZZ?
  3. How do the conditional p.d.f.'s f1(yw)f_1(y|w) and f2(zw)f_2(z|w) contribute to the final result?
  4. Why is the exponential term ez(2+x1++x5)e^{-z(2 + x_1 + \dots + x_5)} important in this conditional distribution?
  5. How would this method generalize if there were more than five observations?

Tip: When working with conditional probabilities, always consider how additional information (conditions) can affect the overall distribution and update your probability accordingly using tools like Bayes' theorem.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Conditional Probability
Probability Density Function (p.d.f.)
Bayes' Theorem

Formulas

f2(z|w) = (1/2)(2 + x1 + x2)^3 * z^2 * e^(-z(2 + x1 + x2))
g1(y|z, w) = z^3 * e^(-z(x3 + x4 + x5)) for x3, x4, x5 > 0
f1(y|w) = (60 * (2 + x1 + x2)^3) / (2 + x1 + ... + x5)^6
g2(z|y, w) = (1/120)(2 + x1 + ... + x5)^5 * z^5 * e^(-z(2 + x1 + ... + x5))

Theorems

Bayes' Theorem

Suitable Grade Level

Advanced Undergraduate/Graduate Level