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Let \(f(x)\) and \(g(y)\) be pdf's with corresponding cdf's \(F(x)\) and \(G(y)\), respectively. With \(c\) denoting a numerical constant satisfying \(|c| \leq 1\), consider $$ f(x, y)=f(x) g(y)\\{1+c[2 F(x)-1][2 G(y)-1]\\} $$ Show that \(f(x, y)\) satisfies the conditions necessary to specify a joint pdf for two continuous rv's. What is the marginal pdf of the first variable \(X\) ? Of the second variable \(Y\) ? For what values of \(c\) are \(X\) and \(Y\) independent? If \(f(x)\) and \(g(y)\) are normal pdf's, is the joint distribution of \(X\) and \(Y\) bivariate normal?

Short Answer

Expert verified
The joint pdf is valid for \(|c| \leq 1\), marginals are \(f(x)\) and \(g(y)\), and variables are independent if \(c = 0\). It's not bivariate normal even if marginals are normal.

Step by step solution

01

Checking Joint pdf Conditions

A joint pdf \(f(x, y)\) for two continuous random variables must satisfy two conditions: \(f(x, y)\geq0\) for all \((x, y)\) and \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \, dx \, dy = 1\). These conditions ensure that probabilities are non-negative and sum to one. Given \(f(x, y) = f(x) g(y)\{1 + c [2F(x) - 1][2G(y) - 1] \}\), if \(|c| \leq 1\), then the terms \( [2F(x)-1] \) and \( [2G(y)-1] \) are each between \(-1\) and \(1\), ensuring that the expression stays non-negative. The integral over the full space becomes the production of two separate integrals of valid pdfs, which equals to 1 by definition.
02

Finding Marginal pdf of the First Variable

To find the marginal pdf of \(X\), we integrate the joint pdf over \(y\): \[ f_X(x) = \int_{-\infty}^{\infty} f(x, y) \, dy = f(x) \int_{-\infty}^{\infty} g(y)\{1 + c [2F(x) - 1][2G(y) - 1] \} \, dy \]. Since \(\int_{-\infty}^{\infty} g(y) \, dy = 1\) and \([2G(y) - 1]\) integrates to zero, the result is \(f_X(x) = f(x)\).
03

Finding Marginal pdf of the Second Variable

Similarly, to find the marginal pdf of \(Y\), we integrate the joint pdf over \(x\): \[ f_Y(y) = \int_{-\infty}^{\infty} f(x, y) \, dx = g(y) \int_{-\infty}^{\infty} f(x)\{1 + c [2F(x) - 1][2G(y) - 1] \} \, dx \]. As \(\int_{-\infty}^{\infty} f(x) \, dx = 1\) and \([2F(x) - 1]\) integrates to zero, the result is \(f_Y(y) = g(y)\).
04

Conditions for Independence

Two variables are independent if their joint pdf is the product of their marginal pdfs: \(f(x, y) = f_X(x) g_Y(y)\). From the expression \(f(x, y) = f(x)g(y)\{1 + c [2F(x) - 1][2G(y) - 1] \}\), for independence, \(c\) must be zero because it removes the dependence term \([2F(x) - 1][2G(y) - 1]\), thus establishing \(c = 0\) as the condition for independence.
05

Checking if the Distribution is Bivariate Normal

The bivariate normal distribution has a specific form that captures correlation through a unique dependence structure, not provided in \(f(x, y)=f(x)g(y)\{1+c[2 F(x)-1][2 G(y)-1]\}\). Thus, even if \(f(x)\) and \(g(y)\) are normal pdf's, the presence of the term \{1+c[2 F(x)-1][2 G(y)-1]\} makes \(f(x, y)\) not bivariate normal.

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Key Concepts

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

Marginal PDF
Marginal probability density functions (marginal pdfs) are crucial when dealing with joint probability distributions. Essentially, a marginal pdf describes the probability distribution of one variable within a joint distribution while integrating out the influence of the other variables.

To find the marginal pdf of a variable in a joint distribution, you integrate the joint pdf over the range of the other variable(s). Take the marginal for the first variable, denoted as \( f_X(x) \). You compute it by integrating \( f(x,y) \) over all possible values of \( y \).
  • Formula for first variable: \( f_X(x) = \int_{-\infty}^{\infty} f(x, y) \, dy \)
  • For the second variable: \( f_Y(y) = \int_{-\infty}^{\infty} f(x, y) \, dx \)
This integration process effectively 'sums out' the dependency of \( y \) from the joint pdf, providing the standalone pdf of \( x \), and vice-versa.

Simply put, knowing the marginal pdfs \( f_X(x) \) and \( f_Y(y) \) helps evaluate how each variable behaves individually in the joint context.
Continuous Random Variables
Continuous random variables are variables that can take on an infinite number of possible values. These are typically values within a certain range or interval.

Since each specific value for a continuous random variable has a probability of zero, we use probability density functions (pdfs) to describe their distributions. Pdfs offer a way to define probabilities over an interval by integrating the function over that range.
  • Example: Suppose \( X \) is a continuous random variable, \( f(x) \) is its pdf, then \( P(a < X < b) = \int_{a}^{b} f(x) \, dx \).
Understanding continuous random variables is fundamental when studying joint probability scenarios. It helps in grasping how individual probability distributions interact and function within a combined distribution without dictating exact outcomes initially.
Independence Condition
A critical concept in probability and statistics is the idea of independence. Two random variables \( X \) and \( Y \) are independent if the occurrence of one does not influence the probability of the occurrence of the other.

For continuous random variables to be independent, their joint pdf \( f(x, y) \) must be the product of their individual marginal pdfs \( f_X(x) \) and \( g_Y(y) \). That is:
  • Independence condition: \( f(x, y) = f_X(x) g_Y(y) \)
In the specific exercise above, the independence condition is met when the constant \( c \) is zero. This nullifies the interaction term \([2F(x) - 1][2G(y) - 1]\), leaving \( f(x, y) \) to be precisely \( f(x)g(y) \).

The independence of variables plays a vital role in simplifying analysis and calculations dealing with multiple random variables within probability and statistical frameworks.
Bivariate Normal Distribution
The bivariate normal distribution is a specific type of joint distribution for two continuous variables that are both normally distributed. It includes a correlation structure between the two variables.

A bivariate normal distribution has a set particular form, incorporating parameters for variances and the correlation coefficient that define how variables relate. They look different from a simple product of two independent normal distribution functions. Components of a bivariate normal distribution include:
  • Mean vectors and covariance matrices that describe each variable and their relationship.
  • Correlation which is not zero and provides insight into how one variable changes concerning the other.
Despite \( f(x) \) and \( g(y) \) being normal pdfs individually in this exercise, the joint distribution \( f(x, y) \) is not bivariate normal due to the added term \{ \( 1+c[2F(x) -1][2G(y)-1] \} \). This prevents the distribution from fitting the strict parameters defining bivariate normality.

Understanding how variations affect overall statistical behavior in complex distributions is key when studying bivariate and other multivariate distributions.

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Most popular questions from this chapter

Let \(X_{1}\) and \(X_{2}\) be quantitative and verbal scores on one aptitude exam, and let \(Y_{1}\) and \(Y_{2}\) be corresponding scores on another exam. If \(\operatorname{Cov}\left(X_{1}, Y_{1}\right)=5, \operatorname{Cov}\left(X_{1}, Y_{2}\right)=1, \operatorname{Cov}\left(X_{2}, Y_{1}\right)=2\), and \(\operatorname{Cov}\left(X_{2}, Y_{2}\right)=8\), what is the covariance between the two total scores \(X_{1}+X_{2}\) and \(Y_{1}+Y_{2}\) ?

a. Use the rules of expected value to show that \(\operatorname{Cov}(a X+b, c Y+d)=a c \operatorname{Cov}(X, Y)\). b. Use part (a) along with the rules of variance and standard deviation to show that \(\operatorname{Corr}(a X+b, c Y+d)=\operatorname{Corr}(X, Y)\) when \(a\) and \(c\) have the same sign. c. What happens if \(a\) and \(c\) have opposite signs?

Two components of a computer have the following joint pdf for their useful lifetimes \(X\) and \(Y\) : $$ f(x, y)=\left\\{\begin{array}{cc} x e^{-x(1+y)} & x \geq 0 \text { and } y \geq 0 \\ 0 & \text { otherwise } \end{array}\right. $$ a. What is the probability that the lifetime \(X\) of the first component exceeds 3 ? b. What are the marginal pdf's of \(X\) and \(Y\) ? Are the two lifetimes independent? Explain. c. What is the probability that the lifetime of at least one component exceeds \(3 ?\)

Let \(X\) denote the number of Canon digital cameras sold during a particular week by a certain store. The pmf of \(X\) is $$ \begin{array}{l|lllll} x & 0 & 1 & 2 & 3 & 4 \\ \hline p_{X}(x) & .1 & .2 & .3 & .25 & .15 \end{array} $$ Sixty percent of all customers who purchase these cameras also buy an extended warranty. Let \(Y\) denote the number of purchasers during this week who buy an extended warranty. a. What is \(P(X=4, Y=2)\) ? [Hint: This probability equals \(P(Y=2 \mid X=4) \cdot P(X=4)\); now think of the four purchases as four trials of a binomial experiment, with success on a trial corresponding to buying an extended warranty.] b. Calculate \(P(X=Y)\) c. Determine the joint pmf of \(X\) and \(Y\) and then the marginal pmf of \(Y\).

A friend of ours takes the bus five days per week to her job. The five waiting times until she can board the bus are a random sample from a uniform distribution on the interval from 0 to \(10 \mathrm{~min}\). a. Determine the pdf and then the expected value of the largest of the five waiting times. b. Determine the expected value of the difference between the largest and smallest times. c. What is the expected value of the sample median waiting time? d. What is the standard deviation of the largest time?

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