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Chapter 9: Problem 519

Show, by altering the joint density of \(\mathrm{X}\) and \(\mathrm{Y}\) in the previous problem, that it is not always possible to construct a unique joint distribution from a pair of given marginal distributions.

Short Answer

Expert verified
We demonstrated that it is not always possible to construct a unique joint distribution from a pair of given marginal distributions by providing two different joint distributions, each compatible with the marginal distributions \(p(x)\) and \(p(y)\) given. Both joint distributions have the same marginal distributions, yet they are different, implying that the joint distribution cannot always be uniquely determined by marginal distributions alone.

Step by step solution

01

Definition of Marginal Distribution

The marginal distribution of a random variable X, denoted as p(x), is obtained by summing the joint probabilities of X and Y over all the possible values of Y. Similarly, the marginal distribution of a random variable Y, denoted as p(y), is obtained by summing the joint probabilities of X and Y over all the possible values of X.
02

Construct two different joint distributions with the same given marginal distributions

Let's consider the following marginal distributions for X and Y: \[p(x) = \begin{cases} \frac{1}{2} & x=0 \\ \frac{1}{2} & x=1 \\ \end{cases}\] \[p(y) = \begin{cases} \frac{1}{2} & y=0 \\ \frac{1}{2} & y=1 \\ \end{cases}\] Now, let's construct two different joint distributions that have these marginal distributions. 1) Joint distribution 1: \[p(x, y) = \begin{cases} \frac{1}{4} & x=0, y=0 \\ \frac{1}{4} & x=0, y=1 \\ \frac{1}{4} & x=1, y=0 \\ \frac{1}{4} & x=1, y=1 \\ \end{cases}\] 2) Joint distribution 2: \[p(x, y) = \begin{cases} \frac{3}{8} & x=0, y=0 \\ \frac{1}{8} & x=0, y=1 \\ \frac{1}{8} & x=1, y=0 \\ \frac{3}{8} & x=1, y=1 \\ \end{cases}\]
03

Verify that the joint distributions have the given marginal distributions

For joint distribution 1: - Summing up for y (0 and 1) for x=0: \(\frac{1}{4} + \frac{1}{4} = \frac{1}{2}\). So, p(X=0) = \(\frac{1}{2}\). - Summing up for y (0 and 1) for x=1: \(\frac{1}{4} + \frac{1}{4} = \frac{1}{2}\). So, p(X=1) = \(\frac{1}{2}\). - Summing up for x (0 and 1) for y=0: \(\frac{1}{4} + \frac{1}{4} = \frac{1}{2}\). So, p(Y=0) = \(\frac{1}{2}\). - Summing up for x (0 and 1) for y=1: \(\frac{1}{4} + \frac{1}{4} = \frac{1}{2}\). So, p(Y=1) = \(\frac{1}{2}\). For joint distribution 2: - Summing up for y (0 and 1) for x=0: \(\frac{3}{8} + \frac{1}{8} = \frac{1}{2}\). So, p(X=0) = \(\frac{1}{2}\). - Summing up for y (0 and 1) for x=1: \(\frac{1}{8} + \frac{3}{8} = \frac{1}{2}\). So, p(X=1) = \(\frac{1}{2}\). - Summing up for x (0 and 1) for y=0: \(\frac{3}{8} + \frac{1}{8} = \frac{1}{2}\). So, p(Y=0) = \(\frac{1}{2}\). - Summing up for x (0 and 1) for y=1: \(\frac{1}{8} + \frac{3}{8} = \frac{1}{2}\). So, p(Y=1) = \(\frac{1}{2}\). We can see that both joint distributions have the same marginal distributions as given.
04

Conclusion

We have constructed two different joint distributions with the same given marginal distributions. This shows that it is not always possible to construct a unique joint distribution from a pair of given marginal distributions.

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Key Concepts

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

Marginal Distribution
Understanding marginal distribution is key to grasping this problem. Suppose you have two random variables, X and Y, and you know their joint probability distribution. The marginal distribution focuses on just one of these variables, either X or Y, but not both together.

To find the marginal distribution of X, you sum the joint probabilities across all values of Y. This gives you the probability distribution of X alone. Similarly, to find the marginal distribution of Y, you sum across all values of X. Marginal distributions offer insights into individual behaviors of each variable without considering their interaction in the joint space.
  • Marginal distribution of X: denoted as \( p(x) \)
  • Marginal distribution of Y: denoted as \( p(y) \)
  • Calculated by summing joint probabilities for each respective variable
These distributions help us analyze one variable independently in the context of probability distributions in discrete mathematics.
Random Variables
Random variables are fundamental concepts in statistics and probability. They can take on different values, each associated with a probability. In our context, X and Y are random variables.

A random variable does not represent one specific number. Instead, it denotes a potential range of values it might assume in any random trials or experiments. Think of it as a mapping of possible outcomes onto numerical values, which allows us to perform mathematical evaluations and predictions.
  • Each random variable can have a discrete (countable) set of outcomes.
  • Random variables typically model real-world situations in probability studies.
  • Combines with distributions to describe and predict behavior.
Understanding what random variables are and how they work helps in evaluating complex probability models like joint and marginal distributions.
Joint Probability Distribution
A joint probability distribution provides the probability of two random variables happening at the same time. In mathematical terms, it is represented as \( p(x, y) \).

This distribution considers all possible combinations of values of X and Y. In the problem, two different joint distributions were constructed based on the same marginal distributions, illustrating that such combinations can vary while still satisfying individual marginal characteristics.

Let's break down its components:
  • Combines probabilities for each pair of outcomes from random variables X and Y.
  • Shows how likely combinations of these outcomes are to occur.
  • Key in understanding interactions between variables in probability theory.
The joint probability distribution paints a complete picture of how two random variables are related through their combined behaviors.
Discrete Mathematics
Discrete mathematics underpins much of probability theory and the concept of distributions we discuss here. It involves a branch of mathematics that deals with discrete objects鈥攍ike integers, graphs, and statements in logic.

Probability, especially involving countable outcomes鈥攁s with our random variables X and Y鈥攊s a fundamental area of discrete mathematics. It emphasizes techniques to compute probabilities when variables take distinct, separate values rather than continuous ranges.
  • Deals with distinct, separate values (like flipping a coin or rolling a die).
  • Crucial for understanding probability with finite outcomes.
  • Widely applied in computer science, cryptography, and algorithm design.
In summary, in the context of this problem, discrete mathematics helps to structure and solve problems involving random variables and their distributions in a clear, logical manner.

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Most popular questions from this chapter

A sports magazine reports that the people who watch Monday night football games on television are evenly divided between men and women. Out of a random sample of 400 people who regularly watch the Monday night game, 220 are men. Using a \(.10\) level of significance, can be conclude that the report is false?

Let \(\mathrm{Y}=\) the Rockwell hardness of a particular alloy of steel. Assume that \(\mathrm{Y}\) is a continuous random variable that can take on any value between 50 and 70 with equal probability. Find the expected Rockwell hardness.

Use the Kolmogorov-Smirnov Statistic to find a \(95 \%\) confidence interval for \(\mathrm{F}(\mathrm{x}) . \mathrm{F}(\mathrm{x})\) is the cumulative distribution function of a population from which the following ordered samples was taken: \(8.2,10.4,10.6,11.5,12.6,12.9\), \(13.3,13.3,13.4,13.4,13.6,13.8,14.0,14.0,14.1,14.2\) \(14.6,14.7,14.9,15.0,15.4,15.6,15.9,16.0,16.2,16.3\) 17.2,17.4,17.7,18.1 .$

Let the random variable \(\mathrm{X}\) represent the number of defective radios in a shipment of four radios to a local appliance store. Assume that each radio is equally likely to be defective or non-defective, hence the probability that a radio is defective is \(\mathrm{p}=1 / 2\). Also assume whether or not each radio is defective or non-defective is indipendent of the status of the other radios. Find the expected number of defective radios.

Consider a probability distribution for random orientations in which the probability of an observation in a region on the surface of the unit hemisphere is proportional to the area of that region. Two angles, \(u\) and \(v\), will determine the position of an observation. It can be shown that the position of an observation is jointly distributed with density function $$ \begin{array}{r} \mathrm{f}(\mathrm{u}, \mathrm{v})=[\\{\sin \mathrm{u}\\} /\\{2 \pi\\}] \quad 0<\mathrm{u}<2 \pi \\ 0<\mathrm{u}<\pi / 2 . \end{array} $$ Two new variables, \(\mathrm{X}\) and \(\mathrm{Y}\) are defined, where $$ \mathrm{X}=\sin \mathrm{u} \cos \mathrm{v} $$ $$ \mathrm{Y}=\sin \mathrm{u} \sin \mathrm{v} $$ Find the joint density function of \(\mathrm{X}\) and \(\mathrm{Y}\).
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