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Chapter 3: Problem 1

If \(X\) and \(Y\) are both discrete, show that \(\sum_{x} p_{X \mid Y}(x \mid y)=1\) for all \(y\) such that \(p_{Y}(y)>0\)

Short Answer

Expert verified
Using the definition of conditional probability \(p_{X \mid Y}(x \mid y) = \frac{p_{X, Y}(x, y)}{p_Y(y)}\), we aim to show that \(\sum_x p_{X \mid Y}(x \mid y) = 1\). Multiplying both sides by \(p_Y(y)\) gives \(\sum_x p_{X, Y}(x, y) = p_Y(y)\). This equation is a result of the Law of Total Probability applied to the joint probability of \(X\) and \(Y\), which confirms that the sum of the conditional probabilities over all possible values of \(x\) is equal to \(1\) for all values of \(y\) where \(p_Y(y) > 0\).

Step by step solution

01

Definition of Conditional Probability

To show the given equation, let's remember the formula for the conditional probability: \[p_{X \mid Y}(x \mid y) = \frac{p_{X, Y}(x, y)}{p_Y(y)}\] where \(p_{X \mid Y}(x \mid y)\) denotes the conditional probability of \(X\) given \(Y\), \(p_{X, Y}(x, y)\) denotes the joint probability of \(X\) and \(Y\), and \(p_Y(y)\) denotes the probability of \(Y\).
02

Using the Law of Total Probability

We have to show that the sum over all possible values of \(x\) is equal to \(1\), i.e., \[\sum_x p_{X \mid Y}(x \mid y) = 1\] Using the definition of conditional probability from Step 1: \[\sum_x \frac{p_{X, Y}(x, y)}{p_Y(y)} = 1\] Since \(p_Y(y) > 0\), we can multiply both sides by \(p_Y(y)\) without changing the equality: \[\sum_x p_{X, Y}(x, y) = p_Y(y)\]
03

The Law of Total Probability for Joint Probabilities

Now let's focus on the left side of the equation above. The sum of the joint probabilities of \(X\) and \(Y\) over all the possible values of \(x\) for a given value of \(y\) is equal to the probability of \(Y\), i.e., \(p_Y(y)\). This relation comes from the Law of Total Probability, applied to the joint probability of \(X\) and \(Y\): \[p_Y(y) = \sum_x p_{X, Y}(x, y)\] Comparing this with the equation at the end of step 2, we see that they are the same: \[\sum_x p_{X, Y}(x, y) = p_Y(y)\]
04

Conclusion

Finally, we have shown that the sum of the conditional probabilities \(p_{X \mid Y}(x \mid y)\) over all possible values of \(x\) is equal to \(1\) for all the values of \(y\) such that \(p_Y(y) > 0\). This is done by using the definition of conditional probability and applying the Law of Total Probability.

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Most popular questions from this chapter

An urn contains \(n\) balls, with ball \(i\) having weight \(w_{i}, i=1, \ldots, n .\) The balls are withdrawn from the urn one at a time according to the following scheme: When \(S\) is the set of balls that remains, ball \(i, i \in S\), is the next ball withdrawn with probability \(w_{i} / \sum_{j \in S} w_{j} .\) Find the expected number of balls that are withdrawn before ball \(i, i=1, \ldots, n\).

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