Math Problem Statement

prove step by step

prove step by step

Solution

The equation in the image represents an application of the law of total probability. Here is the step-by-step proof:

We are given the equation:

P(xz)=yP(xy,z)P(yz)P(x|z) = \sum_y P(x|y,z) P(y|z)

This formula relates the conditional probability P(xz)P(x|z) to other conditional probabilities and marginalizes over the variable yy. Let's break it down and prove it:

Step 1: Start with the Definition of Conditional Probability

By the definition of conditional probability:

P(xz)=P(x,z)P(z)P(x|z) = \frac{P(x, z)}{P(z)}

This gives the probability of xx given zz, which can be rewritten in terms of joint probabilities.

Step 2: Apply the Law of Total Probability

We apply the law of total probability. The law of total probability states that if yy is a variable that partitions the sample space, we can express the joint probability of xx and zz as a sum over the possible values of yy:

P(x,z)=yP(x,y,z)P(x, z) = \sum_y P(x, y, z)

This sums over all possible values of yy, and expresses the joint probability in terms of the marginal and conditional probabilities.

Step 3: Use the Chain Rule for Joint Probability

Next, we apply the chain rule of probability. The chain rule allows us to break down the joint probability P(x,y,z)P(x, y, z) as follows:

P(x,y,z)=P(xy,z)P(y,z)P(x, y, z) = P(x|y, z) P(y, z)

Now, using the chain rule again for P(y,z)P(y, z), we can decompose it into:

P(y,z)=P(yz)P(z)P(y, z) = P(y|z) P(z)

Thus, the joint probability P(x,y,z)P(x, y, z) becomes:

P(x,y,z)=P(xy,z)P(yz)P(z)P(x, y, z) = P(x|y, z) P(y|z) P(z)

Step 4: Substitute Back into the Total Probability Expression

Substituting this back into the expression for P(x,z)P(x, z), we get:

P(x,z)=yP(xy,z)P(yz)P(z)P(x, z) = \sum_y P(x|y, z) P(y|z) P(z)

Step 5: Simplify the Conditional Expression

Now, we divide both sides by P(z)P(z) to isolate P(xz)P(x|z):

P(xz)=1P(z)yP(xy,z)P(yz)P(z)P(x|z) = \frac{1}{P(z)} \sum_y P(x|y, z) P(y|z) P(z)

Since P(z)P(z) appears on both sides, it cancels out:

P(xz)=yP(xy,z)P(yz)P(x|z) = \sum_y P(x|y, z) P(y|z)

This completes the proof.

Summary

We have proved that:

P(xz)=yP(xy,z)P(yz)P(x|z) = \sum_y P(x|y,z) P(y|z)

This is a classic application of the law of total probability, where we marginalize over the variable yy to find the conditional probability P(xz)P(x|z).

Do you want more details or have any questions?


Related Questions:

  1. What is the law of total probability and when is it used?
  2. How does the chain rule of probability work in general?
  3. Can this proof be extended to cases with more than three variables (e.g., x,y,z,wx, y, z, w)?
  4. How is this formula used in Bayesian networks?
  5. What is the difference between marginalization and conditioning in probability?

Tip:

In probability, the chain rule allows us to decompose joint probabilities into simpler conditional probabilities. It's an essential tool for working with complex probability distributions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Conditional Probability
Law of Total Probability
Chain Rule of Probability

Formulas

P(x|z) = Σ_y P(x|y,z) P(y|z)
P(x|z) = P(x,z) / P(z)
P(x, y, z) = P(x|y,z) P(y,z)
P(y,z) = P(y|z) P(z)

Theorems

Law of Total Probability
Chain Rule of Probability

Suitable Grade Level

University Level