Math Problem Statement

prove step by step

Prove step by step

Solution

The equation you provided looks like an application of Bayes' Theorem with conditional probabilities, but in a slightly more complex form involving multiple variables. The goal is to prove this equation step-by-step:

p(xy,z)=p(yx,z)p(xz)p(yz)p(x | y, z) = \frac{p(y | x, z) p(x | z)}{p(y | z)}

We will use the rules of conditional probability to prove this equation.

Step-by-step proof:

  1. Start with the definition of conditional probability:

    Conditional probability is defined as: p(AB)=p(AB)p(B)p(A | B) = \frac{p(A \cap B)}{p(B)} This can also be written in terms of the joint probability: p(AB)=p(AB)p(B)p(A \cap B) = p(A | B) p(B)

  2. Express p(xy,z)p(x | y, z):

    Using the definition of conditional probability, we can write: p(xy,z)=p(x(yz))p(yz)p(x | y, z) = \frac{p(x \cap (y \cap z))}{p(y \cap z)} Since p(x,y,z)p(x, y, z) represents the joint probability of all three variables, we have: p(xy,z)=p(x,y,z)p(y,z)p(x | y, z) = \frac{p(x, y, z)}{p(y, z)}

  3. Factor p(x,y,z)p(x, y, z) into conditional probabilities:

    Now, we can factor the joint probability p(x,y,z)p(x, y, z) as follows: p(x,y,z)=p(yx,z)p(x,z)p(x, y, z) = p(y | x, z) p(x, z) Since p(x,z)=p(xz)p(z)p(x, z) = p(x | z) p(z), we substitute this into the equation: p(x,y,z)=p(yx,z)p(xz)p(z)p(x, y, z) = p(y | x, z) p(x | z) p(z)

  4. Simplify the expression:

    Substituting this back into the equation for p(xy,z)p(x | y, z), we get: p(xy,z)=p(yx,z)p(xz)p(z)p(y,z)p(x | y, z) = \frac{p(y | x, z) p(x | z) p(z)}{p(y, z)}

  5. Simplify the denominator:

    Now, we simplify p(y,z)p(y, z) using the marginal probability: p(y,z)=p(yz)p(z)p(y, z) = p(y | z) p(z)

  6. Substitute into the main equation:

    Substituting this expression for p(y,z)p(y, z) into the main equation: p(xy,z)=p(yx,z)p(xz)p(z)p(yz)p(z)p(x | y, z) = \frac{p(y | x, z) p(x | z) p(z)}{p(y | z) p(z)}

  7. Cancel out p(z)p(z):

    Since p(z)p(z) appears in both the numerator and denominator, it cancels out: p(xy,z)=p(yx,z)p(xz)p(yz)p(x | y, z) = \frac{p(y | x, z) p(x | z)}{p(y | z)}

Thus, we have derived the original equation: p(xy,z)=p(yx,z)p(xz)p(yz)p(x | y, z) = \frac{p(y | x, z) p(x | z)}{p(y | z)}

This completes the proof.


Would you like more details on any specific step? Or if you have any further questions, feel free to ask.

Here are 5 related questions to expand on this:

  1. What is Bayes' theorem, and how does it apply to conditional probabilities in general?
  2. How does this equation change if variables xx, yy, and zz are independent?
  3. What are some real-world applications of this form of Bayes' theorem?
  4. How does the marginalization process work in probability theory?
  5. Can this equation be extended to more than three variables?

Tip: In Bayesian inference, understanding how to manipulate joint and conditional probabilities is crucial, especially when dealing with multiple variables. Always ensure that you are clear on how to factor these probabilities properly.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Conditional Probability
Bayes' Theorem

Formulas

p(A | B) = p(A ∩ B) / p(B)
p(A ∩ B) = p(A | B) * p(B)
p(x | y, z) = (p(y | x, z) * p(x | z)) / p(y | z)

Theorems

Bayes' Theorem

Suitable Grade Level

Undergraduate