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Suppose \(p(x, y, z)\), the joint probability mass function of the random variables \(X\), \(Y\), and \(Z\), is given by $$ \begin{array}{ll} p(1,1,1)=\frac{1}{8}, & p(2,1,1)=\frac{1}{4} \\ p(1,1,2)=\frac{1}{8}, & p(2,1,2)=\frac{3}{16} \\ p(1,2,1)=\frac{1}{16}, & p(2,2,1)=0 \\ p(1,2,2)=0, & p(2,2,2)=\frac{1}{4} \end{array} $$ \text { What is } E[X \mid Y=2] ? \text { What is } E[X \mid Y=2, Z=1] ?

Short Answer

Expert verified
In conclusion, the conditional expectations are: $$ E[X | Y=2] = \frac{9}{5} $$ $$ E[X | Y=2, Z=1] = 1 $$

Step by step solution

01

Find the conditional probability mass function p(X | Y=2)

To find the conditional probability mass function p(X | Y=2), we need to calculate the probabilities p(X=x | Y=2) for x=1,2. We can do this using the formula: $$ p(X=x | Y=2) = \frac{p(X=x, Y=2)}{p(Y=2)} $$ First, we need to find the probabilities p(Y=2) and p(X=x, Y=2) given in the joint probability mass function table: $$ p(Y=2) = p(1,2,1) + p(1,2,2) + p(2,2,1) + p(2,2,2) = \frac{1}{16}+0+0+\frac{1}{4} = \frac{5}{16} $$ Now, we can calculate: $$ p(X=1 | Y=2) = \frac{p(1,2,1) + p(1,2,2)}{p(Y=2)} = \frac{\frac{1}{16}+0}{\frac{5}{16}} = \frac{1}{5} $$ $$ p(X=2 | Y=2) = \frac{p(2,2,1) + p(2,2,2)}{p(Y=2)} = \frac{0+\frac{1}{4}}{\frac{5}{16}} = \frac{4}{5} $$ So, the conditional probability mass function p(X | Y=2) is given by: $$ p(X=1 | Y=2) = \frac{1}{5}, \quad p(X=2 | Y=2) = \frac{4}{5} $$
02

Calculate E[X | Y=2]

Now, use the conditional probability mass function p(X | Y=2) to compute the conditional expectation E[X | Y=2] using the formula: $$ E[X | Y=2] = \sum_x x \cdot p(X=x | Y=2) $$ Substitute the values of the conditional probability mass function: $$ E[X | Y=2] = 1 \cdot \frac{1}{5} + 2 \cdot \frac{4}{5} = \frac{1}{5}+ \frac{8}{5} = \frac{9}{5} $$ Therefore, the conditional expectation E[X | Y=2] is given by: $$ E[X | Y=2] = \frac{9}{5} $$
03

Find the conditional probability mass function p(X | Y=2, Z=1)

To find the conditional probability mass function p(X | Y=2, Z=1), we need to calculate the probabilities p(X=x | Y=2, Z=1) for x=1,2. We can do this using the formula: $$ p(X=x | Y=2, Z=1) = \frac{p(X=x, Y=2, Z=1)}{p(Y=2, Z=1)} $$ First, we need to find the probabilities p(Y=2, Z=1) and p(X=x, Y=2, Z=1) given in the joint probability mass function table: $$ p(Y=2, Z=1) = p(1,2,1) + p(2,2,1) = \frac{1}{16}+0 = \frac{1}{16} $$ Now, we can calculate: $$ p(X=1 | Y=2, Z=1) = \frac{p(1,2,1)}{p(Y=2, Z=1)} = \frac{\frac{1}{16}}{\frac{1}{16}} = 1 $$ $$ p(X=2 | Y=2, Z=1) = \frac{p(2,2,1)}{p(Y=2, Z=1)} = \frac{0}{\frac{1}{16}} = 0 $$ So, the conditional probability mass function p(X | Y=2, Z=1) is given by: $$ p(X=1 | Y=2, Z=1) = 1, \quad p(X=2 | Y=2, Z=1) = 0 $$
04

Calculate E[X | Y=2, Z=1]

Now, use the conditional probability mass function p(X | Y=2, Z=1) to compute the conditional expectation E[X | Y=2, Z=1] using the formula: $$ E[X | Y=2, Z=1] = \sum_x x \cdot p(X=x | Y=2, Z=1) $$ Substitute the values of the conditional probability mass function: $$ E[X | Y=2, Z=1] = 1 \cdot 1 + 2 \cdot 0 = 1 $$ Therefore, the conditional expectation E[X | Y=2, Z=1] is given by: $$ E[X | Y=2, Z=1] = 1 $$ In conclusion, the conditional expectations are: $$ E[X | Y=2] = \frac{9}{5} $$ $$ E[X | Y=2, Z=1] = 1 $$

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Key Concepts

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

Joint Probability Mass Function
The joint probability mass function, often abbreviated as jpmf, is a fundamental concept in probability that helps assess the probability of combinations of two or more discrete random variables. It essentially tells us how the probabilities are distributed over the possible pairs or sets of values these variables can take simultaneously.
In our example, we have three random variables \(X\), \(Y\), and \(Z\), and their jpmf is denoted as \(p(x, y, z)\). The table provided illustrates the probability of each possible combination of \(X\), \(Y\), and \(Z\).
Hence, if you're asked, \(p(1, 1, 1)\), indicates there's a \(\frac{1}{8}\) chance that \(X=1\), \(Y=1\), and \(Z=1\) occur together. Similarly, \(p(2, 2, 2)\) conveys a \(\frac{1}{4}\) probability for \(X=2\), \(Y=2\), and \(Z=2\) happening at the same time.
This function assists in evaluating conditional and marginal probabilities, setting the stage for understanding the relationships between different variables.
Random Variables
Random variables are the entities in probability models that account for the different outcomes of a random phenomenon. Each outcome of a random variable is associated with a numerical value, which can be either discrete or continuous. In our case, they're discrete.
In simpler terms, a random variable is a function that assigns a numerical value to each possible outcome in a sample space. With \(X\), \(Y\), and \(Z\) as our random variables, each can take specific values such as 1 or 2, defined within the joint probability mass function.
These variables are the building blocks of probability models, which enable us to calculate measures like variance and expectation. An understanding of random variables and their distributions is essential in predicting outcomes and making probabilistic inferences within a given model.
Conditional Probability
Conditional probability is a key concept that describes the probability of an event occurring given that another event has already happened. It is usually denoted as \(P(A | B)\), which is read as the probability of \(A\) given \(B\).
In our example, the conditional probabilities \(p(X=x \mid Y=2)\) or \(p(X=1 \mid Y=2, Z=1)\) are crucial. They tell us about the probability distribution of \(X\) when specific conditions about \(Y\) or both \(Y\) and \(Z\) are met.
By using values from the jpmf table, these probabilities help narrow down the potential outcomes of a random variable when some known data about another variable is available. This concept simplifies real-world decision-making under uncertainty by providing more targeted probabilities.
Probability Models
Probability models play a vital role in understanding and predicting outcomes in various situations by providing a structured way to consider chance and uncertainty. They encompass various tools and methods to systematically compute probabilities concerning random variables.
Our model here involves discussing the conditional expectation, which uses the joint probabilities and random variable distributions to compute an expected value given specific conditions. A probability model helps visualize and calculate how likely different outcomes are based on predefined rules or functions.
By constructing such models, whether for simple experiments or complex phenomena, we can better explore the nature of uncertainty and randomness, allowing for more informed decisions in engineering, science, finance, and daily life.

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Most popular questions from this chapter

Let \(X_{i}, i \geqslant 0\) be independent and identically distributed random variables with probability mass function $$ p(j)=P\left[X_{i}=i\right\\}, \quad j=1, \ldots, m, \quad \sum_{j=1}^{m} P(j)=1 $$ Find \(E[N]\), where \(N=\min \left[n>0: X_{n}=X_{0}\right\\}\)

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

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.

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There are three coins in a barrel. These coins, when flipped, will come up heads with respective probabilities \(0.3,0.5,0.7 .\) A coin is randomly selected from among these three and is then flipped ten times. Let \(N\) be the number of heads obtained on the ten flips. (a) Find \(P[N=0\\}\). (b) Find \(P[N=n\\}, n=0,1, \ldots, 10\) (c) Does \(N\) have a binomial distribution? (d) If you win \(\$ 1\) each time a head appears and you lose \(\$ 1\) each time a tail appears, is this a fair game? Explain.

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