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Let \(X\) and \(Y\) have the bivariate normal distribution with zero means, unit variances, and correlation \(\rho\). Find the joint density function of \(X+Y\) and \(X-Y\), and their marginal density functions.

Short Answer

Expert verified
The joint density is \( \frac{1}{4\pi} e^{-\frac{u^2 + v^2}{4}} \), and the marginals are \( \frac{1}{\sqrt{4\pi}} e^{-\frac{u^2}{4}} \) for both \(U\) and \(V\).

Step by step solution

01

Recognize the Problem Type

We have two random variables, \(X\) and \(Y\), which are jointly normally distributed. We need to find the joint density function of \(X+Y\) and \(X-Y\), and their marginal density functions.
02

Understand the Variance and Covariance

Since \(X\) and \(Y\) are standard normal variables with mean 0 and variance 1, and they have a correlation \(\rho\), the covariance can be expressed as \(\text{Cov}(X,Y) = \rho\), and the covariance matrix is \(\boldsymbol{\Sigma} = \begin{pmatrix} 1 & \rho \ \rho & 1 \end{pmatrix}\).
03

Define New Variables

Let \(U = X + Y\) and \(V = X - Y\). We need to find their joint distribution. The transformation can be described by \(\begin{pmatrix} U \ V \end{pmatrix} = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix} \begin{pmatrix} X \ Y \end{pmatrix}\).
04

Calculate the Variance and Covariance of U and V

Calculate the covariance matrix of \(U\) and \(V\) using the transformation matrix \(A = \begin{pmatrix} 1 & 1 \ 1 & -1 \end{pmatrix}\): \(\text{Cov}(U,V) = A \boldsymbol{\Sigma} A^T = \begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}\). This implies \(U\) and \(V\) are uncorrelated and have variances equal to 2.
05

Determine the Joint Density Function

Since the covariance matrix of \(U\) and \(V\) is \(\begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}\), \(U\) and \(V\) are independent normal variables with mean zero and variance 2 each. Hence, the joint density function is the product of individual densities: \\( f_{U,V}(u,v) = \frac{1}{2\pi \sqrt{2 \cdot 2}} \cdot e^{-\frac{1}{8}(u^2 + v^2)} = \frac{1}{4\pi} e^{-\frac{u^2 + v^2}{4}} \).
06

Find Marginal Density Functions

Because \(U\) and \(V\) are independent, their marginal density functions are the same as their individual density functions. For example, \\( f_U(u) = \frac{1}{\sqrt{4\pi}}e^{-\frac{u^2}{4}} \) and similarly for \(f_V(v)\). Both are normal distributions with mean 0 and variance 2.

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Key Concepts

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

Joint Density Function
In probability theory, the joint density function is a fundamental concept that describes the density of probability for two random variables occurring together. For the bivariate normal distribution, the joint density function provides insights into how likely two correlated variables are to simultaneously take on a given pair of values.

In our case, let's consider the random variables \( U = X + Y \) and \( V = X - Y \). Their joint density function can be particularly interesting because it incorporates the individual probabilities of \( X \) and \( Y \), as well as their relationship, dictated by the correlation coefficient \( \rho \).

After recognizing the nature of the transformation, it is established that \( U \) and \( V \) are independent. This independence is key, as it means their combined probability is simply the product of their individual probabilities. The resulting joint density function is given by\[ f_{U,V}(u,v) = \frac{1}{4\pi} e^{-\frac{u^2 + v^2}{4}},\]which tells us how likely \( U \) and \( V \) are to simultaneously occupy certain values.
Covariance Matrix
The covariance matrix is an essential tool in understanding the relationship between two or more random variables in a multivariate distribution. For the bivariate normal distribution, the covariance matrix not only reflects the variances of individual variables but also captures their covariance, shedding light on their linear interdependence.

For two random variables \( X \) and \( Y \), the covariance is \( \text{Cov}(X,Y) = \rho \), where \( \rho \) is the correlation coefficient outlining the strength and direction of the linear relationship. Their covariance matrix is:\[\boldsymbol{\Sigma} = \begin{pmatrix} 1 & \rho \ \rho & 1 \end{pmatrix}.\]This matrix encapsulates both their variances (on the diagonal) and their covariance (off-diagonal), setting the stage for further transformations.

Upon transforming the variables to \( U = X + Y \) and \( V = X - Y \), the covariance matrix becomes\[\begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix}.\]This transformation results in \( U \) and \( V \) being uncorrelated and having equal variances, highlighting the power of manipulating covariance matrices to reveal new, independent dimensions within data.
Marginal Density Function
Understanding the marginal density function is crucial for examining the behavior of individual variables within a multivariate set, excluding the interactions with others. It signifies the statistical distribution for a single random variable from a joint distribution.

For \( U = X + Y \) and \( V = X - Y \), because they were determined to be independent, discovering their marginal density simplifies to considering their individual normal densities. Both \( U \) and \( V \) are normal distributions described by\[ f_U(u) = \frac{1}{\sqrt{4\pi}} e^{-\frac{u^2}{4}} \]and similarly for \( V \).This result underscores their behavior independently of each other. Each function reflects a normal distribution with a mean of 0 and a variance of 2, demonstrating the simplified analysis possible when working with uncorrelated transformations.
  • Mean: 0
  • Variance: 2
Understanding these marginal densities allows statisticians and data scientists to derive meaningful insights about individual components of dependent systems.

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Most popular questions from this chapter

Three points \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are chosen independently at random on the circumference of a circle. Let \(b(x)\) be the probability that at least one of the angles of the triangle ABC exceeds \(x \pi\). Show that $$ b(x)= \begin{cases}1-(3 x-1)^{2} & \text { if } \frac{1}{3} \leq x \leq \frac{1}{2} \\ 3(1-x)^{2} & \text { if } \frac{1}{2} \leq x \leq 1\end{cases} $$ Hence find the density and expectation of the largest angle in the triangle.

Find the density function of \(Y=a X\), where \(a>0\), in terms of the density function of \(X\). Show that the continuous random variables \(X\) and \(-X\) have the same distribution function if and only if \(f_{X}(x)=f_{X}(-x)\) for all \(x \in \mathbb{R}\)

Buffon's needle revisited. Two grids of parallel lines are superimposed: the first grid contains lines distance \(a\) apart, and the second contains lines distance \(b\) apart which are perpendicular to those of the first set. A needle of length \(r(<\min |a, b|)\) is dropped at random. Show that the probability it intersects a line equals \(r(2 a+2 b-r) /(\pi a b)\)

For what values of the parameters are the following functions probability density functions? (a) \(f(x)=C\\{x(1-x)\\}^{-\frac{1}{2}}, 0

Importance sampling. We wish to estimate \(l=\int g(x) f_{X}(x) d x=\mathrm{E}(g(X))\), where either it is difficult to sample from the density \(f x\), or \(g(X)\) has a very large variance. Let \(f y\) be equivalent to \(f_{x}\), which is to say that, for all \(x, f x(x)=0\) if and only if \(f y(x)=0\). Let \(\left[Y_{i}: 0 \leq i \leq n \mid\right.\) be independent random variables with density function \(f y\), and define $$ J=\frac{1}{n} \sum_{r=1}^{n} \frac{g\left(Y_{r}\right) f_{X}\left(Y_{r}\right)}{f_{Y}\left(Y_{Y}\right)} $$ Show that: (a) \(\mathrm{E}(J)=I=\mathbb{E}\left[\frac{g(Y) f_{X}(Y)}{f_{Y}(Y)}\right]\) (b) \(\operatorname{var}(J)=\frac{1}{n}\left[\mathbb{E}\left(\frac{g(Y)^{2} f_{X}(Y)^{2}}{f_{Y}(Y)^{2}}\right)-I^{2}\right]\) (c) \(J \stackrel{\text { ?.S }}{\longrightarrow} I\) as \(n \rightarrow \infty\). (See Chapter 7 for an account of convergence.) The idea here is that \(f_{y}\) should be easy to sample from, and chosen if possible so that var \(J\) is much smaller than \(n^{-1}\left[\mathbb{E}\left(g(X)^{2}\right)-I^{2}\right\\}\). The function \(f_{Y}\) is called the importance density.

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