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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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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 probability theory because they model situations where outcomes can be counted. For example, when we toss a fair die, the number of dots that land face up is a discrete random variable because it can take on the values 1 through 6, which are finite and countable.

These variables can also take on an infinite sequence of values, like the number of times you would need to flip a coin before it lands heads. However, despite the potential for an infinite set, we still consider the variable discrete since the outcomes are distinct and countable, with no intermediate values possible.

Understanding discrete random variables is crucial in grasping other concepts like joint probability and conditional probability. They serve as the building blocks for constructing probability distributions, which give us the probabilities associated with each possible value of the random variable.
Joint Probability
Joint probability refers to the likelihood of two or more events happening at the same time. It is denoted as \(p_{X, Y}(x, y)\) and is a measure of the probability that event X and event Y occur together. For instance, consider rolling two dice simultaneously; the joint probability would involve getting a specific number on one die AND a specific number on the other at the same time.

Joint probability is the foundation of understanding relationships between different events. When involving discrete random variables, it requires considering the combined outcomes of these variables. If the variables are independent, the joint probability is simply the product of their individual probabilities. However, when variables are not independent, as is often the case, the calculation of joint probabilities becomes more complex and the concept of conditional probability comes into play.
Law of Total Probability
The law of total probability is a critical concept within probability theory that helps in computing the probability of an event based on several distinct conditions or partitions of the sample space. When you have a set of mutually exclusive and exhaustive outcomes, the probability of an event can be found by adding up the probabilities of intersecting that event with the various outcomes.

The magic of this law is that it allows us to break complex problems into simpler parts. For example, if you have four different, non-overlapping paths to reach a destination, the probability of reaching the destination via any path is the sum of the probabilities of reaching it through each individual path. In the context of joint probabilities and discrete variables, this law helps link the probabilities of separate occurrences and find the overall likelihood of an event.
Probability Theory
Probability theory is the mathematical framework that deals with the analysis of random phenomena. It is the backbone of statistics and enables us to make sense of uncertainty and randomness. From the seemingly unpredictable outcomes of rolling dice to the complex algorithms powering machine learning models, probability theory provides the tools to assess and predict the likelihood of events.

In probability theory, we are often concerned with random variables, events, outcomes, and their respective probabilities. We seek to understand the behavior of systems governed by chance. Key principles like the addition rule, multiplication rule, conditional probability, and the law of total probability comes handy when tackling a wide array of problems, from determining the fairness of a game to formulating strategies in risk assessment and decision-making processes.

Whenever we’re working with probability, clear and precise definitions are necessary so that the theoretical concepts can be effectively applied to real-world situations. This involves methodologies for calculating probabilities and characterizing the distributions of random variables, both discrete and continuous.

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Most popular questions from this chapter

You have two opponents with whom you alternate play. Whenever you play \(A\), you win with probability \(p_{A}\); whenever you play \(B\), you win with probability \(p_{B}\), where \(p_{B}>p_{A}\). If your objective is to minimize the expected number of games you need to play to win two in a row, should you start with \(A\) or with \(B\) ? Hint: Let \(E\left[N_{i}\right]\) denote the mean number of games needed if you initially play \(i\). Derive an expression for \(E\left[N_{A}\right]\) that involves \(E\left[N_{B}\right] ;\) write down the equivalent expression for \(E\left[N_{B}\right]\) and then subtract.

Let \(X_{1}\) and \(X_{2}\) be independent geometric random variables having the same parameter \(p\). Guess the value of $$ P\left\\{X_{1}=i \mid X_{1}+X_{2}=n\right\\} $$ Hint: Suppose a coin having probability \(p\) of coming up heads is continually flipped. If the second head occurs on flip number \(n\), what is the conditional probability that the first head was on flip number \(i, i=1, \ldots, n-1 ?\) Verify your guess analytically.

If \(X_{i}, i=1, \ldots, n\) are independent normal random variables, with \(X_{i}\) having mean \(\mu_{i}\) and variance 1, then the random variable \(\sum_{i=1}^{n} X_{i}^{2}\) is said to be a noncentral chi-squared random variable. (a) if \(X\) is a normal random variable having mean \(\mu\) and variance 1 show, for \(|t|<1 / 2\), that the moment generating function of \(X^{2}\) is $$ (1-2 t)^{-1 / 2} e^{\frac{t \mu^{2}}{1-2 t}} $$ (b) Derive the moment generating function of the noncentral chi-squared random variable \(\sum_{i=1}^{n} X_{i}^{2}\), and show that its distribution depends on the sequence of means \(\mu_{1}, \ldots, \mu_{n}\) only through the sum of their squares. As a result, we say that \(\sum_{i=1}^{n} X_{i}^{2}\) is a noncentral chi-squared random variable with parameters \(n\) and \(\theta=\sum_{i=1}^{n} \mu_{i}^{2}\) (c) If all \(\mu_{i}=0\), then \(\sum_{i=1}^{n} X_{i}^{2}\) is called a chi- squared random variable with \(n\) degrees of freedom. Determine, by differentiating its moment generating function, its expected value and variance. (d) Let \(K\) be a Poisson random variable with mean \(\theta / 2\), and suppose that conditional on \(K=k\), the random variable \(W\) has a chi-squared distribution with \(n+2 k\) degrees of freedom. Show, by computing its moment generating function, that \(W\) is a noncentral chi-squared random variable with parameters \(n\) and \(\theta\). (e) Find the expected value and variance of a noncentral chi-squared random variable with parameters \(n\) and \(\theta\).

Let \(X\) be uniform over \((0,1) .\) Find \(E\left[X \mid X<\frac{1}{2}\right]\).

(a) From the results of Section \(3.6 .3\) we can conclude that there are \(\left(\begin{array}{c}n+m-1 \\ m-1\end{array}\right)\) nonnegative integer valued solutions of the equation \(x_{1}+\cdots+x_{m}=n\) Prove this directly. (b) How many positive integer valued solutions of \(x_{1}+\cdots+x_{m}=n\) are there? Hint: Let \(y_{i}=x_{i}-1\). (c) For the Bose-Einstein distribution, compute the probability that exactly \(k\) of the \(X_{i}\) are equal to 0 .

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