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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
To prove that \(\sum_{x} p_{X \mid Y}(x \mid y) = 1\) for all \(y\) such that \(p_Y(y)>0\), we start by writing down the definition of conditional probability, \( p_{X \mid Y}(x \mid y) = \frac{p_{X, Y}(x, y)}{p_Y(y)} \). We then substitute the definition into the sum, factor out the constant denominator, and recognize the sum of joint probabilities over x to find that \(\sum_{x} p_{X, Y}(x, y) = p_Y(y)\). Finally, we substitute the probability of Y and simplify to show that \(1 = 1\), thus proving that the sum of the conditional probabilities equals 1 for all y such that \(p_Y(y)>0\).

Step by step solution

01

Write down the definition of conditional probability

The definition of conditional probability is given as: \( p_{X \mid Y}(x \mid y) = \frac{p_{X, Y}(x, y)}{p_Y(y)} \), where \(p_{X, Y}(x, y)\) is the joint probability of X and Y, and \(p_Y(y)\) is the probability of Y. Remember that we must show that the sum of the conditional probabilities equals 1, that is: \( \sum_{x} p_{X \mid Y}(x \mid y) = 1 \)
02

Substitute the definition into the sum

Replace the conditional probability in our target formula with its definition: \( \sum_{x} \frac{p_{X, Y}(x, y)}{p_Y(y)} = 1 \)
03

Factor out the constant denominator

Since the denominator \(p_Y(y)\) doesn't depend on the summation variable x, we can factor it out of the sum: \( \frac{1}{p_Y(y)} \sum_{x} p_{X, Y}(x, y) = 1 \)
04

Recognize the sum of joint probabilities over x

The sum of the joint probabilities of X and Y over all values of x is equal to the probability of Y, that is: \( \sum_{x} p_{X, Y}(x, y) = p_Y(y) \) This is because we are summing over all possible values of X, given a specific value of Y.
05

Substitute the probability of Y and simplify

Using the result from Step 4, substitute the expression for the sum of joint probabilities into our formula: \( \frac{1}{p_Y(y)} p_Y(y) = 1 \) Since \(p_Y(y)>0\), we can safely divide both sides by \(p_Y(y)\), which simplifies the expression to: \( 1 = 1 \) Thus, we have proved that \(\sum_{x} p_{X \mid Y}(x \mid y) = 1\) for all \(y\) such that \(p_Y(y)>0\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Discrete Random Variables
Discrete random variables are fundamental in the study of probability. They are variables that can take on a countable number of distinct values, each with an associated probability. An example of a discrete random variable could be the number of heads observed when flipping a coin three times.

Each potential outcome of a discrete random variable has a probability that ranges between 0 and 1, and the sum of all these probabilities must equal 1. This concept of discrete random variables is essential when we dive into the realm of conditional probabilities and joint distributions, as it underscores how probabilities are quantified and interpreted in discrete settings.
Joint Probability
Joint probability is the probability of two events occurring simultaneously and is a crucial aspect of probability theory. It is denoted as pX, Y(x, y), the probability that event X occurs with a particular value alongside event Y also occurring with a specific value. Joint probability distribution for discrete random variables is the collection of probabilities for all possible combinations of their outcomes.

Understanding the joint probability helps in calculating the conditional probabilities, as seen in the given problem. The concept showcases the interconnectedness between different events or random variables.
Probability Theory
Probability theory is a mathematical framework for quantifying uncertainty. It entails rules and axioms that allow us to compute the likelihood of various events or outcomes.

In probability theory, the concepts of discrete random variables and joint probabilities are tightly interwoven. The rules include the sum of probabilities, which dictates that the probabilities for all possible outcomes of a random variable should add up to 1, and the foundations of conditional probabilities, which focus on understanding the probability of an event given that another event has occurred. The example in the exercise relies on these rules to derive that the sum of conditional probabilities given a discrete random variable is always 1.
Sum of Probabilities
The sum of probabilities principle states that the total probability across all possible outcomes of a discrete random variable must sum to 1. This is a core tenant of probability theory that ensures consistency and coherency in the calculation of probabilities.

When dealing with joint probabilities, the principle applies to the sum of probabilities across all combinations of X and Y. In conditional probability, like in the exercise, it guarantees that when considering all potential outcomes for X given Y, their probabilities still sum to 1. This is what makes probability distributions useful models for predicting event likelihoods in varied scenarios.

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Most popular questions from this chapter

The joint density of \(X\) and \(Y\) is given by $$f(x, y)=\frac{e^{-y}}{y}, \quad 0

A rat is trapped in a maze. Initially it has to choose one of two directions. If it goes to the right, then it will wander around in the maze for three minutes and will then return to its initial position. If it goes to the left, then with probability \(\frac{1}{3}\) it will depart the maze after two minutes of traveling, and with probability \(\frac{2}{3}\) it will return to its initial position after five minutes of traveling. Assuming that the rat is at all times equally likely to go to the left or the right, what is the expected number of minutes that it will be trapped in the maze?

A prisoner is trapped in a cell containing three doors. The first door leads to a tunnel that returns him to his cell after two days of travel. The second leads to a tunnel that returns him to his cell after three days of travel. The third door leads immediately to freedom. (a) Assuming that the prisoner will always select doors 1,2, and 3 with probabilities \(0.5,0.3,0.2\), what is the expected number of days until he reaches freedom? (b) Assuming that the prisoner is always equally likely to choose among those doors that he has not used, what is the expected number of days until he reaches freedom? (In this version, for instance, if the prisoner initially tries door 1 , then when he returns to the cell, he will now select only from doors 2 and 3.) (c) For parts (a) and (b) find the variance of the number of days until the prisoner reaches freedom.

In Example \(3.25\) show that the conditional distribution of \(N\) given that \(U_{1}=y\) is the same as the conditional distribution of \(M\) given that \(U_{1}=1-y\). Also, show that $$ E\left[N \mid U_{1}=y\right]=E\left[M \mid U_{1}=1-y\right]=1+e^{y} $$

Consider a gambler who on each bet either wins 1 with probability \(18 / 38\) or loses 1 with probability \(20 / 38\). (These are the probabilities if the bet is that a roulette wheel will land on a specified color.) The gambler will quit either when he or she is winning a total of 5 or after 100 plays. What is the probability he or she plays exactly 15 times?
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