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Chapter 6: Problem 44

The joint density of \(X\) and \(Y\) is $$ f(x, y)=c\left(x^{2}-y^{2}\right) e^{-x} \quad 0 \leq x<\infty,-x \leq y \leq x $$ Finel the conditional distribution of \(Y\), given \(X=x\).

Short Answer

Expert verified
The conditional distribution of \(Y\), given \(X=x\), is: \[ f_{Y|X}(y | x) = \frac{c\left(x^{2}-y^{2}\right) e^{-x}}{ \int_{-x}^{x} c\left(x^{2}-y^{2}\right) e^{-x} \, dy} \] for \(-x \leq y \leq x\), where \(c\) is the normalization constant.

Step by step solution

01

Verify the Joint Density

We need to verify that the given function satisfies the properties of a joint probability density function. The properties are: 1. \(f(x, y) \ge 0\) for all \((x, y)\) in their respective domain. 2. \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x, y) \, dxdy = 1\) Since \(x^{2}-y^{2} \geq 0\) in the given domain \((0 \leq x < \infty, -x \leq y \leq x)\) and the exponential function is always non-negative, we have \(f(x, y) \geq 0\). Now, let's determine the constant \(c\) such that the integral of \(f(x, y)\) over its domain equals 1: \[ \int_{0}^{\infty} \int_{-x}^{x} c\left(x^{2}-y^{2}\right) e^{-x} \, dy dx = 1 \]
02

Find Marginal Density of X

To find the marginal density of \(X\), we need to integrate the joint density function with respect to Y: \[ f_X(x) = \int_{-x}^{x} f(x, y) \, dy = \int_{-x}^{x} c\left(x^{2}-y^{2}\right) e^{-x} \, dy \]
03

Calculate Conditional Distribution

Finally, we can calculate the conditional distribution of \(Y\), given \(X=x\), using the definition: \[ f_{Y|X}(y | x) = \frac{f_{X, Y}(x, y)}{f_X(x)} \] Plugging in the expressions for \(f(x, y)\) and \(f_X(x)\), we have, \[ f_{Y|X}(y | x) = \frac{c\left(x^{2}-y^{2}\right) e^{-x}}{ \int_{-x}^{x} c\left(x^{2}-y^{2}\right) e^{-x} \, dy} \] This is the conditional distribution of \(Y\), given \(X=x\). Note that the integral in the denominator is simply the normalization constant. This distribution is valid for \(-x \leq y \leq x\).

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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 provides a way to describe the probability distribution of two continuous random variables simultaneously. In our case, the function given is \( f(x, y) = c(x^2 - y^2) e^{-x} \), which is defined for \( 0 \leq x < \infty \) and \( -x \leq y \leq x \). Understanding this function hinges on a few key points:
  • **Non-negativity:** The joint density must be non-negative across its entire domain, meaning \( f(x, y) \geq 0 \). The function \( x^2 - y^2 \) ensures this condition as it's non-negative within the given limits.

  • **Normalization:** The integral of the joint density over its entire domain must equal 1. This step involves integrating the function \( \int_{0}^{\infty} \int_{-x}^{x} c(x^2 - y^2) e^{-x} \, dy \, dx = 1 \) to find the constant \( c \) ensuring the total probability is 1.

This concept allows us to represent the probability of \( X \) and \( Y \) occurring together, forming the basis to derive other probability measures.
Marginal Density
Marginal density functions are derived from joint density functions to describe the probability distribution of a single random variable irrespective of others. For example, to find the marginal density of \( X \), \( f_X(x) \), we integrate the joint density with respect to \( Y \). The process is illustrated as follows:
  • The formula for the marginal density, \( f_X(x) = \int_{-x}^{x} c(x^2 - y^2) e^{-x} \, dy \), simplifies the joint density by removing dependence on \( y \).

  • This technique sums up all probabilities along \( y \), providing a comprehensive distribution solely in terms of \( x \).

Marginal density functions are pivotal when focusing on a single variable in multi-dimensional probability scenarios, helping in simplifying and analyzing specific data aspects without losing the complete context.
Probability Distribution
A probability distribution describes how probabilities are allocated over the values of a random variable. In this exercise, the focus is on finding the conditional distribution of \( Y \) given \( X = x \), which represents the distribution of \( Y \) given a specific value of \( X \).
  • The formula for conditional distribution \( f_{Y|X}(y | x) = \frac{f(x, y)}{f_X(x)} \) decomposes the joint density using the marginal density, emphasizing how special conditions affect our probability insights.

  • Understanding this conditional distribution helps us determine how \( Y \)'s behavior changes in response to different \( x \) values, allowing us to see the interdependence between the variables.

  • This approach is vital for situations where knowing the "conditional" scenario shifts our understanding of the probability spread.

Probability distributions, whether joint, marginal, or conditional, form the cornerstone of statistical analysis by modeling data and aiding in inferential predictions across various domains.

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Most popular questions from this chapter

Let \(X\) and \(Y\) be independent continuous random variables with respective hazard rate functions \(\lambda_{X}(t)\) and \(\lambda_{Y}(t)\), and set \(W=\min (X, Y)\). (a) Determine the distribution function of \(W\) in terms of those of \(X\) and \(Y\). (b) Show that \(\lambda_{W}(t)\), the hazard rate function of \(W\), is given by $$ \lambda_{W}(t)=\lambda_{X}(t)+\lambda_{Y}(t) $$

Let \(X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)}\) be the ordered values of \(n\) independent uniform \((0,1)\) random variables. Prove that for \(1 \leq k \leq n+1\), $$ P\left\\{X_{(k)}-X_{(k-1)}>t\right\\}=(1-t)^{n} $$ where \(X_{0} \equiv 0, X_{n+1} \equiv t\).

Derive the distribution of the range of a sample of size 2 from a distribut having density function \(f(x)=2 x, 0

An ambulance travels back and forth, at a constant speed, along a road of length \(L\). At a certain moment of time an accident occurs at a point uniformly distributed on the road. [That is, its distance from one of the fixed ends of the road is uniformly distributed over \((0, L)\).] Assuming that the ambulance's location at the moment of the accident is also uniformly distributed, compute, assuming independence, the distribution of its distance from the accident.

If \(X\) and \(Y\) have joint density function $$ f(x, y)=\frac{1}{x^{2} y^{2}} \quad x \geq 1, y \geq 1 $$ (a) Compute the joint density function of \(U=X Y, V=X / Y\). (b) What are the marginal densities?
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