/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 The random variables \(X\) and \... [FREE SOLUTION] | 91Ó°ÊÓ

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The random variables \(X\) and \(Y\) are said to have a bivariate normal distribution if their joint density function is given by $$ \begin{aligned} f(x, y)=& \frac{1}{2 \pi \sigma_{x} \sigma_{y} \sqrt{1-\rho^{2}}} \exp \left\\{-\frac{1}{2\left(1-\rho^{2}\right)}\right.\\\ &\left.\times\left[\left(\frac{x-\mu_{x}}{\sigma_{x}}\right)^{2}-\frac{2 \rho\left(x-\mu_{x}\right)\left(y-\mu_{y}\right)}{\sigma_{x} \sigma_{y}}+\left(\frac{y-\mu_{y}}{\sigma_{y}}\right)^{2}\right]\right\\} \end{aligned} $$ for \(-\infty0, \sigma_{y}>0,-\infty<\mu_{x}<\infty,-\infty<\mu_{y}<\infty\) (a) Show that \(X\) is normally distributed with mean \(\mu_{x}\) and variance \(\sigma_{x}^{2}\), and \(Y\) is normally distributed with mean \(\mu_{y}\) and variance \(\sigma_{y}^{2}\). (b) Show that the conditional density of \(X\) given that \(Y=y\) is normal with mean \(\mu_{x}+\left(\rho \sigma_{x} / \sigma_{y}\right)\left(y-\mu_{y}\right)\) and variance \(\sigma_{x}^{2}\left(1-\rho^{2}\right)\) The quantity \(\rho\) is called the correlation between \(X\) and \(Y\). It can be shown that $$ \begin{aligned} \rho &=\frac{E\left[\left(X-\mu_{x}\right)\left(Y-\mu_{y}\right)\right]}{\sigma_{x} \sigma_{y}} \\ &=\frac{\operatorname{Cov}(X, Y)}{\sigma_{x} \sigma_{y}} \end{aligned} $$

Short Answer

Expert verified
To summarize, we integrated the joint PDF of X and Y to find their marginal PDFs and verified that X and Y are normally distributed with means \(\mu_x\) and \(\mu_y\) and variances \(\sigma_x^2\) and \(\sigma_y^2\), respectively. Furthermore, we found the conditional PDF of X given Y=y and verified that it is normally distributed with mean \(\mu_X^{*} = \mu_{x}+\left(\rho \sigma_{x} / \sigma_{y}\right)\left(y-\mu_{y}\right)\) and variance \(\sigma_{X^{*}}^{2} = \sigma_{x}^{2}\left(1-\rho^{2}\right)\).

Step by step solution

01

(Marginal PDF of X and Y)

: First, we will find the marginal PDFs for X and Y by integrating their joint PDF. Marginal PDF of X can be found by integrating the joint PDF with respect to y: \(f_X(x) = \int_{-\infty}^{\infty} f(x, y) \, dy \)
02

(Verify Normal Distribution of X)

: To show that X is normally distributed with mean \(\mu_x\) and variance \(\sigma_x^2\), we need to verify that the marginal PDF of X is of the normal distribution form: \(f_X(x) = \frac{1}{\sqrt{2\pi\sigma_x^2}}\exp\left\{-\frac{1}{2}\left(\frac{x-\mu_x}{\sigma_x}\right)^2\right\}\) Similarly, Marginal PDF of Y can be found by integrating the joint PDF with respect to x: \(f_Y(y) = \int_{-\infty}^{\infty} f(x, y) \, dx \)
03

(Verify Normal Distribution of Y)

: To show that Y is normally distributed with mean \(\mu_y\) and variance \(\sigma_y^2\), we need to verify that the marginal PDF of Y is of the normal distribution form: \(f_Y(y) = \frac{1}{\sqrt{2\pi\sigma_y^2}}\exp\left\{-\frac{1}{2}\left(\frac{y-\mu_y}{\sigma_y}\right)^2\right\}\)
04

(Conditional PDF of X given Y=y)

: Now, we will find the conditional PDF of X given that Y=y. By the definition of conditional probability, we can find the conditional PDF of X given Y=y as: \(f_{X|Y}(x|y) = \frac{f(x, y)}{f_Y(y)}\)
05

(Verify Normal Distribution of X given Y=y)

: To show that the conditional PDF of X given that Y=y is normal with mean \(\mu_X^{*} = \mu_{x}+\left(\rho \sigma_{x} / \sigma_{y}\right)\left(y-\mu_{y}\right)\) and variance \(\sigma_{X^{*}}^{2} = \sigma_{x}^{2}\left(1-\rho^{2}\right)\), we need to verify that the conditional PDF of X given Y=y is of the normal distribution form: \(f_{X|Y}(x|y) = \frac{1}{\sqrt{2\pi\sigma_{X^{*}}^2}}\exp\left\{-\frac{1}{2}\left(\frac{x-\mu_{X^{*}}}{\sigma_{X^{*}}}\right)^2\right\}\)

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Key Concepts

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

Joint Density Function
The joint density function is a fundamental concept when dealing with multivariate distributions, like the bivariate normal distribution. This function describes the likelihood of two continuous random variables, say \(X\) and \(Y\), occurring simultaneously. Imagine it as a mathematical representation that provides a complete picture of how \(X\) and \(Y\) behave together.
In the case of a bivariate normal distribution, the joint density function is given by:
\ \[ f(x, y) = \frac{1}{2 \pi \sigma_{x} \sigma_{y} \sqrt{1-\rho^{2}}} \exp \left\{-\frac{1}{2\left(1-\rho^{2}\right)}\right.\] \ \[ \left.\left[\left(\frac{x-\mu_{x}}{\sigma_{x}}\right)^{2} - \frac{2 \rho\left(x-\mu_{x}\right)\left(y-\mu_{y}\right)}{\sigma_{x} \sigma_{y}} + \left(\frac{y-\mu_{y}}{\sigma_{y}}\right)^{2}\right]\right\} \]
Here:
  • \(\mu_x\) and \(\mu_y\) are the means of \(X\) and \(Y\), respectively.
  • \(\sigma_x\) and \(\sigma_y\) are the standard deviations of \(X\) and \(Y\).
  • \(\rho\) represents the correlation coefficient between \(X\) and \(Y\).
The function captures not only the individual behaviors of \(X\) and \(Y\), but also how they influence each other. Understanding this interplay is crucial for fields ranging from statistics to finance, where outcomes are rarely influenced by a single variable.
Marginal and Conditional Distribution
Marginal and conditional distributions help us gain deeper insights into the behavior of multiple random variables. Let's explore each one:
In a joint distribution, a marginal distribution pertains to one variable, ignoring the presence of others. To find the marginal distribution of a variable, we integrate the joint density function over the range of the other variable. For instance, to find the marginal probability density function (PDF) of \(X\), we integrate out \(Y\):
\[ f_X(x) = \int_{-\infty}^{\infty} f(x, y) \, dy \]
Similarly, to find the marginal PDF of \(Y\), integrate out \(X\):
\[ f_Y(y) = \int_{-\infty}^{\infty} f(x, y) \, dx \]
These marginal distributions help us see whether each variable follows a normal distribution individually.
The conditional distribution, on the other hand, provides insights on one variable, say \(X\), given that we know the value of another variable, \(Y\). It is like focusing a lens; you concentrate on specific circumstances by fixing another variable:
\[ f_{X|Y}(x|y) = \frac{f(x, y)}{f_Y(y)} \]
This equation tells us the probability distribution of \(X\) when \(Y\) is fixed at \(y\). In our bivariate normal example, the conditional distribution also turns out to be a normal distribution. Its mean and variance, however, are adjusted to reflect the dependency on \(Y\). The mean is shifted to \(\mu_{x} + \left(\rho \frac{\sigma_{x}}{\sigma_{y}}\right)(y - \mu_{y})\), and the variance is reduced to \(\sigma_{x}^{2}(1 - \rho^{2})\).
These concepts are pivotal for understanding how one variable behaves when you have partial information about another, such as stock prices or customer preferences.
Correlation Coefficient
The correlation coefficient, denoted as \(\rho\), plays a vital role in understanding the relationship between two variables in a bivariate normal distribution. It quantifies the strength and direction of a linear relationship between two random variables, in our scenario, \(X\) and \(Y\).
Think of \(\rho\) as a number that tells you how tightly \(X\) and \(Y\) are linked. Its value ranges between -1 and 1:
  • A \(\rho\) of 1 implies a perfect positive linear relationship; as \(X\) increases, \(Y\) increases in a perfectly predictable way.
  • Conversely, a \(\rho\) of -1 implies a perfect negative linear relationship; here, \(X\) goes up and \(Y\) predictably goes down.
  • A \(\rho\) near 0 suggests a weak or no linear relationship, meaning \(X\) and \(Y\) move independently.
Mathematically, it's calculated as:
\[ \rho = \frac{\operatorname{Cov}(X, Y)}{\sigma_{x} \sigma_{y}} \]
Where \(\operatorname{Cov}(X, Y)\) represents the covariance, a measure of how much two random variables vary together.
A key point is that a higher absolute value of \(\rho\) signifies a stronger linear relationship. In practical terms, knowing \(\rho\) gives you the power to anticipate how changes in one variable could affect the other. It's a cornerstone in predictive analytics, finance, and the sciences.

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Most popular questions from this chapter

A deck of \(n\) cards, numbered 1 through \(n\), is randomly shuffled so that all \(n !\) possible permutations are equally likely. The cards are then turned over one at a time until card number 1 appears. These upturned cards constitute the first cycle. We now determine (by looking at the upturned cards) the lowest numbered card that has not yet appeared, and we continue to turn the cards face up until that card appears. This new set of cards represents the second cycle. We again determine the lowest numbered of the remaining cards and turn the cards until it appears, and so on until all cards have been turned over. Let \(m_{n}\) denote the mean number of cycles. (a) Derive a recursive formula for \(m_{n}\) in terms of \(m_{k}, k=1, \ldots, n-1\). (b) Starting with \(m_{0}=0\), use the recursion to find \(m_{1}, m_{2}, m_{3}\), and \(m_{4}\). (c) Conjecture a general formula for \(m_{n}\). (d) Prove your formula by induction on \(n\). That is, show it is valid for \(n=1\), then assume it is true for any of the values \(1, \ldots, n-1\) and show that this implies it is true for \(n\). (e) Let \(X_{i}\) equal 1 if one of the cycles ends with card \(i\), and let it equal 0 otherwise, \(i=1, \ldots, n\). Express the number of cycles in terms of these \(X_{i}\). (f) Use the representation in part (e) to determine \(m_{n}\). (g) Are the random variables \(X_{1}, \ldots, X_{n}\) independent? Explain. (h) Find the variance of the number of cycles.

The opponents of soccer team \(\mathrm{A}\) are of two types: either they are a class 1 or a class 2 team. The number of goals team A scores against a class \(i\) opponent is a Poisson random variable with mean \(\lambda_{i}\), where \(\lambda_{1}=2, \lambda_{2}=3\). This weekend the team has two games against teams they are not very familiar with. Assuming that the first team they play is a class 1 team with probability \(0.6\) and the second is, independently of the class of the first team, a class 1 team with probability \(0.3\), determine (a) the expected number of goals team A will score this weekend. (b) the probability that team \(\mathrm{A}\) will score a total of five goals.

Independent trials, resulting in one of the outcomes \(1,2,3\) with respective probabilities \(p_{1}, p_{2}, p_{3}, \sum_{i=1}^{3} p_{i}=1\), are performed. (a) Let \(N\) denote the number of trials needed until the initial outcome has occurred exactly 3 times. For instance, if the trial results are \(3,2,1,2,3,2,3\) then \(N=7\) Find \(E[N]\). (b) Find the expected number of trials needed until both outcome 1 and outcome 2 have occurred.

An individual whose level of exposure to a certain pathogen is \(x\) will contract the disease caused by this pathogen with probability \(P(x) .\) If the exposure level of a randomly chosen member of the population has probability density function \(f\), determine the conditional probability density of the exposure level of that member given that he or she (a) has the disease. (b) does not have the disease. (c) Show that when \(P(x)\) increases in \(x\), then the ratio of the density of part (a) to that of part (b) also increases in \(x\).

In the list example of Section \(3.6 .1\) suppose that the initial ordering at time \(t=0\) is determined completely at random; that is, initially all \(n !\) permutations are equally likely. Following the front-of-the-line rule, compute the expected position of the element requested at time \(t\). Hint: To compute \(P\left\\{e_{j}\right.\) precedes \(e_{i}\) at time \(\left.t\right\\}\) condition on whether or not either \(e_{i}\) or \(e_{j}\) has ever been requested prior to \(t\).

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